Coordinate
Geometry
With Sessionwise Theory & Exercises
Coordinate
Geometry
With Sessionwise Theory & Exercises
Dr. SK Goyal
ARIHANT PRAKASHAN (Series), MEERUT
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PREFACE
IF YOU CONTINUOUSLY PUT YOUR EFFORTS ON AN ASPECT, YOU HAVE VERY
GOOD CHANCE TO GET POSITIVE OUTCOME i.e. SUCCESS
It is a matter of great pride and honour for me to have received such an overwhelming response
to the previous editions of this book from the readers. In a way, this has inspired me to revise this
book thoroughly as per the changed pattern of JEE Main & Advanced. I have tried to make the
contents more relevant as per the needs of students, many topics have been re-written, a lot of
new problems of new types have been added in etcetc. All possible efforts are made to remove
all the printing errors that had crept in previous editions. The book is now in such a shape that
the students would feel at ease while going through the problems, which will in turn clear their
concepts too.
A Summary of changes that have been made in Revised & Enlarged Edition
—
Theory has been completely updated so as to accommodate all the changes made in JEE Syllabus &
Pattern in recent years.
—
The most important point about this new edition is, now the whole text matter of each chapter has
been divided into small sessions with exercise in each session. In this way the reader will be able to
go through the whole chapter in a systematic way.
—
Just after completion of theory, Solved Examples of all JEE types have been given, providing the
students a complete understanding of all the formats of JEE questions & the level of difficulty of
questions generally asked in JEE.
—
Along with exercises given with each session, a complete cumulative exercises have been given at
the end of each chapter so as to give the students complete practice for JEE along with the
assessment of knowledge that they have gained with the study of the chapter.
—
Last 13 Years questions asked in JEE Main & Adv, IIT-JEE & AIEEE have been covered in all the
chapters.
However I have made the best efforts and put my all Coordinate Geometry teaching
experience in revising this book. Still I am looking forward to get the valuable suggestions and
criticism from my own fraternity i.e. the fraternity of JEE teachers.
I would also like to motivate the students to send their suggestions or the changes that they
want to be incorporated in this book. All the suggestions given by you all will be kept in prime
focus at the time of next revision of the book.
Dr. SK Goyal
CONTENTS
1. COORDINATE SYSTEM AND COORDINATES
LEARNING PART
Session 1
— Introduction
— Coordinate Axes
— Rectangular Cartesian Coordinates
of a Point
— Polar Coordinates of a Point
— Relation between the Polar and Cartesian
Coordinates
Session 2
— Distance between Two Points
— Choice of Axes
— Distance between Two Points in Polar
Coordinates
Session 3
— Section formulae
— Centroid of a Triangle
— Incentre
— Some Standard Results
— Area of Triangle
Session 4
— Locus and Its Equation
— Change of Axes the Transformations of Axes
— Removal of the Term xy from F(x,y) = ax2 +
2hxy + by2 without Changing the Origin
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
73-190
2. THE STRAIGHT LINES
LEARNING PART
Session 1
— Definition
— Angle of Inclination of a Line
— Slope or Gradient of a Line
— Angle Between Two Lines
— Lines Parallel to Coordinate Axes
— Intercepts of a Line on Axes
— Different Forms of the Equation of a
Straight Line
— Reduction of General Equation to
Standard Form
1-71
— The Distance Form or Symmetric Form or
Parametric Form of a Line
Session 2
— Position of Two Points Relative to a
Given Line
— Position of a Point which lies Inside a Triangle
— Equations of Lines Parallel and Perpendicular
to a Given Line
— Distance Between Two Parallel Lines
— Distance of a Point From a Line
— Area of Parallelogram
Session 3
— Points of Intersection of Two Lines
— Concurrent Lines
— Family of Lines
— How to Find Circumcentre and Orthocentre by
Slopes
Session 4
— Equations of Straight Lines Passing Through a
Given Point and Making a Given Angle with a
Given Line
— A Line Equally Inclined with Two Lines
— Equation of the Bisectors
— Bisector of the Angle Containing
The Origin
— Equation of that Bisector of the Angle Between
Two Lines Which Contains a Given Point
— How to Distinguish the Acute (Internal) and
Obtuse (External) Angle Bisectors
Session 5
— The Foot of Perpendicular Drawn from the
Point (x1 , y1 ) to the Line ax + by + c = 0
— Image or Reflection of a Point (x1 , y1 )
about a Line Mirror
— Image or Reflection of a Point (x1 , y1 ) in
Different Cases
— Use of Image or Reflection
Session 6
— Reflection of Light
— Refraction of Light
— Condition of Collineirty If Three Given Points
in Cyclic Order
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
191-239
3. PAIR OF STRAIGHT LINES
LEARNING PART
Session 1
— Introduction
— Homogeneous Equation in Two Variables
Session 2
— Angle between the Pair of Lines ax2+2hxy+by2
Session 3
— Bisectors of the Angle between the Lines
Given by a Homogeneous Equation
Session 4
— General Equation of Second Degree
— Important Theorems
Session 5
— To Find the Point of Intersection of Lines
Represented by ax2 + 2hxy + by2 + 2gx + 2fy +
c = 0 with the Help of Partial Differentiation
— Removal of First Degree Term
— Equation of the Lines Joining the Origin to the
Points of Intersection of a Given Line and a
Given Curve
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
4. CIRCLE
LEARNING PART
Session 1
— Definition
— Equation of Circles in Different Forms
— Locus of the Mid-point of the Chords of the
Circle that Subtends an Angle of 2q at its Centre
Session 2
— Diametric Form of a Circle
— Equation of Circle Passing Through Three NonCollinear Points
Session 3
— Intercepts Made on the Axes by a Circle
— Different Forms of the Equations of a Circle
— Position of a Point with Respect to Circle
— Maximum and Minimum Distance of a Point
from the Circle
Session 4
— Intersection of a Line and a Circle
— Product of the Algebraical Distances PA and
PB is Constant when from P, A Secant be
Drawn to Cut the circle in the Point A and B
— The Length of Intercept Cut-off from a Line by
a Circle
— Tangent to a Circle at a Given Point
— Normal to a Circle at a Given Point
241-362
Session 5
— Tangents from a Point to the Circle
— Length of the Tangent from a Point to a Circle
— Power of a Point with Respect to a Circle
— Chord of Contact
— Chord Bisected at a Given Point
— Pair of Tangents
— Director Circle
Session 6
— Diameter of a Circle
— Two Circles Touching Each Other
— Common Tangents to Two Circles
— Common Chord of Two Circles
— Family of Circles
Session 7
— Angle of Intersection of Two Circles
— Radical Axis
— Radical Centre
— Co-axial System of Circles
— Limiting Point
— Image of the Circle by the Line Mirror
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
5. PARABOLA
363-459
LEARNING PART
— Intersection of a Line and a Parabola
Session 1
— Introduction
— Conic Section
— Section of a Right Circular Cone by Different
Planes
— Conic Section : Definition
— Equation of Conic Section
— Recognisation of Conics
— How to Find the Centre of Conics
— Parabola : Definition
— Standard Equation of Parabola
— Some Terms Related to Parabola
— Other forms of Parabola with
Latusrectum 4a
— Smart Table
— General Equation of a Parabola
— Equation of Parabola if Equation of
axis, Tangent at Vertex and Latusrectum
are given
2
— The Generalised form (y-k) = 4a (x-h)
— Parabolic Curve
— Equation of Tangent in Different Forms
Session 2
— Position of a Point (x1 , y1 ) with respect to a
Parabola y2 = 4ax
— Parametric Relation between the Coordinates
of the Ends of a Focal Chord of a Parabola
— Point of Intersection of Tangents at any Two
—
—
—
—
—
Points on the Parabola
Equation of Normals in Different Forms
Point of Intersection of Normals at any Two
Points on the Parabola
Relation Between ‘t1’ and ‘t2’ if Normal at ‘t1’
meets the Parabola Again at ‘t2’
Co-normal Points
Circle Through Co-normal Points
Session 3
— Pair of Tangents SS1 = T 2
— Chord of Contact
— Equation of the Chord Bisected at a
—
—
—
—
—
Given Point
Diameter
Lengths of Tangent, Subtangent, Normal and
Subnormal
Some Standard Properties of the Parabola
Reflection Property of a Parabola
Study of Parabola of the Form
2
(ax + by) + 2gx + 2fy + c = 0
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
6. ELLIPSE
461-553
LEARNING PART
— Smart Table
Session 1
— Ellipse Definition
— Standard Equation of Ellipse
— The Foci and Two Directrices of an Ellipse
— Tracing of the Ellipse
— Some Terms Related to an Ellipse
— Focal Distances of a Point
— The Shape of the Ellipse,
x2
y2
— when b>a
——
+ ——
= 1,
2
b2
— Mechanical Construction ofaan Ellipse
— Smart Table
Session 3
— Pair of Tangents
— Chord of Contact
— Chord Bisected at a Given Point
— Diameter
— Conjugate Diameters
— Properties of Conjugate Diameters
— Equi-Conjugate Diameters
— Director Circle
— Sub-Tangent and Sub-Normal
— Concyclic Points
— Some Standard Properties of the Ellipse
— Reflection Property of an Ellipse
— Equation of an Ellipse Referred to Two
Perpendicular Lines
Session 2
— Position of a Point with Respect to an Ellipse
— Intersection of a Line and an Ellipse
— Equation of Tangent in Different Forms
— Equations of Normals in Different Forms
— Properties of Eccentric Angles of the Conormal Points
— Co-normal Points Lie on a Fixed Curve
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
7. HYPERBOLA
555-638
LEARNING PART
Session 3
Session 1
— Hyperbola : Definition
— Standard Equation of Hyperbola
— The Foci and Two Directrices of a Hyperbola
— Tracing of the Hyperbola
— Some Terms Related to Hyperbola
— Focal Distances of a Point
— Conjugate Hyperbola
— Position of a Point with Respect to a
Hyperbola
— Intersection of a Line and a Hyperbola
— Diameter
Session 2
— Equations of Tangents in Different Forms
— Equations of Normals in Different Forms
— Pair of Tangents
— Chord of Contact
— Equation of the Chord Bisected at a
Given Point
— Conjugate Diameters
— Properties of Hyperbola
— Intersection of Conjugate Diameters and
—
—
—
—
—
—
—
—
Hyperbola
Director Circle
Asymptotes
Rectangular Hyperbola
The Rectangular Hyperbola xy = c2
Study of Hyperbola xy = c2
Properties of Rectangular Hyperbola xy= c2
Reflection Property of a Hyperbola
Equation of a Hyperbola Referred to Two
Perpendicular Lines
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
SYLLABUS
JEE MAIN
Cartesian system of rectangular coordinates in a plane, distance formula, section
formula, locus and its equation, translation of axes, slope of a line, parallel and
perpendicular lines, intercepts of a line on the coordinate axes.
Straight Lines
Various forms of equations of a line, intersection of lines, angles between two
lines, conditions for concurrence of three lines, distance of a point from a line,
equations of internal and external bisectors of angles between two lines,
coordinates of centroid, orthocentre and circumcentre of a triangle, equation
of family of lines passing through the point of intersection of two lines.
Circles, Conic Sections
Standard form of equation of a circle, general form of the equation of a circle, its
radius and centre, equation of a circle when the end points of a diameter are
given, points of intersection of a line and a circle with the centre at the origin and
condition for a line to be tangent to a circle, equation of the tangent. Sections of
cones, equations of conic sections (parabola, ellipse and hyperbola) in standard
forms, condition for y=mx + c to be a tangent and point (s) of tangency.
JEE ADVANCED
Analytical Geometry
Two Dimensions Cartesian Coordinates, distance between two points, section
formulae, shift of origin.
Equation of a straight line in various forms, angle between two lines, distance of
a point from a line. Lines through the point of intersection of two given lines,
equation of the bisector of the angle between two lines, concurrency of lines,
centroid, orthocentre, incentre and circumcentre of a triangle.
Equation of a circle in various forms, equations of tangent, normal and chord.
Parametric equations of a circle, intersection of a circle with a straight line or a
circle, equation of a circle through the points of intersection of two circles and
those of a circle and a straight line.
Equations of a parabola, ellipse and hyperbola in standard form, their foci,
directrices and eccentricity, parametric equations, equations of tangent and
normal.
CHAPTER
01
Coordinate System
and Coordinates
Learning Part
Session 1
● Introduction
● Rectangular Cartesian Coordinates of a Point
● Relation between the Polar and Cartesian Coordinates
Session 2
● Distance between Two Points
● Distance between Two Points in Polar Coordinates
Session 3
● Section formulae
● Incentre
● Area of Triangle
●
●
●
●
●
Coordinate Axes
Polar Coordinates of a Point
Choice of Axes
Centroid of a Triangle
Some Standard Results
Session 4
● Locus and Its Equation
● Change of Axes or the Transformations of Axes
2
● Removal of the Term xy from F ( x, y ) = ax
+ 2hxy + by 2 without Changing the Origin
●
Position of a Point Which Lies Inside a Triangle
Practice Part
●
●
JEE Type Examples
Chapter Exercises
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Session 1
Introduction, Coordinate Axes, Rectangular Cartesian
Coordinates of a Point, Polar Coordinates of a Point,
Relation between the Polar and Cartesian Coordinates
Introduction
The great philospher and mathematician of France Rane
Descartes (1596-1665) published a book ‘La Geometric’
in 1637.
Descartes gave a new idea i.e. each point in a plane is
expressed by an ordered pair of algebraic real numbers
like ( x , y ), (r , θ ) etc., called coordinates of the point.
The point ( x , y ) is called cartesian coordinates and (r , θ ) is
called polar coordinates of the point. Then represents
different forms of equations which are developed for all
types of straight lines and curves.
Thus the Coordinate Geometry (or Analytical Geometry) is
that branch of mathematics in which geometrical problems
are solved with the help of Algebra.
Rectangular Cartesian
Coordinates of a Point
Let X ′ OX and Y ′ OY be two perpendicular axes in the
plane of paper intersecting at O. Let P be any point in the
plane of the paper. Draw PM perpendicular to OX. Then
the lengths OM and PM are called the rectangular
cartesian coordinates or briefly the coordinates of P.
Y
P (x, y)
y
X′
Coordinate Axes
O
90°
x-axis
X
IV
quadrant
III
quadrant
ax
is
Y
ue
I (α ≠ 90°)
quadrant
α
X
O x-axis
IV
quadrant
liq
II
quadrant
ob
X′
y-axis
Y
I
quadrant
X′
III
quadrant
Y′
M
X
Let
OM = x and MP = y
Then, the position of the point P in the plane with respect
to the coordinate axes is represented by the ordered pair
(x, y). The ordered pair (x, y) is called the coordinates of
point P.
i.e.
OM = x -coordinate or abscissa of the point P
and
MP = y -coordinate or ordinate of the point P.
Remarks
1.
2.
3.
4.
The ordinate of every point on X-axis is 0.
The abscissa of every point on Y-axis is 0.
The abscissa and ordinate of the origin O(0, 0) are both zero.
The abscissa and ordinate of a point are at perpendicular
distance from Y-axis and X-axis respectively.
5. Table for conversion sign of coordinates :
Quadrants
Sign of x coordinates
Sign of y coordinates
Y′
(a) Rectangular axes
x
Y′
The position of a point in a plane is determined with
reference to two intersecting straight lines called the
coordinate axes and their point of intersection is called
the origin of coordinates.
If these two axes of reference (generally we call them x
and y axes) cut each other at right angle, they are called
rectangular axes otherwise they are called oblique
axes. The axes divide the coordinate plane in four
quadrants.
II
quadrant
O
(b) Oblique axes
Sign of ( x, y )
XOY
(I)
X ′OY
(II)
X ′OY ′
(III)
XOY ′
(IV)
+
−
−
+
+
+
−
−
( +, + )
( −, + )
( −, − )
( +, −)
6. Equation of X-axis, y = 0 and equation of Y-axis, x = 0.
Chap 01 Coordinate System and Coordinates
Polar Coordinates of a Point
OP = r
∠XOP = θ
If
and
(radius vector)
(vectorial angle)
P (r, θ)
r
θ
O
(Pole)
X
Initial line
Q (r, – θ)
Relation between the Polar
and Cartesian Coordinates
Let P ( x , y ) be the cartesian coordinates with respect to
axes OX and OY and (r , θ ) be its polar coordinates with
respect to pole O and initial line OX.
It is clear from figure
OM = x = r cos θ
… (i)
and
…(ii)
MP = y = r sin θ
Then, the ordered pair of real numbers (r , θ ) called the
polar coordinates of the point P.
Y
P(x, y)
Remarks
r
1. r may be positive or negative according as θ is measured in
anticlockwise or clockwise direction. θ lies between − π to π i.e.
− π < θ ≤ π. If it is greater than π, then we subtract 2π from it
and if it is less than − π, then we add 2π, to it. It is also known
as principal value of P.
2. Always taken θ in radian.
y Example 1. Draw the polar coordinates
π 
π
π
 π 

 2 ,  ,  −2 ,  ,  −2 , −  and  2 , −  on the plane.
 3 

3
3 
3
Sol.
π
S –2, –
3
2
X
O
2
2
π
R 2, –
3
π
Q –2,
3
5π
y Example 2. Draw the polar coordinate  3,  on
 4
the plane.
Sol. Here,
θ=
5π
>π
4
O
3
P
X
3π
4
3, 5π or 3, – 3π
4
4
3π
5π
− 2π = −
4
4
3π 
 5π 

∴ 3,  is same 3, − 
 4

4
then,
θ − 2π =
x
M
X
Squaring and adding Eqs. (i) and (ii), we get
x2 +y2 =r2
or r = ( x 2 + y 2 )
Dividing Eqs. (ii) by (i), then
y 
y 
tan θ =   or θ = tan −1  
x
x
and
2
y
θ
O
i.e.
π
P 2,
3
3
(r cos θ, r sin θ ) ⇒ ( x , y )

2
2
−1  y  
 ( x + y ), tan    ⇒ (r , θ )
 x 

…(iii)
…(iv)
If r and θ are known then we can find ( x , y ) from Eq. (iii)
and if x and y are known then we can find (r , θ ) from
Eq. (iv).
y 
Q
θ = tan −1  
x
y
α = tan −1
x
If
Then, values of θ in four quadrants
Quadrant
I
II
III
IV
θ
α
π −α
−π + α
−α
y Example 3. Find the cartesian coordinates of the
points whose polar coordinates are
π


 4 
(i)  5 , π − tan −1   
(ii)  5 2 , 

 3

4
 4
Sol. (i) Given, r = 5, θ = π − tan −1  
 3
4
Textbook of Coordinate Geometry
Now,
(ii) Given, x = − 3, y = 4

 4
x = r cos θ = 5 cos  π − tan −1   
 3


 4
= − 5 cos  tan −1   
 3

Hence, cartesian coordinates of the given point will be
( −3, 4 ).
π
(ii) Given, r = 5 2, θ =
4
1
 π
Now,
x = r cos θ = 5 2 cos   = 5 2 ×
=5
 4
2
and
y
 = tan −1
 4
 = tan −1  4 
α = tan −1
 3
x
−3
 4
θ = π − α = π − tan −1  
 3
Hence, polar coordinates of the given points, will be

−1  4  
5 , π − tan    .

3 

 4
= 5 sin  tan −1   
 3

4
 4
  = 5 × = 4
 5
5
and
∴

 4
y = r sin θ = 5 sin  π − tan −1   
 3


= 5 sin sin −1

r = ( x 2 + y 2 ) = (9 + 16) = 5
Since, point ( −3, 4 ) lies in II quadrant
3

 3 
= − 5 cos  cos −1    = − 5 × = − 3
 5 
5

and
∴
y Example 5. Transform the equation r 2 = a 2 cos 2θ
into cartesian form.
r = (x 2 + y 2 )
Sol. Q
r 2 = (x 2 + y 2 )
or
1
 π
y = r sin θ = 5 2 sin   = 5 2 ×
=5
 4
2
y
x
y2 

x2 
y2 

x2 
This is the required equation in cartesian form.
Aliter :
r 2 = a 2 cos 2θ
r = (x 2 + y 2 ) = (4 + 4) = 2 2
and
= (2, 2) ≠ ( −2, − 2)
tanθ =
( x 2 + y 2 )2 = a 2 ( x 2 − y 2 )
or
Sol. (i) Given, x = − 2 , y = − 2
Remark : If we find θ, from the equation,
y −2
=1
tanθ = =
x −2
π
then,
θ = and then
4
1
1

( x , y ) = (r cos θ, r sin θ ) = 2 2 ×
,2 2 ×


2
2
and

1 −
( x 2 + y 2 ) = a2 

1 +

or
y Example 4. Find the polar coordinates of the points
whose cartesian coordinates are
(i) (−2 , − 2)
(ii) (−3 , 4)
 −2
 = tan −1 1 = π
y 
 = tan −1
α = tan −1
4
 −2
 x
Since, point ( −2, − 2) lies in III quadrant.
π
3π
∴
θ= −π +α = −π + = −
4
4
Hence, polar coordinates of the given points will be
3π 

2 2, −  .

4
y 
θ = tan −1  
x
 1 − tan 2 θ 
Given, r 2 = a 2 cos 2θ = a 2 

 1 + tan 2 θ 
Hence, cartesian coordinates of the given point,
will be (5, 5).
∴
and
or
r 2 = a 2 (cos 2 θ − sin 2 θ )
Q
and
x = r cosθ and y = r sinθ
r 2 = x2 + y2
then
 x2 y2
r 2 = a2  2 − 2 
r 
r
or
r 4 = a2 ( x 2 − y 2 )
or
( x 2 + y 2 )2 = a 2 ( x 2 − y 2 ).
y Example 6. Transform the equation x 2 + y 2 = ax into
polar form.
Sol.
Q
Given,
⇒
x = r cosθ, y = r sinθ
x 2 + y 2 = ax
r 2 = a (r cos θ )
or
r = a cosθ
This is the required equation in polar form.
Chap 01 Coordinate System and Coordinates
Exercise for Session 1
1.
The polar coordinates of the point whose cartesian coordinates are ( −1, − 1) is
π
(a)  2, 

4
2.
(b) (−12, 5)
(c) (−12, − 5)
(d) (12, − 5)
The transform equation of r 2 cos 2 θ = a 2 cos 2θ to cartesian form is ( x 2 + y 2 ) x 2 = a 2λ, then value of λ is
(a) y 2 − x 2
4.
3π 
(d)  2, −


4

 5 
The cartesian coordinates of the point whose polar coordinates are 13 , π − tan−1    is
 12 

(a) (12, 5)
3.
π
(c)  2, − 

4
3π 
(b)  2,


4
(b) x 2 − y 2
(c) xy
(d) x 2 y 2
The coordinates of P′ in the figure is
P –3, – π
3
O
X
–π
3
P′
π
(a)  3, 
 3
5.
π
(b)  3, − 

3
π
(c)  −3, − 

3
π
(d)  −3, 

3
The cartesian coordinates of the point Q in the figure is
π
P 2,
3
Q
2
2
2
π/3
O –π/6
2
X
π
S 2, –
6
R
–2, π
3
(a) ( 3, 1)
(b) (− 3, 1)
(c) (− 3, − 1)
(d) ( 3, − 1)
6.
A point lies on X-axis at a distance 5 units from Y-axis. What are its coordinates ?
7.
A point lies on Y-axis at a distance 4 units from X-axis. What are its coordinates ?
8.
A point lies on negative direction of X-axis at a distance 6 units from Y-axis. What are its coordinates ?
9.
Transform the equation y = x tan α to polar form.
10.
Transform the equation r = 2 a cos θ to cartesian form.
5
Session 2
Distance between Two Points, Choice of Axes, Distance
between Two Points in Polar Coordinates
Distance Between Two Points
Theorem : The distance between two points P ( x 1 , y 1 )
and Q ( x 2 , y 2 ) is given by
| PQ | = ( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2
Proof : Let P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) be any two points in
the plane. Let us assume that the points P and Q are both
in 1st quadrant (for the sake of exactness).
Y
Notations : We shall denote the distance between two
points P and Q of the coordinate plane, either by | PQ | or
by PQ.
Corollary 1 : The above formula is true for all positions
of the points (i.e. either point or both points are not in the
1st quadrant) keeping in mind, the proper signs of their
coordinates.
Corollary 2 : The distance of the point P ( x , y ) from the
origin O(0, 0 ) is given by
Q (x2, y2)
(x1, y1)
P
| OP | = ( x − 0 ) 2 + (y − 0 ) 2 = ( x 2 + y 2 )
y2–y1
x2– x1
R
Corollary 3 : The above formula can also be used as
( x 1 − x 2 ) 2 + (y 1 − y 2 ) 2
y2
y1
X′
O
x1
L
M
X
x2
Y′
From P and Q draw PL and QM perpendiculars to X-axis.
From P draw PR perpendicular to QM and join PQ. Then
OL = x 1 , OM = x 2 , PL = y 1 , QM = y 2
∴
PR = LM = OM − OL = x 2 − x 1
and QR = QM − RM = QM − PL = y 2 − y 1
Since, PRQ is a right angled triangle, therefore by
pythagoras theorem.
Corollary 4 : (i) If PQ is parallel to X-axis, then y 1 = y 2
and so
| PQ | = ( x 2 − x 1 ) 2 = | x 2 − x 1 |
(ii) If PQ is parallel to Y-axis, then x 1 = x 2 and so
| PQ | = (y 2 − y 1 ) 2 = | y 2 − y 1 |
Corollary 5 : If distance between two points is given, then
use ± sign.
Remarks
1. If three points A( x1, y1 ), B( x 2, y2 ) and C( x 3, y3 ) are collinear,
then
AB ± BC = AC
C
( PQ ) 2 = ( PR ) 2 + (QR ) 2
∴
B
| PQ | = ( PR ) 2 + (QR ) 2 (Q PQ is always positive)
= ( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2
A
∴ The distance PQ between the points P ( x 1 , y 1 ) and
Q ( x 2 , y 2 ) is given by ( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2
or
or
B
C
(difference in x coordinates) 2
+ (difference in y coordinates) 2
(difference of abscissaes) 2
+ (difference of ordinates) 2
A
2. When three points are given and it is required to :
(i) an Isosceles triangle, show that two of its sides
(or two angles) are equal.
Chap 01 Coordinate System and Coordinates
(ii) an Equilateral triangle, show that its all sides are equal
or each angle is of 60°.
(iii) a Right angle triangle, show that the sum of the
squares of two sides is equal to the square of the third
side.
(iv) an Isosceles right angled triangle, show that two of its
sides are equal and the sum of the squares of two equal
sides is equal to the square of the third side.
(v) a Scalene triangle, show that its all sides are unequal.
3. When four points are given and it is required to
(i) a Square, show that the four sides are equal and the
diagonals are also equal.
(ii) a Rhombus, (or equilateral trapezium) show that the
four sides are equal and the diagonals are not equal.
(iii) a Rectangle, show that the opposite sides are equal and
the diagonals are also equal.
(iv) a Parallelogram, show that the opposite sides are equal
and the diagonals are not equal.
(v) a Trapezium, show that the two sides are parallel and
the other two sides are not parallel.
(vi) An Isosceles Trapezium, show that the two sides are
parallel and the other two sides are not parallel but equal.
4. If A, B, C be the vertices of a triangle and we have to find the
coordinates of the circumcentre then, let the circumcentre be
P( x, y ) and use PA2 = PB2 and PA2 = PC2 this will give two
equations in x and y then solve these two equations and ( x, y ).
| PQ | = (a cos α − a cos β )2 + (a sin α − a sin β )2
then
= a 2 {(cos α − cos β )2 + (sin α − sin β )2 }
=
a 2 {cos 2 α + cos 2 β − 2 cos α cos β + sin 2 α
+ sin 2 β − 2 sin α sin β }
= a 2 {1 + 1 − 2(cos α cos β + sin α sin β )}
= a 2 (2 − 2 cos(α − β ))
= 2a 2 (1 − cos(α − β ))

α − β
= 2a 2 ⋅ 2 sin 2 

 2 

α − β
= 4a 2 sin 2 

 2 

 α − β 
=2a sin 

 2 


= 2asin

 α − β 


 2 
y Example 9. If the point ( x , y ) be equidistant from the
points (6, − 1) and (2, 3) , prove that x − y = 3 .
By the given condition, PA = PB
(i) If ( x 1 , y 1 ) and ( x 2 , y 2 ) are the ends of the hypotenuse
of a right angled isosceles triangle, then the third
vertex is given by
 ( x 1 + x 2 ) ± (y 1 − y 2 ) (y 1 + y 2 ) m ( x 1 − x 2 ) 
,




2
2
(ii) If two vertices of an equilateral triangle are ( x 1 , y 1 )
and ( x 2 , y 2 ), then coordinates of the third vertex are
 x 1 + x 2 m 3 (y 2 − y 1 ) y 1 + y 2 ± 3 ( x 2 − x 1 )
,


2
2


y Example 7. Prove that the distance of the point
(a cos α , a sin α ) from the origin is independent of α.
⇒
( x − 6) 2 + ( y + 1) 2 = ( x − 2) 2 + ( y − 3) 2
or
( x − 6) 2 + ( y + 1) 2 = ( x − 2) 2 + ( y − 3) 2
or
x 2 − 12x + 36 + y 2 + 2y + 1
= x 2 − 4 x + 4 + y 2 − 6y + 9
8x − 8y = 24
x −y =3
or
or
y Example 10. Using distance formula, show that the
points (1, 5), (2, 4 ) and ( 3, 3) are collinear.
Sol. Let A ≡ (1, 5), B ≡ (2, 4 ) and C ≡ (3, 3) be the given points,
then
| AB| = (1 − 2)2 + (5 − 4 )2 = 2
Sol. Let P ≡ (a cos α, a sin α ) and O ≡ (0, 0)
then
| OP | = (a cos α − 0) + (a sin α − 0)
2
(Qa > 0)
Sol. Let P ≡ ( x , y ) , A ≡ (6, − 1) and B ≡ (2, 3)
Important Remarks for
Objective Questions
2
C
2
B
= (a 2 cos 2 α + a 2 sin 2 α )
2
A
= a 2 (cos 2 α + sin 2 α ) = a 2
= | a | , which is independent of α.
y Example 8. Find the distance between the points
(a cos α , a sin α ) and (a cos β, a sin β), where a > 0.
Sol. Let P ≡ (a cos α, a sin α ) and Q ≡ (a cos β, a sin β )
7
| BC | = (2 − 3)2 + ( 4 − 3)2 = 2
and
| AC | = (1 − 3)2 + (5 − 3)2 = 2 2
Clearly,
| AB| + | BC | = 2 + 2 = 2 2 = | AC |.
Hence, A, B, C are collinear.
8
Textbook of Coordinate Geometry
y Example 11. An equilateral triangle has one vertex at
the point (0, 0) and another at ( 3, 3 ) . Find the
coordinates of the third vertex.
| BC | = (3 − 5)2 + (6 − 10)2 = 4 + 16 = 2 5
Sol. Let O ≡ (0, 0) and A ≡ (3, 3 ) be the given points and
let B ≡ ( x , y ) be the required point. Then
| AD | = (1 − 3)2 + ( −2 − 2)2 = 4 + 16 = 2 5
| CD | = (5 − 3)2 + (10 − 2)2 = 4 + 64 = 2 17
| AC | = (1 − 5)2 + ( −2 − 10)2 = 16 + 144 = 4 10
OA = OB = AB
and
Y
| BD | = (3 − 3)2 + (6 − 2)2 = 4
D
C
(x, y)B
A(3, 3)
A
X′
X
O
(0, 0)
Y′
⇒
Clearly, | AB| = | CD |, | BC | = | AD | and | AC | ≠ | BD |
Hence, ABCD is a parallelogram.
B(x, y)
y Example 13. Let the opposite angular points of a
square be ( 3, 4 ) and (1, − 1). Find the coordinates of the
remaining angular points.
(OA )2 = (OB )2 = ( AB )2
⇒ ( 3 − 0) 2 + ( 3 − 0) 2 = ( x − 0) 2 + ( y − 0) 2
= ( x − 3) 2 + ( y − 3 ) 2
⇒
12 = x 2 + y 2 = x 2 + y 2 − 6x − 2 3y + 12
Taking first two members then
x 2 + y 2 = 12
…(i)
and taking last two members, then
6x + 2 3y = 12 or y = 3(2 − x )
…(ii)
From Eqs. (i) and (ii), we get
x 2 + 3 (2 − x )2 = 12
or
B
Sol. Let A(3, 4 ) and C(1, − 1) be the given angular points of a
square ABCD and let B( x , y ) be the unknown vertex. Then
AB = BC
⇒
( AB )2 = ( BC )2
⇒
⇒
⇒
( x − 3) 2 + ( y − 4 ) 2 = ( x − 1) 2 + ( y + 1) 2
4 x + 10y − 23 = 0
 23 − 10y 
x=



4
…(i)
A(3, 4)
4 x 2 − 12x = 0
D
⇒
x = 0, 3
Putting x = 0, 3 in Eq. (ii), we get y = 2 3, − 3
O
Hence, the coordinates of the third vertex B are (0, 2 3 ) or
(3, − 3 ).
Short Cut Method : According to important note :
 x 1 + x 2 m 3( y 2 − y 1 ) y 1 + y 2 ± 3( x 2 − x 1 ) 
,


2
2


B(x, y)
C(1, –1)
Also, in ∆ ABC,
( AB )2 + ( BC )2 = ( AC )2
i.e.
 0 + 3 m 3( 3 − 0) 0 + 3 ± 3( 3 − 0) 
,


2
2


or
3 m 3 3 ± 3 3
,


2

 2
⇒
⇒
(0, 2 3 ) or (3, − 3 )
Substituting the value of x from Eqs. (i) into (ii), we get
y Example 12. Show that four points (1, − 2) , ( 3, 6 ) ,
( 5 , 10) and ( 3 , 2) are the vertices of a parallelogram.
Sol. Let A ≡ (1, − 2), B ≡ (3, 6), C ≡ (5, 10) and D ≡ (3, 2) be the
given points. Then
| AB| = (1 − 3)2 + ( −2 − 6)2 = 4 + 64 = 2 17
⇒
( x − 3) 2 + ( y − 4 ) 2 + ( x − 1) 2 + ( y + 1) 2
= ( 3 − 1) 2 + ( 4 + 1) 2
x 2 + y 2 − 4 x − 3y − 1 = 0
2
 23 − 10y 
 23 − 10y 
2
 − 3y − 1 = 0
 +y − 4 





4
4
⇒
4y 2 − 12y + 5 = 0
or
(2y − 1)(2y − 5) = 0
5
1
or
y=
2
2
∴
…(ii)
Chap 01 Coordinate System and Coordinates
1
9
in Eq. (i), we get x = ,
2
2
5
1
and putting y = in Eq. (i), we get x = −
2
2
Putting y =
Also,
| AB | = (a − c )2 + (b − d )2
= a 2 + b 2 + c 2 + d 2 − 2ac − 2bd
 9 1
Hence, the required vertices of the square are  ,  and
 2 2
 1 5
− ,  .
 2 2
= r12 + r 22 − 2 (ac + bd ) [from Eqs. (i) and (ii)]
Y
A (a , b )
y Example 14. Find the circumcentre of the triangle
whose vertices are ( −2, − 3) , ( −1 , 0) and (7, − 6 ) . Also
find the radius of the circumcircle.
Sol. Let A ≡ ( −2, − 3), B ≡ ( −1, 0) and C ≡ (7, − 6) .
r1
B(c, d)
θ
X′
Let P ≡ ( x , y ) be the circumcentre of ∆ABC.
X
O
By using Cosine formula in ∆ AOB
(–2, –3)A
P (x,
cos θ =
y)
C(7, –6)
Since, P is the circumcentre
∴
| PA | = | PB| = | PC | ⇒ ( PA )2 = ( PB )2 = ( PC )2
( x + 2) 2 + ( y + 3) 2 = ( x + 1) 2 + ( y − 0) 2
= ( x − 7 ) + ( y + 6)
2
= x + y − 14 x + 12y + 85
Taking first two members, we get
x + 3y + 6 = 0
and taking 1st and last member then, we get
3x − y − 12 = 0
Solving Eqs. (i) and (ii), we get
x = 3, y = − 3
Hence, circumcentre is (3, − 3).
Radius of the circumcircle
…(i)
…(ii)
= PB = (3 + 1)2 + ( −3 − 0)2
= 16 + 9 = 5 units
y Example 15. If the line segment joining the points
A(a, b ) and B(c , d ) subtends an angle θ at the origin O,
prove that
ac + bd
cos θ =
(a 2 + b 2 )(c 2 + d 2 )
or
OA .OB cos θ = ac + bd
Sol. Let OA = r1 and OB = r 2
Now,
r1 = | OA | = (a 2 + b 2 )
…(i)
and
r 2 = | OB| = (c + d )
…(ii)
2
2
r12 + r 22 − (r12 + r 22 − 2 (ac + bd ))
2r1 r 2
=
2(ac + bd ) (ac + bd )
=
r1r 2
2r1 r 2
=
2
(OA )2 + (OB )2 − ( AB )2
2 ⋅ OA . OB
=
=
2
x 2 + y 2 + 4 x + 6y + 13 = x 2 + y 2 + 2x + 1
2
r2
Y′
(–1, 0)B
⇒
9
(ac + bd )
(a + b 2 ) (c 2 + d 2 )
2
…(iii)
[from Eqs. (i) and (ii)]
(ac + bd )
(a + b 2 ) (c 2 + d 2 )
2
Also from Eq. (iii),
r1r 2 cosθ = ac + bd
or OA .OB cosθ = ac + bd
Choice of Axes
For simplification we carefully choose the axes or the
origin. Some situations are given below :
(i) If two lines are perpendicular then point of
intersection is taken as origin and these lines must be
taken as the coordinate axes.
(ii) If two fixed points A and B are given then we take
| AB| = 2a and the mid-point of AB as origin ‘O’, line
AOB as X-axis and the line perpendicular to AB
through O is taken as Y-axis then the coordinates of
the fixed points are ( ±a, 0 ). Similarly if AOB as Y-axis
and the line perpendicular to AB through O is taken
as X-axis then the coordinates of the fixed points are
(0, ± a ).
(iii) If there is a symmetry of any kind then take the
coordinates of the points in a general way i.e. ( x i , y i ),
i = 1, 2, 3, … etc.
10
Textbook of Coordinate Geometry
y Example 16. Show that the triangle, the coordinates of
whose vertices are given by integers, can never be an
equilateral triangle.
Sol. Let A ≡ (0, 0), B ≡ (a , 0) and C ≡ (b, c ) be the vertices of
equilateral triangle ABC where a, b, c are integers then,
| AB| = | BC | = | CA |
⇒
( AB )2 = ( BC )2 = (CA )2
⇒
y Example 18. Let ABCD be a rectangle and P be any
point in its plane. Show that PA 2 + PC 2 = PB 2 + PD 2
Sol. Let A as the origin and AB and AD as the X and Y-axes
respectively. Let AB = a and AD = b, then
B ≡ (a , 0) , D ≡ (0, b ) and C ≡ (a , b )
Let
P ≡ ( α, β )
Y
P(α, β)
a 2 = (a − b )2 + c 2 = b 2 + c 2
…(i)
and taking first and third members, then
b2 + c 2 = a2
…(ii)
From Eqs. (i) and (ii) we get
a = 2b
From Eq. (ii), b 2 + c 2 = (2b )2
c = 3b
2
or
X′
Y′
= ( α − 0) 2 + ( β − 0) 2 + ( α − a ) 2 + ( β − b ) 2
= 2α 2 + 2β 2 − 2aα − 2bβ + a 2 + b 2
and
y Example 17. In any triangle ABC, show that
AB 2 + AC 2 = 2 ( AD 2 + BD 2 )
Distance between Two Points
in Polar Coordinates
A(b, c)
Let O be the pole and OX be the initial line. Let P and Q be
two given points whose polar coordinates are (r1 , θ 1 ) and
(r2 , θ 2 ) respectively.
X
P(r1, θ1)
Y′
LHS = AB 2 + AC 2
r1
= ( b + a ) + ( c − 0) + ( b − a ) + ( c − 0)
2
2
= 2 (a 2 + b 2 + c 2 )
and
2
2
–θ
(θ 1
…(i)
RHS = 2 ( AD 2 + BD 2 )
= 2 {(b − 0)2 + (c − 0)2 + a 2 }
= 2 (a 2 + b 2 + c 2 )
From Eqs. (i) and (ii), we get
AB 2 + AC 2 = 2 ( AD 2 + BD 2 )
…(ii)
From Eqs. (i) and (ii), we get
PA 2 + PC 2 = PB 2 + PD 2
Y
C(a, 0)
…(i)
2
= 2α 2 + 2β 2 − 2aα − 2bβ + a 2 + b 2
Sol. Let D as the origin and DC and DY as the X and Y-axes
respectively. Let BC = 2a, then
B ≡ ( −a , 0), C ≡ (a , 0) and let A ≡ (b, c )
D (0, 0)
RHS = PB + PD
2
= ( α − a ) 2 + ( β − 0) 2 + ( α − 0) 2 + ( β − b ) 2
where, D is the middle point of BC.
Now,
X
Now, LHS = PA + PC 2
2
B(–a, 0)
B(a, 0)
2
which is impossible, since b and c are integers.
X′
A(0, 0)
(Qa ≠ 0)
c = ±b 3
or
C(a, b)
(0, b)D
From first two members, we get
b 2 + c 2 = 2ab
O
(Pole)
…(ii)
Then,
and
∴
θ 1 θ2
Q(r2, θ2)
)
2
r2
X Initial line
OP = r1 , OQ = r2
∠ POX = θ 1 , ∠ QOX = θ 2
∠ POQ = θ 1 − θ 2
Chap 01 Coordinate System and Coordinates
By using Cosine formula in ∆POQ,
or
cos( ∠ POQ ) =
(OP ) + (OQ ) − ( PQ )
2(OP )(OQ )
cos(θ 1 − θ 2 ) =
r12 + r22 − ( PQ ) 2
2 r1 r2
2
2
2
| PQ | = (r12 + r22 − 2 r1 r2 cos(θ 1 − θ 2 ))
∴
11
∴
| AB| = | BC | = | CA |
Hence, points A , B, C are the vertices of an equilateral
triangle.
Aliter :
π
Q
∠ BAX =
2
π
and
∠ CAX =
6
B 3, π
2
Remark
Always taking θ1 and θ2 in radians.
π
y Example 19. Prove that the points (0, 0) ,  3 ,  and
 2
π


 3 ,  are the vertices of an equilateral triangle.
 6
3
π/3
 π
 π
Sol. Let A ≡ (0, 0), B ≡ 3,  and C ≡ 3, 
 2
 6
π/6

π
| AB| = (02 + 32 − 2 ⋅ 0 ⋅ 3 cos  − 0 = 3 units

2

 π π
| BC | = 32 + 32 − 2 ⋅ 3 ⋅ 3 cos  −  
 2 6 

π

= 18 − 18 sin  = (18 − 9 ) = 3 units

6
and

π

| CA | = 32 + 02 − 2 ⋅ 3 ⋅ 0 cos  − 0  = 3 units
6


X
A(0, 0)
Here, given coordinates are in polar form
∴
C, 3 π
6
3
∴
Q In ∆ ABC
∴
∴
or
Hence,
∠ BAC =
π π π
− =
2 6 3
AB = AC
∠ ACB = ∠ ABC = α
π
α +α + = π
3
π
α=
3
| AB| = | BC | = | CA |.
(say)
12
Textbook of Coordinate Geometry
Exercise for Session 2
1.
If the distance between the points (a, 2) and (3, 4) be 8, then a equals to
(a) 2 + 3 15
2.
(b) 2 − 3 15
(b) 10
(b) a =
1
,2
2
(b) 6
(c) 3, − 5
(d) −3, − 5
(c) a = 2, 1
1
(d) a = − , 2
2
(c) 10
(d) 12
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
(b) 2
(c) 0
(d) 3
The point on the axis of y which its equidistant from ( −1, 2) and (3 , 4) , is
(a) (0, 3)
9.
(d) 14
If A ≡ (3, 4) and B is a variable point on the lines | x | = 6. If AB ≤ 4 then the number of positions of B with integral
coordinates is
(a) 1
8.
(d) None of these
The points (a + 1, 1) , (2a + 1, 3) and (2a + 2 , 2a ) are collinear, if
(a) 5
7.
(c) 12
(b) −3, 5
(a) a = − 1, 2
6.
(c) a right angled triangle
Let A(6, − 1), B (1, 3) and C( x , 8) be three points such that AB = BC, then the value of x are
(a) 3, 5
5.
(b) an equilateral triangle
 π
 5π 
The distance between the points 3 ,  and 7,  is
 4
 4
(a) 8
4.
(d) 3 ± 2 15
The three points ( −2 , 2), (8, − 2) and ( −4, − 3) are the vertices of
(a) an isosceles triangle
3.
(c) 2 ± 3 15
(b) (0, 4)
(c) (0, 5)
(d) (0, − 6)
Find the distance between the points (at12, 2 at1) and (at 22, 2 at 2 ) , where t1 and t 2 are the roots of the equation
x 2 − 2 3x + 2 = 0
and
a > 0.
10.
2a 
1
1
a
If P (at 2, 2 at ), Q  2 , −  and S(a, 0) be any three points, show that
is independent of t.
+
t
t 
SP SQ
11.
Prove that the points (3, 4), (8, − 6) and (13, 9) are the vertices of a right angled triangle.
12.
Show that the points (0, − 1), (6, 7), ( −2 , 3) and (8, 3) are the vertices of a rectangle.
13.
Find the circumcentre and circumradius of the triangle whose vertices are ( −2, 3), (2 , − 1) and (4, 0) .
14.
The vertices of a triangle are A(1, 1), B(4, 5) and C(6, 13) . Find cos A.
15.
Two opposite vertices of a square are (2 , 6) and (0,−2) . Find the coordinates of the other vertices.
16.
If the point ( x , y ) is equidistant from the points (a + b , b − a ) and (a − b , a + b ), prove that bx = ay .
17.
If a and b are real numbers between 0 and 1 such that the points (a , 1), (1, b ) and (0, 0) form an equilateral
triangle, find a and b.
18.
An equilateral triangle has one vertex at (3, 4) and another at ( −2 , 3). Find the coordinates of the third vertex.
19.
If P be any point in the plane of square ABCD, prove that
PA2 + PC 2 = PB 2 + PD 2
Session 3
Section Formula, Centroid of a Triangle, Incentre,
Some Standard Results, Area of Triangle
Section Formula
Y
B
P
B
A
A
P
(x
,y
A
AP Positive, in internally division
=
PB Negative, in externally division
(i) Formula for Internal Division
Theorem : If the point P ( x , y ) divides the line segment
joining the points A ( x 1 , y 1 ) and B( x 2 , y 2 ) internally in
the ratio m : n, then prove that
mx 2 + nx 1
x=
m +n
y=
A
(x1,y1)
B
Note
my 2 + ny 1
m +n
Proof : The given points are A ( x 1 , y 1 ) and B ( x 2 , y 2 ). Let
us assume that the points A and B are both in 1st quadrant
(for the sake of exactness). Since P ( x , y ) divides AB
internally in the ratio m : n i.e. AP : PB = m : n. From A, B
and P draw AL, BM and PN perpendiculars to X-axis.
From A and P draw AH and PJ perpendiculars to PN and
BM respectively, then
OL = x 1 , ON = x , OM = x 2 , AL = y 1 , PN = y and BM = y 2
∴
AH = LN = ON − OL = x − x 1
PJ = NM = OM − ON = x 2 − x
PH = PN − HN = PN − AL = y − y 1
and
BJ = BM − JM = BM − PN = y 2 − y
P
J
m
P
Also, if P be any point on the line AB but not between A
and B ( P may be to the right or the left of the points A, B)
then P divides AB externally in the ratio AP : PB
B(x2, y2)
n
)
Definition : If P be any point on the line AB between A
and B then we say that P divides segment AB internally in
the ratio AP : PB.
X′
O
H
L
N
M
X
Y′
Clearly, the ∆ s AHP and PJB are similar and therefore,
their sides are proportional
AH PH AP
=
=
∴
PJ
BJ PB
x − x1 y − y1 m
or
=
=
x2 − x y2 − y n
(i)
(ii)
(iii)
From Eqs. (i) and (iii), we have
x − x1 m
=
x2 − x n
⇒
nx − nx 1 = mx 2 − mx
⇒
(m + n ) x = mx 2 + nx 1
mx 2 + nx 1
∴
x=
m +n
and from Eqs. (ii) and (iii), we have
y − y1 m
=
y2 − y n
⇒
ny − ny 1 = my 2 − my
⇒
(m + n ) y = my 2 + ny 1
my 2 + ny 1
y=
m +n
∴
Thus, the coordinates of P are
 mx 2 + nx 1 my 2 + ny 1 
,

.
 m +n
m +n 
14
Textbook of Coordinate Geometry
Corollary 1 : The above section formula is true for all
positions of the points (i.e. either point or both points are
not in the 1st quadrant), keeping in mind, the proper signs
of their coordinates.
Corollary 2 : If P is the mid-point of AB then m = n, the
coordinates of the middle-point of AB are
 x1 + x2 y1 + y2 
,



2
2 
Remarks
1. If P ( α, β ) be the mid-point of AB and if coordinates of A are
( λ, µ ) then the coordinates of B are ( 2α − λ, 2β − µ ), i.e.
(Double the x-co-ordinate of mid point – x-coordinate of given
point, Double the y-co-ordinate of mid point – y-coordinate of
given point).
2. The following diagram will help to remember the section
formula.
m
A
(x1, y1)
y Example 20. Find the coordinates of the point which
divides the line segment joining the points ( 5, − 2) and
(9, 6 ) in the ratio 3 : 1.
Sol. Let the required point be (x , y ), then
 3 × 9 + 1 × 5
x=
 =8
3+1


and
Thus, the coordinates of the required point are (8, 4 ).
y Example 21. Find the length of median through A of
a triangle whose vertices are A ( −1, 3), B (1, − 1) and
C ( 5, 1) .
Sol. Let D be the mid-point of BC, then coordinates of D are
 1 + 5 − 1 + 1
,

 i.e. (3, 0)
 2
2 
n
P (x, y)
A
(–1, 3)
B
(x2, y2)
C(5, 1)
D(3, 0)
B(1, –1)
∴
Median AD = (3 + 1)2 + (0 − 3)2
= 16 + 9 = 25
= 5 units
c=
0
3. For finding ratio, use ratio λ : 1, then coordinates of P are
 x1 + λx 2 y1 + λy2 
,

 . If λ is positive then divides internally and
1+ λ 
 1+ λ
if λ is negative, then divides externally.
4. The straight line ax + by + c = 0 divides the joint of points
A ( x1, y1 ) and B ( x 2, y2 ) in the ratio
( ax1 + by1 + c )
AP λ
=
=−
PB 1
( ax 2 + by2 + c )
 3 × 6 + 1 × ( − 2) 
y=
 =4
3+1


ax +
by +
y Example 22. Determine the ratio in which
y − x + 2 = 0 divides the line joining ( 3, − 1) and (8, 9 ) .
λ
A
(x1, y1)
1
P
B
(x2, y2)
If ratio is positive, then divides internally and if ratio is negative
then divides externally.
 x + λx 2 y1 + λy2 
Proof : Coordinates of P are  1
,

1+ λ 
 1+ λ
QP lies on the line ax + by + c = 0, then
 y1 + λy2 
 x + λx 2 
a 1
 + c =0
 + b
 1+ λ 
 1+ λ 
or
or
( ax1 + by1 + c ) + λ ( ax 2 + by2 + c ) = 0
( ax1 + by1 + c )
λ
=−
( ax 2 + by2 + c )
1
5. The line joining the points ( x1, y1 ) and ( x 2, y2 ) is divided by the
y
x
X-axis in the ratio − 1 and by Y-axis in the ratio − 1 .
y2
x2
6. In square, rhombus, rectangle and parallelogram diagonals
bisect to each other.
Sol. Suppose the line y − x + 2 = 0 divides the line segment
joining A (3, − 1) and B (8, 9 ) in the ratio λ : 1 at point P,
 8λ + 3 9 λ − 1
then the coordinates of the point P are 
,
.
 λ +1 λ +1
But P lies on y − x + 2 = 0 therefore
 9 λ − 1  8λ + 3
 +2=0

 −
 λ +1  λ +1
⇒
9 λ − 1 − 8λ − 3 + 2λ + 2 = 0
2
⇒
3λ − 2 = 0 or λ =
3
2
So, the required ratio is : 1, i.e. 2 : 3 (internally) since here
3
λ is positive.
Shortcut method
According to Remark 4 :
 − 1 − 3 + 2 2
λ=−
=
 9 −8+2 3
or
λ :1 = 2:3
Chap 01 Coordinate System and Coordinates
y Example 23. The coordinates of three consecutive
vertices of a parallelogram are (1, 3), ( −1, 2) and (2 , 5).
Then find the coordinates of the fourth vertex.
Sol. Let the fourth vertex be D (α, β ). Since ABCD is a
parallelogram, the diagonals bisect to each other.
i.e. mid-point of BD = mid-point of AC
∴
 α − 1 β + 2  2 + 1 5 + 3
,
,


 =
 2
2   2
2 
or
 α − 1 β + 2  3 
,

 =  , 4
2  2 
 2
,
C(2
15
Shortcut Method
According to Remark 5 :
λ
y
− ( − 3) 1
=− 1 =
=
y2
1
6
2
1
∴ The ratio is : 1 i.e. 1 : 2 (internally)
2
y Example 25. The mid-points of the sides of a triangle
are (1, 2), (0, − 1) and (2 , − 1). Find the coordinates of the
vertices of a triangle with the help of two unknowns.
Sol. Let D (1, 2), E (0, − 1) and F (2, − 1) be the mid-points of
BC , CA and AB respectively.
Let the coordinates of A be (α, β ) then coordinates of B and
C are ( 4 − α, − 2 − β ) and ( −α, − 2 − β ) respectively
(see note 1)
Q D is the mid-point of B and C
5)
C
(–α,–2–β)
A(1, 3)
D(1, 2)
B
(4–α,–2–β)
B(–1, 2)
On equating abscissaes and ordinates, we get
α −1 3
or α − 1 = 3 or α = 4
=
2
2
β +2
and
= 4 or β + 2 = 8 or β = 6
2
Hence, the coordinates of the fourth vertex D (α, β ) is ( 4, 6).
y Example 24. In what ratio does X-axis divide the line
segment joining (2 , − 3) and ( 5 , 6 )?
Sol. Let the given points be A (2, − 3) and B (5, 6). Let AB be
divided by the X-axis at P ( x , 0) in the ratio λ : 1
internally. Considering the ordinate of P, then
λ × 6 + 1 × ( − 3)
0=
λ +1
or
λ=
1
2
1
∴ The ratio is : 1 i.e. 1 : 2
2
(Internally)
B(5, 6)
1
λ
O P(x, 0)
A(2, –3)
X
E
(0, –1)
F
(2, –1)
A (α, β)
4 −α −α
2
⇒
1 = 2 − α or α = 1
−2 − β − 2 − β
and
2=
2
⇒
2 = − 2 − β or β = − 4
Hence, coordinates of A , B and C are
(1, − 4 ), (3, 2) and ( −1, 2) respectively.
then
1=
y Example 26. Prove that in a right angled triangle
the mid-point of the hypotenuse is equidistant from
its vertices.
Sol. Let the given right angled triangle be ABC, with right
angled at B. We take B as the origin and BA and BC as the
X andY -axes respectively.
Let BA = a and BC = b
then A ≡ (a , 0) and C ≡ (0, b )
Let M to be the mid-point of the hypotenuse AC, then
a b
coordinates of M are  , 
 2 2
16
Textbook of Coordinate Geometry
(ii) Formula for External Division
Y
Theorem : If the point P ( x , y ) divides the line joining the
points A ( x 1 , y 1 ) and B ( x 2 , y 2 ) externally in the ratio m : n
then prove that
mx 2 − nx 1
my 2 − ny 1
x=
,y =
m −n
m −n
C(0, b)
M
B(0, 0)
A(a, 0)
X
Y′
2
2
(a 2 + b 2 )
a
b


| AM | = a −  + 0 −  =


2
2
2
2
2
(a 2 + b 2 )
a
b


| BM | = 0 −  + 0 −  =


2
2
2
Y
…(ii)
P(x, y)
)
2
(a 2 + b 2 )
a
b


| CM | = 0 −  + b −  =


2
2
2
y2
2
and
... (i)
…(iii)
2,
∴
Proof : The given points are A ( x 1 , y 1 ) and B ( x 2 , y 2 ). Let
us assume that the points A and B are both in the 1st
quadrant (for the sake of exactness). Let P ( x , y ) be the
point which divides AB externally in the ratio m : n, so that
AP m
= .
BP n
(x
X′
a b
2 2
From Eqs. (i), (ii) and (iii), we get
X′
y Example 27 Show that the line joining the mid-points
of any two sides of a triangle is half the third side.
Sol. We take O as the origin and OC and OY as the X and Y-axes
respectively.
Y
A(b, c)
b–a c
F
2 2
B(–a, 0)
E
O
a+b c
2 2
C(a, 0)
X
Y′
Let BC = 2a, then B ≡ ( −a , 0), C ≡ (a , 0)
Let A ≡ (b, c ), if E and F are the mid-points of sides AC and
AB respectively.
a + b c 
b − a c 
Then,
,  and F ≡ 
, 
E≡
 2 2
 2 2
2
Now,
2
 a + b b − a
c c 
FE = 
−
 +  −  =a
 2
 2 2
2 
1
1
(2a ) = ( BG )
2
2
Hence, the line joining the mid-points of any two sides of a
triangle is half the third side.
=
S
A
(x1,y1)
| AM | = | BM | = | CM |
X′
B
L
O
R
M
N
X
Y′
From A, B and P draw AL, BM and PN perpendiculars on
X-axis. Also, from A and B draw AR and BS perpendiculars
on PN ,
then
AR = LN = ON − OL = x − x 1
BS = MN = ON − OM = x − x 2
PR = PN − RN = PN − AL = y − y 1
and
PS = PN − SN = PN − BM = y − y 2
Clearly, the ∆s APR and BPS are similar and therefore their
sides are proportional.
AP AR PR
∴
=
=
PB BS PS
m x − x1 y − y1
or
=
=
n x − x2 y − y2
(i)
(ii)
(iii)
From Eqs. (i) and (ii), we have
m x − x1
=
n x − x2
⇒
⇒
or
mx − mx 2 = nx − nx 1
(m − n ) x = mx 2 − nx 1
mx 2 − nx 1
x=
m −n
Chap 01 Coordinate System and Coordinates
Also, from Eqs. (i) and (iii), we have
m y − y1
=
n y − y2
⇒
⇒
Remarks
1. The following diagram will help to remember the section
formula
+
my − my 2 = ny − ny 1
(m − n ) y = my 2 − ny 1
my 2 − ny 1
y=
m −n
or
–
n
A(x1, y1)
and this can be thought of as the coordinates of the point
dividing AB internally in the ratio m : − n
AP m
=
PB n
P
(x, y)
A
(x1, y1)
B
(x2, y2)
AP
m
−1= −1
PB
n
AP − PB m − n
or
=
PB
n
AB m − n
or
=
PB
n
Now, we can say that B divides AP in the ratio m − n : n
internally.
(m − n ) x + nx 1
mx 2 − nx 1
i.e.
x2 =
⇒ x=
(m − n ) + n
m −n
or
and
y2 =
(m − n ) y + ny 1
(m − n ) + n
⇒ y=
B(x2, y2)
m
 mx − nx 1 my 2 − ny 1 
Thus, the coordinates of P are  2
,
.
 m −n
m −n 
(Here, m ≠ n)
Corollary 1 : The above formula is true for all positions of
the points, keeping in mind, the proper signs of their
coordinates.
Corollary 2 : The above coordinates can also be expressed
as
 mx 2 + ( −n ) x 1 my 2 + ( −n ) y 1 
,


 m + ( −n )
m + ( −n ) 
Corollary 3 : Q
17
2. Let
 mx 2 − nx1 my2 − ny1 
m
,
= λ, then 

m−n 
n
 m−n
or
 mx − x m y − y 
1
2
1
 n 2
, n
 m

m

−1
−1 
 n

n
 λx 2 − x1 λy2 − y1 
,


λ −1 
 λ −1
or
y Example 28. Find the coordinates of a point which
divides externally the line joining (1, − 3) and ( −3, 9 ) in
the ratio 1 : 3.
Sol. Let the coordinates of the required point be P ( x , y ),
Then,
 1 × ( − 3) − 3 × 1
x=

1−3


and
 1 × 9 − 3 × ( − 3) 
y=

1−3


i.e.
x = 3 and
y = −9
Hence, the required point is (3, − 9 ).
y Example 29. The line segment joining A (6, 3) to
B ( −1, − 4 ) is doubled in length by having its length
added to each end. Find the coordinates of the new
ends.
Sol. Let P and Q be the required new ends
Let the coordinates of P be ( x 1, y1 )
Given,
AB = 2AP
AB 2
⇒
=
AP 1
i.e. A divides BP internally in the ratio 2 : 1.
P(x1, y1)
A(6, 3)
my 2 − ny 1
m −n
Corollary 4 : (for proving A, B and C are collinear)
If A, B, C three points are collinear then let C divides AB in
the ratio λ : 1 internally.
If λ = + ve rational, then divide internally
and if λ = − ve rational, then divide externally.
B(–1, –4)
Q(x2, y2)
Then,
6=
2 × x 1 + 1 × ( − 1)
2+1
18
Textbook of Coordinate Geometry
19
2
2 × y 1 + 1 × ( −4 )
3=
2+1
⇒
y Example 30. Using section formula show that the
points (1, − 1), (2 , 1) and (4, 5) are collinear.
19 = 2x 1 or x 1 =
and
Sol. Let A ≡ (1, − 1), B ≡ (2, 1) and C ≡ ( 4, 5)
Suppose C divides AB in the ratio λ : 1 internally, then
13
2
19
13


∴ Coordinates of P are  ,  .
 2 2
13 = 2y1 or y1 =
⇒
4=
i.e. B divides AQ internally in the ratio 2 : 1
2 × x2 + 1 × 6
Then
−1 =
2+1
−9 = 2x 2 or
and
−4 =
x2 = −
C(4, 5)
⇒
4 λ + 4 = 2λ + 1
3
or
λ=−
2
i.e. C divides AB in the
ratio 3 : 2 (externally).
Hence, A , B,C are collinear.
Also, let coordinates of Q be ( x 2 , y 2 )
AB 2
Given,
=
AB = 2BQ ⇒
BQ 1
⇒
λ ×2+1×1
λ +1
9
2
2 × y2 + 1 × 3
2+1
A(1, –1)
y Example 31. Find the ratio in which the point (2 , y )
divides the line segment joining (4, 3) and (6, 3) and
hence find the value of y.
Sol. Let A ≡ ( 4, 3), B ≡ (6, 3) and P ≡ (2, y )
Let P divides AB internally in the ratio λ : 1
6λ + 4
then,
⇒ 2λ + 2 = 6λ + 4
2=
λ +1
15
⇒
−15 = 2y 2 or y 2 = −
2
 9 15
∴ Coordinates of Q are  − , − 
 2
2
AB = 2AP
2
AB 2
AB
⇒
⇒
=
+1= +1
1
AP 1
AP
AB + AP 3
BP 3
⇒
=
=
⇒
AP
1
AP 1
∴ P divides AB externally in the ratio 1 : 3
1 × ( −1) − 3 × 6 19
Then,
=
x1 =
1−3
2
B(2, 1)
⇒
−4 λ = 2 or λ = −
1
2
Aliter : Q
and
y1 =
1 × ( −4 ) − 3 × 3 13
=
1−3
2
 19 13
∴ Coordinates of P are  , 
 2 2
Also,
⇒
⇒
AB = 2BQ
2
AB 2
AB
= ⇒
+1= +1
1
BQ 1
BQ
AB + BQ 3
AQ 3
=
= ⇒
BQ
1
BQ 1
∴ Q divides AB externally in the ratio 3 : 1
3 × ( − 1) − 1 × 6
9
then, x 2 =
=−
3−1
2
and
3 × ( −4 ) − 1 × 3
15
y2 =
=−
3−1
2
 9 15
∴ Coordinates of Q are  − , −  .
 2
2
A(2, 3)
B(4, 3)
C(6, 3)
∴ P divides AB externally in the ratio 1 : 2 (Qλ is negative)
Now, y =
1×3−2×3
=3
1−2
(iii) Harmonic Conjugates
If four points in a line, then the system is said to form a
range. Let four points say P , Q , R, S .
If the range ( PQ , RS ) has a cross ratio equal to −1, then it is
called harmonic.
PR
SP
PR SQ
(say)
i.e.
=−
=λ
.
= −1⇒
RQ
SQ
RQ SP
∴
PR λ
= ⇒ PR : RQ = λ : 1
RQ 1
(internally)
and
λ
SP
= − ⇒ PS : SQ = λ : 1
1
SQ
(externally)
S
R
Q
P
Hence, R and S are called the harmonic conjugates to
each other with respect to the points P and Q.
Chap 01 Coordinate System and Coordinates
 x + x 3 y2 + y 3 
D ≡ 2
,


2
2 
y Example 32 Find the harmonic conjugates of the
point R ( 5, 1) with respect to the points P (2 , 10) and
Q (6, − 2).
 x + x1 y 3 + y1 
,
E ≡ 3


2
2 
Sol. Let S (α , β ) (be the harmonic conjugates of the point R(5, 1 )).
Suppose R divides PQ in the ratio λ : 1 internally, then S
divides PQ in the ratio λ : 1 externally, then
19
 x + x2 y1 + y2 
,
F ≡ 1


2
2 
and
P(2, 10)
Y
A(x1, y1)
E
R(5, 1)
x1+x2 y1+y2
F
2
2
Q(6, –2)
∴
Also,
λ + 1 = − 2λ + 10 ⇒ 3λ = 9
∴
Now,
and
X′
6λ + 2
⇒ 5λ + 5 = 6λ + 2
λ +1
λ =3
−2λ + 10
1=
λ +1
λ =3
3×6−1×2
=8
α=
3−1
β=
3 × ( −2) − 1 × 10
3−1
= −8
Hence, harmonic conjugates of R(5, 1) is S(8, − 8).
Centroid of a Triangle
Definition : The point of intersection of the medians of
a triangle is called the centroid of the triangle and it
divides the median internally in the ratio 2 : 1.
Theorem : Prove that the coordinates of the centroid of
the triangle whose vertices are ( x 1 , y 1 ),( x 2 , y 2 ) and
( x 3 , y 3 ) are
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
Also, deduce that the medians of a triangle are concurrent.
Proof : Let A ≡ ( x 1 , y 1 ), B ≡ ( x 2 , y 2 ) and C ≡ ( x 3 , y 3 ) be
the vertices of the triangle ABC. Let us assume that the
points A, B and C are in the 1st quadrant (for the sake of
exactness) whose medians are AD , BE and CF respectively
so D , E and F are respectively the mid-points of BC , CA
and AB then the coordinates of D , E , F are
G
C(x3, y3)
D x2+x3 y2+y3
2
2
S(α, β)
5=
x3+x1 y3+y1
2
2
B(x2, y2)
X
O
Y′
The coordinates of a point dividing AD in the ratio 2 : 1 are

  x2 + x 3 
y2 + y 3 
 2⋅
 + 1⋅ x 1 2 ⋅ 
 + 1⋅y1 

 2 
2

 
,


2 +1
2 +1




or
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
and the coordinates of a point dividing BE in the ratio 2 : 1
are

  x 3 + x1 
y 3 + y1 
2 ⋅ 
 + 1⋅ x 2 2 ⋅ 
 + 1⋅y2 

 2 
2

 
,


2 +1
2 +1




or
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
Similarly the coordinates of a point dividing CF in the
ratio 2 : 1 are
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
∴ The common point which divides AD , BE and CF in the
ratio 2 : 1 is
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
Hence, medians of a triangle are concurrent and the
coordinates of the centroid are
 x1 + x2 + x 3 y1 + y2 + y 3 
,




3
3
20
Textbook of Coordinate Geometry
Important Theorem
Centroid of the triangle obtained by joining the middle
points of the sides of a triangle is the same as the centroid
of the original triangle.
Or
If (a 1 , b 1 ), (a 2 , b 2 ) and (a 3 , b 3 ) are the mid-points of the
sides of a triangle, then its centroid is given by
 a1 + a2 + a 3 b1 + b2 + b 3 
,




3
3
Proof : Let D , E , F are the mid-points of BC , CA and AB
respectively now let coordinates of A are (α, β) then
coordinates of B and C are (2a 3 − α, 2b 3 − β) and
(2a 2 − α, 2b 2 − β) are respectively.
A(α, β)
B
D(a1, b1)
C
Q D (a 1 , b 1 ) is the mid-point of B and C, then
2a 1 = 2a 3 − α + 2a 2 − α
∴
Area of ∆ABC ( BC ) 2
=
Area of ∆DEF ( EF ) 2
=
4 { (a 2 − a 3 ) 2 + (b 2 − b 3 ) 2 }
{(a 2 − a 3 ) 2 + (b 2 − b 3 ) 2 }
=4
∴ Area of ∆ ABC = 4 × Area of ∆DEF
i.e. Area of a triangle is four times the area of the triangle
formed by joining the mid-points of its sides.
y Example 33. Two vertices of a triangle are ( −1, 4 ) and
( 5 , 2) . If its centroid is (0, − 3) , find the third vertex.
E(a2, b2)
(a 3 , b 3 ) F
Corollary 2 : If two vertices of a triangle are ( x 1 , y 1 ) and
( x 2 , y 2 ) and the coordinates of centroid are (α, β), then
coordinates of the third vertex are
(3α − x 1 − x 2 , 3β − y 1 − y 2 )
Corollary 3 : According to important theorem ∆s ABC
and DEF are similar
⇒ α = a2 + a 3 − a1
and
2b 1 = 2b 3 − β + 2b 2 − β ⇒ β = b 2 + b 3 − b 1
Now, coordinates of B are (2a 3 − α, 2b 3 − β)
or
(a 3 + a 1 − a 2 , b 3 + b 1 − b 2 )
and coordinates of C are (2a 2 − α, 2b 2 − β)
or
(a 2 + a 1 − a 3 , b 2 + b 1 − b 3 )
Hence, coordinates of A, B and C are
A ≡ (a 2 + a 3 − a 1 , b 2 + b 3 − b 1 ),
B ≡ (a 3 + a 1 − a 2 , b 3 + b 1 − b 2 )
and
C ≡ (a 2 + a 1 − a 3 , b 2 + b 1 − b 3 )
∴ Coordinates of centroid of triangle ABC are
 a1 + a2 + a 3 b1 + b2 + b 3 
,




3
3
which is same as the centroid of triangle DEF .
Corollary 1 (Finger Rule) : If mid-points of the sides of a
triangle are ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ), then
coordinates of the original triangle are
(x 2 + x 3 − x 1 , y 2 + y 3 − y 1 ) ,
(x 3 + x 1 − x 2 , y 3 + y 1 − y 2 )
and
(x 1 + x 2 − x 3 , y 1 + y 2 − y 3 ) .
Sol. Let the third vertex be ( x , y ) then the coordinates of the
centroid of triangle are
 −1 + 5 + x 4 + 2 + y 
 4 + x 6 + y
,
,
 i.e. 




 3
3
3
3 
 4 + x 6 + y
Now, 
,
 = (0, − 3)
 3
3 
4+x
6+y
=−3
= 0 and
3
3
⇒
4 + x = 0 and y + 6 = − 9
or
x = − 4 and y = − 15
Hence, the third vertex is ( −4, − 15).
⇒
Shortcut Method
According to corollary 2
( x , y ) = (3 × 0 − ( −1) − 5, 3 × ( −3) − 4 − 2)
= ( −4, − 15)
y Example 34. The vertices of a triangle are
(1, 2), (h, − 3) and ( −4, k ). Find the value of
{(h + k ) 2 + (h + 3k ) 2 }. If the centroid of the triangle
be at the point ( 5, − 1) .
1+h − 4
2−3+k
= 5 and
= −1
3
3
then, we get h = 18, k = −2
Sol. Here,
∴
= (h + k )2 + (h + 3k )2
= (18 − 2)2 + (18 − 6)2
= (162 + 122 ) = 20
21
Chap 01 Coordinate System and Coordinates
y Example 35. If D ( −2 , 3), E (4, − 3) and F (4, 5) are the
mid-points of the sides BC , CA and AB of triangle ABC,
then find (| AG | 2 + | BG | 2 − | CG | 2 ) where, G is the
centroid of ∆ ABC.
Hence,
(| AG | 2 + | BG | 2 − | CG | 2 )
16
16
 64

=  × 10 +
× 58 −
× 34
9

9
9
=
Sol. Let the coordinates of A be (α, β )
32
(20 + 29 − 17 )
9
 32
 32
=  × 32 =

9
3
then coordinates of B are (8 − α, 10 − β )
and coordinates of C are (8 − α, − 6 − β )
Q D is the mid-point of BC, then
8−α +8−α
= −2
2
10 − β − 6 − β
and
=3
2
i.e.
α = 10 and β = − 1
y Example 36. If G be the centroid of the ∆ABC and O
be any other point in the plane of the triangle ABC,
then show that
OA 2 + OB 2 + OC 2 = GA 2 + GB 2 + GC 2 + 3GO 2 .
Sol. Let G be the origin and GO be X-axis.
O ≡ (a, 0), A ≡ ( x 1, y1 ), B ≡ ( x 2 , y 2 )
and C ≡ ( x 3 , y 3 )
B(–2, 11)
Now, LHS = OA 2 + OB 2 + OC 2
F(4, 5)
= ( x 1 − a )2 + y12 + ( x 2 − a )2 + y 22 + ( x 3 − a )2 + y 32
(–2, 3)D
= ( x 12 + x 22 + x 32 ) + (y12 + y 22 + y 32 )
A(10, –1)
− 2a ( x 1 + x 2 + x 3 ) + 3a 2
E(4, –3)
C(–2, –5)
∴ Coordinates of A , B, C are (10, − 1), ( −2, 11) and ( −2, − 5)
respectively.
Now, coordinates of centroid
 10 − 2 − 2 −1 + 11 − 5
,
G≡



3
3
i.e.
∴
 5
G ≡ 2, 
 3
∑y12 − 0 + 3a 2
= ∑ x 12 +
∑y12 + 3a 2
x1 + x 2 + x 3


= 0
Q
3


 i.e., x 1 + x 2 + x 3 = 0
...(i)
RHS = GA + GB + GC + 3GO
2
2
2
2
= x 12 + y12 + x 22 + y 22 + x 32 + y 32 + 3 {(a − 0)2 }
=
∑ x 12 + ∑y12 + 3a 2
... (ii)
Hence, from Eqs. (i) and (ii), we get
5

AG = (10 − 2) +  −1 − 

3
2
OA 2 + OB 2 + OC 2 = GA 2 + GB 2 + GC 2 + 3GO 2 .
2
64  8

= 64 +  =
(10)

9 3
and
and
= ∑ x 12 +
2

BG = ( −2 − 2)2 + 11 −

5

3
(28)2 4
= 16 +
=
9
3
(58)
5

CG = ( −2 − 2)2 +  −5 − 

3
400 4

= 16 +
 =

9  3
y Example 37. If G be the centroid of ∆ ABC, show
that
AB 2 + BC 2 + CA 2 = 3 (GA 2 + GB 2 + GC 2 ) .
Sol. We take B as the origin and BC and BY as the X and
Y-axes respectively.
Y
A (h , k )
2
(34 )
G
X′
B(0, 0)
Y′
C(a, 0)
X
22
Textbook of Coordinate Geometry
Let
BC = a, then B ≡ (0, 0) and C ≡ (a , 0)
and let A ≡ (h , k )
then, coordinates of G will be
 h + 0 + a k + 0 + 0
h + a k 
,
, 
 , i.e. 



 3
3
3
3
Incentre
Take ∆ ABC as in 1st quadrant (for the sake of exactness).
Now, LHS = ( AB )2 + ( BC )2 + (CA )2
= ( h − 0) 2 + ( k − 0) 2 + a 2 + ( h − a ) 2 + ( k − 0) 2
= 2h 2 + 2k 2 − 2ah + 2a 2
... (i)
RHS = 3 ((GA ) + (GB ) + (GC ) )
2
2
2
2
2

  a + h
a + h

k

− 0
= 3 
− h +  − k  + 
 3

3

  3

2
Definition : The point of intersection of internal angle
bisectors of triangle is called the incentre of the triangle.
Theorem : Prove that the coordinates of the incentre of a
triangle whose vertices are
A ( x 1 , y 1 ), B ( x 2 , y 2 ), C ( x 3 , y 3 ) are
 ax 1 + bx 2 + cx 3 ay 1 + by 2 + cy 3 
,


a +b +c
a +b +c


where, a, b, c are the lengths of sides BC, CA and AB
respectively.
Also, prove that the internal bisectors of the angles of a
triangle are concurrent.
2
2
2
k
a
h
k
+
  Proof : Given A ≡ ( x 1 , y 1 ), B ( x 2 , y 2 ), C ≡ ( x 3 , y 3 ) be the





− a +  − 0 
+  − 0 + 
  vertices of ∆ ABC and BC = a, CA = b and AB = c . Let AD
3

 3

3

be the bisector of A. We know that the bisector of an angle
3
of a triangle divides the opposite side in the ratio of the
= {(a − 2h )2 + ( −2k )2 + (a + h )2 + k 2
9
sides containing the triangle.
+ (h − 2a )2 + k 2 }
BD AB c
... (i)
=
=
∴
1
2
2
2
DC
AC b
= {6a + 6h + 6k − 6ah }
3
= 2h 2 + 2k 2 − 2ah + 2a 2
A(x1, y1)
... (ii)
Hence, from Eqs. (i) and (ii), we get
AB 2 + BC 2 + CA 2 = 3 (GA 2 + GB 2 + GC 2 )
A/2 A/2
b+c : a
c
F
y Example 38. The vertices of a triangle are (1, a ), (2, b )
and (c 2 , − 3)
(i) Prove that its centroid can not lie on the Y-axis.
(ii) Find the condition that the centroid may lie on the
X-axis.
Sol. Centroid of the triangle is
 3 + c 2 a + b − 3
 1 + 2 + c 2 a + b − 3
,
G≡
,

 i.e. 
3
3
3
 3



(i)Q G will lie on Y-axis, then
3 + c2
=0
3
⇒
c2 = −3
or
c = ±i 3
Q Both values of c are imaginary.
Hence, G can not lie on Y-axis.
a+b−3
(ii) QG will lie on X-axis, then
=0
3
⇒
a+b−3=0
or
a+b=3
b
E
I
B/2
C/2
B/2
B (x2, y2)
C/2
D
c:b
C(x3, y3)
Hence, D divides BC in the ratio c : b
 cx + bx 2 cy 3 + by 2 
∴ Coordinates of D are  3
,

c +b 
 c +b
DC b
DC
b
= or
+1= +1
BD c
BD
c
DC + BD  b + c 
a
b +c 
=
=
 or



BD
c
BD  c 
From Eq. (i),
or
∴
BD =
ac
(b + c )
Also, in ∆ ABD, BI is the bisector of B.
b +c
AI AB
c
Then,
=
=
=
ID BD  ac 
a


b +c 
Chap 01 Coordinate System and Coordinates
∴ I divides AD in the ratio b + c : a
∴ Coordinates of I are
cx 3 + bx 2
cy + by 2

+ a ⋅ x 1 (b + c ) ⋅ 3
+ b ⋅y1
 (b + c ) ⋅
c +b
c +b

,

b +c +a
b +c +a


i.e.






 ax 1 + bx 2 + cx 3 ay 1 + by 2 + cy 3 
,


a +b +c
a +b +c


Similarly we can show that the coordinates of the point
which divides BE internally in the ratio c + a : b and the
coordinates of the point which divides CF internally in the
ratio a + b : c will be each
 ax 1 + bx 2 + cx 3 ay 1 + by 2 + cy 3 
,


a +b +c
a +b +c


and
CE =
BD = BF = β
CD = CE = γ
Also,
a = BC = BD + DC = β + γ
b = CA = CE + AE = γ + α
and
c = AB = AF + BF = α + β
Adding all, we get
a + b + c = 2 (α + β + γ )
or
2s = 2 (α + β + γ )
∴
s =α +β + γ
From Eqs. (i), (ii) and (iii), we get
α = s − a, β = s − b , γ = s − c
Let A ( 4, − 2), B ( −2, 4 ) and C (5, 5) be the vertices of the
given triangle. Then
a = BC = ( −2 − 5)2 + ( 4 − 5)2 = 50 = 5 2
bc
ac
AF =
, BF =
a +b
a +b
b = CA = (5 − 4 )2 + (5 + 2)2 = 50 = 5 2
Thus, the three internal bisectors of the angles of a
triangle meet in a point I.
and
c = AB = ( 4 + 2)2 + ( −2 − 4 )2 = 72 = 6 2
B
(–2, 4)
Corollary 1 : If ∆ ABC is equilateral, then a = b = c
 x + x2 + x 3 y1 + y2 + y 3 
incentre =  1
,
 = centroid


3
3
i.e. incentre and centroid coincide in equilateral, triangle.
Corollary 2 :
AE = AF = s − a
BD = BF = s − b
CD = CE = s − c
A(4, –2)
Let ( x , y ) be the coordinates of incentre of ∆ ABC. Then
ax + bx 2 + cx 3
x= 1
a+b+c
=
A/2 A/2
I
C/2
and
C/2
D
B
C
a +b +c
2
and
| BC | = a, | CA | = b, | AB| = c
Proof : Let AE = AF = α
(Q Lengths of tangents are equal
from a point to a circle)
where,
s=
5 2 × 4 + 5 2 × ( − 2) + 6 2 × 5
5 2 +5 2 +6 2
20 2 − 10 2 + 30 2
16 2
40 5
=
=
16 2
ay + by 2 + cy 3
y= 1
a+b+c
=
E
B/2
b
c
A
B/2
C
(5, 5)
a
 ax + bx 2 + cx 3 ay 1 + by 2 + cy 3 
I ≡ 1
,

a +b +c
a +b +c


F
....(i)
…(ii)
…(iii)
y Example 39. Find the coordinates of incentre of the
triangle whose vertices are (4, − 2) , ( −2 , 4 ) and ( 5 , 5).
Sol.
bc
ab
,
, AE =
(c + a )
(c + a )
23
=
5 2 × ( − 2) + 5 2 × 4 + 6 2 × 5
5 2 +5 2 +6 2
=
40 5
=
16 2
 5 5
∴ The coordinates of the incentre are  , 
 2 2
24
Textbook of Coordinate Geometry
3
3
y Example 40. If  , 0 ,  , 6 and ( −1, 6 ) are
2  2 
mid-points of the sides of a triangle, then find
(i) Centroid of the triangle
(ii) Incentre of the triangle
Sol. Let coordinate of A be (1, 1) and mid-points of AB and AC
are F and E.
∴
F ≡ ( −2, 3) and E ≡ (5, 2)
Y
B(–5, 5)
D(2, 4)
Sol. Let A ≡ (α, β ), then coordinates of B ≡ ( −2 − α, 12 − β ) and
coordinates of C ≡ (3 − α, 12 − β ). But mid-point of BC is
3 
 , 0
2 
C(9, 3)
(–2, 3) F
X¢
A(1, 1)
O
E(5, 2)
X
Y¢
A(–1, 12)
E
F
(–1, 6)
3,
6
2
Hence, coordinates of B and C are (2 × ( −2) − 1, (2 × 3 − 1)
and (2 × 2 − ( −5), 2 × 4 − 5) respectively.
i.e.
B ≡ ( −5 ,5) and C ≡ (9, 3)
5 
 1 − 5 + 9 1 + 5 + 3
Then, centroid is 
,
 i.e.,  , 3
3 


3
3
Also, a = BC = ( −5 − 9 )2 + (5 − 3)2 = 200 = 10 2
B
(–1, 0)
D
3,
0
2
C
(4, 0)
then 3 = − 2 − α + 3 − α
⇒
α = −1
and
0 = 12 − β + 12 − β
⇒
β = 12
∴ Coordinates of vertices are
A ≡ ( −1, 12), B ≡ ( −1, 0) and C ≡ ( 4, 0)
(i) Centroid : The centroid of ∆ ABC is
 x 1 + x 2 + x 3 y1 + y 2 + y 3 
,




3
3
or
2 
 −1 − 1 + 4 12 + 0 + 0
,

 i.e.  , 4
3 


3
3
(ii) Incentre : We have
a = BC = ( −1 − 4 )2 + (0 − 0)2 = 5
b = CA = ( 4 + 1)2 + (0 − 12)2 = 13
and
c = AB = ( −1 + 1) + (12 − 0) = 12
2
2
∴ The incentre of ∆ ABC is
 ax 1 + bx 2 + cx 3 ay1 + by 2 + cy 3 
,



 a+b+c
a+b+c
b = CA = (9 − 1)2 + (3 − 1)2 = 68 = 2 17
and
c = AB = (1 + 5)2 + (1 − 5)2 = 52 = 2 13
Then, incentre is
 10 2 × 1 + 2 17 × ( −5) + 2 13 × 9
,

10 2 + 2 17 + 2 13

10 2 × 1 + 2 17 × 5 + 2 13 × 3

10 2 + 2 17 + 2 13

i.e.
 5 2 − 5 17 + 9 13 5 2 + 5 17 + 3 13 
,


 5 2 + 17 + 13 5 2 + 17 + 13 
y Example 42. If G be the centroid and I be the
incentre of the triangle with vertices A( −36, 7 ), B(20, 7 )
25
and C(0, − 8 ) and GI =
(205) λ, then find the value
3
of λ.
Sol. Coordinates of centroid are
 16 
G ≡  − , 2
 3 
and
a = BC = (20 − 0)2 + (7 + 8)2
= 625 = 25
 5 × ( −1) + 13 × ( −1) + 12 × 4 5 × 12 + 13 × 0 + 12 × 0
or 
,


5 + 13 + 12
5 + 13 + 12

b = CA = (0 + 36)2 + ( − 8 − 7 )2
i.e.
c = AB = ( −36 − 20)2 + (7 − 7 )2
(1, 2)
y Example 41. If a vertex of a triangle be (1, 1) and the
middle points of two sides through it be ( − 2, 3) and
( 5, 2), then find the centroid and the incentre of the
triangle.
= 1521 = 39
= (56)2 = 56
Therefore, the coordinates of incentre are
 25 × ( −36) + 39 × 20 + 56 × 0 25 × 7 + 39 × 7 + 56 × ( −8)
I ≡
,

25 + 39 + 56
25 + 39 + 56


Chap 01 Coordinate System and Coordinates
i.e.,
Y
I ≡ ( − 1, 0)
25
B(2, 8)
2
∴
(205)

 16
GI =  −
+ 1 + (2 − 0)2 =

 3
3
F(3, 4)
25
but given
(205) λ
GI =
3
25
1
∴
(205) =
(205) λ
3
3
1
λ=
⇒
25
D (1, 1)
X¢
E (2, –3)
C (0, –6)
Y¢
Some Standard Results
1. Excentres of a Triangle
(x1
,y
1)
This is the point of intersection of the external bisectors of
the angles of a triangle.
I3
A
X
A (4, 0)
O
I2
then B ≡ (6 − α, 8 − β )
and C ≡ ( 4 − α, − 6 − β )
Also, D is the mid-point of B and C, then
6−α + 4 −α
⇒α = 4
1=
2
8−β −6−β
and 1 =
⇒β = 0
2
∴
A ≡ ( 4, 0), B ≡ (2, 8) and C ≡ (0, − 6), then
a = | BC | = (2 − 0)2 + (8 + 6)2 = 200 = 10 2
(x2, y2)B
C(x3, y3)
I1
The circle opposite to the vertex A is called the escribed
circle opposite A or the circle escribed to the side BC. If I 1
is the point of intersection of internal bisector of ∠ BAC
and external bisector of ∠ ABC and ∠ ACB, then
b = | CA | = (0 − 4 )2 + ( −6 − 0)2 = 52 = 2 13
and
Hence, the coordinates of the excentre opposite to A are
 −ax 1 + bx 2 + cx 3 ay1 + by 2 + cy 3 
,



−a + b + c
−a + b + c 
i.e.,
 ax − bx 2 − cx 3 ay 1 − by 2 − cy 3 
I1 ≡  1
,

a −b −c
a −b −c


or
 − ax 1 + bx 2 + cx 3 − ay 1 + by 2 + cy 3 
I1 ≡ 
,

−a + b + c
−a + b + c


 ax − bx 2 + cx 3 ay 1 − by 2 + cy 3 
Similarly, I 2 ≡  1
,

a −b +c
a −b +c


and
 ax + bx 2 − cx 3 ay 1 + by 2 − cy 3 
,
I3 ≡ 1

a +b −c
a +b −c


where, | BC | = a, | CA| = b and | AB| = c
c = | AB | = ( 4 − 2)2 + (0 − 8)2 = 68 = 2 17
 −10 2 × 4 + 2 13 × 2 + 2 17 × 0
,

−10 2 + 2 13 + 2 17

−10 2 × 0 + 2 13 × 8 + 2 17 × ( −6)

−10 2 + 2 13 + 2 17

or
 −20 2 + 2 13
8 13 − 6 17 
,


 −5 2 + 13 + 17 −5 2 + 13 + 17 
2. Circumcentre of a Triangle
The circumcentre of a triangle is the point of intersection of
the perpendicular bisectors of the sides of a triangle (i.e., the
lines through the mid-point of a side and perpendicular to
it). Let A( x 1 , y 1 ),B( x 2 , y 2 ) and C ( x 3 , y 3 ) be the vertices of
∆ABC and if angles of ∆ABC are given, then coordinates
of circumcentre
A
y Example 43. If the coordinates of the mid-points of
sides BC , CA and AB of triangle ABC are (1, 1), (2, − 3)
and ( 3, 4 ) , then find the excentre opposite to the
vertex A.
Sol. Let D (1, 1), E(2, − 3) and F(3, 4 ) are the mid-points of the
sides of the triangle BC , CA and AB respectively. Let
A ≡ ( α, β )
E
F
O
B
D
C
26
Textbook of Coordinate Geometry
are
 x 1 sin 2 A + x 2 sin 2 B + x 3 sin 2C
,

sin 2 A + sin 2 B + sin 2C

y 1 sin 2 A + y 2 sin 2 B + y 3 sin 2C 

sin 2 A + sin 2 B + sin 2C

Or
 ax 1 cos A + bx 2 cos B + cx 3 cos C
,

a cos A + b cos B + c cos C

ay 1 cos A + by 2 cos B + cy 3 cos C 

a cos A + b cos B + c cos C

where, | BC | = a, | CA| = b and | AB| = c
y Example 44. In a ∆ABC with vertices A(1, 2), B(2, 3)
 1 
and C( 3, 1) and ∠A = ∠B = cos −1 
,
 10 
 4
∠C = cos −1   , then find the circumcentre of ∆ABC .
 5
 1 
Sol. Since, ∠A = ∠B = cos −1 

 10 
1
⇒
cos A = cos B =
10
3
then, sin A = sin B =
10
3
1
3
∴
×
=
sin 2A = sin 2B = 2 ×
10
10 5
 4
and
∠C = cos −1  
 5
4
3
then, sinC =
5
5
3 4 24
sin2C = 2 × × =
∴
5 5 25
Let the circumcenter be ( x , y ), then
x sin 2A + x 2 sin 2B + x 3 sin 2C
x= 1
sin 2A + sin 2B + sin 2C
⇒
cosC =
24
3
3
+2× +3×
25 = 13
5
5
=
3 3 24
6
+ +
5 5 25
y sin 2A + y 2 sin 2B + y 3 sin 2C
y= 1
sin 2A + sin 2B + sin 2C
1×
and
24
3
3
+3× +1×
25 = 11
5
5
=
3 3 24
6
+ +
5 5 25
 13 11
Hence, coordinates of circumcenter are  ,  .
 6 6
2×
Two Important Tricks for Circumcentre
(a) If angles of triangle ABC are not given and the
vertices A( x 1 , y 1 ), B( x 2 , y 2 ) and C ( x 3 , y 3 ) are given,
then the circumcentre of the ∆ABC is given by
 ( x 1 + x 2 ) + λ(y 1 − y 2 ) (y 1 + y 2 ) − λ( x 1 − x 2 ) 
,




2
2
Here, we observe that
x 1 − x 3
P =
x 2 − x 3
→
y1 − y 3 
y 2 − y 3 
→
R1 ⋅ R2
λ=
| P|
∴
(b) If the angle C is given instead of coordinates of the
vertex C and the vertices A( x 1 , y 1 ), B( x 2 , y 2 ) of
∆ABC are given, then the circumcentre of ∆ABC is
given by
 ( x 1 + x 2 ) ± cot C (y 1 − y 2 ) (y 1 + y 2 ) ± cot C ( x 1 − x 2 ) 
,




2
2
Remark
Circumcentre of the right angled triangle ABC, right angled at A
is B + C.
2
y Example 45. Find the circumcentre of the triangle
whose vertices are (2, 2), (4, 2) and (0, 4).
Sol. Let the given points are ( x 1, y1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) respectively.
for the matrix
 x 1 − x 3 y1 − y 3  2 −2
P=

=
x 2 − x 3 y 2 − y 2  4 −2
→
∴
→
R1⋅ R 2
λ=
| P|
=
2 × 4 + ( −2) × ( −2) 12
=
=3
2 × ( − 2) − 4 × ( − 2) 4
∴ Circumcentre of the triangle
 ( x + x 2 ) + λ (y 1 − y 2 ) (y 1 + y 2 ) − λ ( x 1 − x 2 )
,
≡ 1



2
2
 2 + 4 + 3( 2 − 2) 2 + 2 − 3( 2 − 4 ) 
,
≡



2
2
≡ (3, 5)
y Example 46. Find the circumcentre of triangle ABC if
π
A ≡ (7, 4 ), B ≡ ( 3, − 2) and ∠C = .
3
Sol. Here, x 1 = 7, y1 = 4, x 2 = 3, y 2 = − 2 and ∠C =
π
3
Chap 01 Coordinate System and Coordinates
∴ The circumcentre of ∆ABC
 ( x + x 2 ) ± cot C (y1 − y 2 ) (y1 + y 2 ) m cot C ( x 1 − x 2 )
,
≡ 1



2
2
1
1


( 4 + 2) ( 4 − 2) m
( 7 − 3) 
 ( 7 + 3) ±
3
3
,
≡

2
2




2 
2

≡ 5 + 3, 1 −
 or 5 − 3, 1 +




3
3
y Example 47. Find the orthocentre of ∆ABC if
A ≡ (0, 0), B ≡ ( 3, 5) and C ≡ (4, 7 ).
Sol. Here, x 1 = 3, y1 = 5, x 2 = 4, y 2 = 7
∴
λ=
The orthocentre of a triangle is the point of intersection of
altitudes
(i.e., the lines through the vertices and perpendicular to
opposite sides).
Let A( x 1 , y 1 ), B( x 2 , y 2 ) and C ( x 3 , y 3 ) be the vertices of
∆ABC and if angles of ∆ABC are given, then coordinates
of orthocentre are
⇒ Orthocentre of ∆ABC ≡ ( 47(7 − 5), − 47( 4 − 3))
≡ (94, − 47 )
If a circle passing through the feet of perpendiculars (i.e.,
D , E , F ) mid-points of sides BC , CA, AB respectively (i.e.,
H , I , J ) and the n x
mid-points of the line joining the orthocentre O to the
angular points A, B, C (i.e., K , L, M ) thus the nine points
D , E , F , H , I , J , K , L, M all lie on a circle.
F
E
J
H
D
N
L
I
G
M
C
 x 1 tan A + x 2 tan B + x 3 tan C
,

tan A + tan B + tan C

y 1 tan A + y 2 tan B + y 3 tan C 

tan A + tan B + tan C

or
A K
E
O
F
B
3× 4 +5×7
= 47
3×7 − 4 ×5
4. Nine Point Centre of a Triangle
3. Orthocentre of a Triangle
A
27
 ax 1 sec A + bx 2 sec B + cx 3 sec C
,

a sec A + b sec B + c sec C

ay 1 sec A + by 2 sec B + cy 3 sec C 

a sec A + b sec B + c sec C

where, | BC | = a, | CA | = b and | AB | = c
Important trick for orthocentre :
orthocentre of the triangle whose vertices are (0, 0), (x 1 , y 1 )
is given by
( λ(y 2 − y 1 ), − λ ( x 2 − x 1 ))
x x + y 1y 2
where,
λ= 1 2
x 1y 2 − x 2y 1
Remarks
1. The orthocentre of a triangle having vertices ( α, β ), ( β, α ) and
( α, α ) is ( α, α ).


1
2. The orthocentre of a triangle having vertices is  −
, − αβγ 
 αβγ

3. The orthocentre of right angled triangle ABC, right angled at A
is A.
B
D H
C
This circle is known as nine point circle and its centre is
called the nine point centre. The nine-point centre of a
triangle is collinear with the circumcentre and the
orthocentre and bisects the segments joining them and
radius of nine point circle of a triangle is half the radius of
the circumcircle.
Corollary 1 : The orthocentre, the nine point centre, the
centroid and the circumcentre therefore all lie on a
straight line.
Corollary 2 : If O is orthocentre, N is nine point centre, G
is centroid and C is circumcentre, then to remember it see
ONGC (i.e. Oil Natural Gas Corporation) in left of G are 2
and in right is 1, therefore G divides O and C in the ratio 2
: 1 (internally).
Corollary 3 : N is the mid-point of O and C
1
Corollary 4 : Radius of nine point circle = × Radius of
2
circumcircle
Remarks
1. The distance between the orthocentre and circumcentre in an
equilateral triangle is zero.
2. If the circumcentre and centroid of a triangle are respectively
( α, β ) ( γ, δ) then orthocentre will be ( 3γ − 2α, 3δ − 2β ).
28
Textbook of Coordinate Geometry
y Example 48. If a triangle has its orthocentre at (1, 1)
 3 3
and circumcentre at  ,  , then find the centroid
 2 4
and nine point centre.
Sol. Since, centroid divides the orthocentre and circumcentre in
the ratio 2 : 1 (internally) and if centroid G ( x , y ), then
2
1
3 3
O (1, 1)  G ( x , y )  C  , 
2 4
3
2× +1×1
4
2
=
x=
2+1
3
and
 4 5
∴ Centroid is  ,  and nine point centre is the mid-point
 3 6
of orthocentre and circumcentre.
3
3

1 + 1 + 
2,
4  , i.e.  5 , 7  .
∴ Nine point centre is 
 4 8
2 
 2


y Example 49. The vertices of a triangle are
A (a, a tan α ) , B (b , b tan β) and C (c , c tan γ ) . If the
circumcentre of ∆ ABC coincides with the origin and
H ( x , y ) is the orthocentre, then show that
y  sin α + sin β + sin γ 
=
.
x  cos α + cos β + cos γ 
Sol. If R be the circumradius and O be the circumcentre
∴
OA = OB = OC = R
or
Since, G divides H and O in the ratio 2 : 1 (internally), then
2⋅ 0 + 1⋅ x
R
(cos α + cos β + cos γ ) =
3
2+1
or
3
2× +1×1
5
4
=
y=
2+1
6
and
Similarly,
b tan β = R sin β and c tan γ = R sin γ
∴ Coordinates of the vertices of a triangle are :
A ( R cos α, R sin α ), B ( R cos β, R sin β )
and C ( R cos γ , R sin γ )
∴ Centroid
 R (cos α + cos β + cos γ ) R (sin α + sin β + sin γ )
(G ) ≡ 
,



3
3
(a 2 + a 2 tan 2 α ) = (b 2 + b 2 tan 2 β )
= (c 2 + c 2 tan 2 γ ) = R
a sec α = b sec β = c sec γ = R
a = R cos α, b = R cos β, c = R cos γ
sin α
then, a tan α = R cos α ⋅
= R sin α
cos α
or
or
or
R
x
(cos α + cos β + cos γ ) =
3
3
R
2⋅0 + 1⋅y
(sin α + sin β + sin γ ) =
3
2+1
R
y
(sin α + sin β + sin γ ) =
3
3
y  sin α + sin β + sin γ 
=

x  cos α + cos β + cos γ 
Area of a Triangle
Theorem : The area of a triangle, the coordinates of
whose vertices are ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) is
1
| x 1 (y 2 − y 3 ) + x 2 (y 3 − y 1 ) + x 3 (y 1 − y 2 ) |
2
x1
1
y1
1
or
| x2
1|
y2
2
x3 y3
1
Proof : Let ABC be a triangle with vertices A ( x 1 , y 1 ),
B ( x 2 , y 2 ) and C ( x 3 , y 3 ). Let us assume that the points
A, B and C are in 1st quadrant (for the sake of exactness).
Draw AL, BM and CN perpendicular on X-axis. Let ∆ be
the required area of the triangle ABC, then
Y
A(x1, y1)
R
H
(x, y) 2 G
1 O (0, 0)
C(x3, y3)
B
(x2,y2)
R
X′
B
(b,b tan β)
D
E
...(ii)
Dividing Eqs. (ii) by (i), then we get
A(a,a tan α)
R
...(i)
C
(c,c tan γ)
O
Y′
M
L
N
X
Chap 01 Coordinate System and Coordinates
∆ = Area of triangle ABC
= [Area of trapezium ABML + Area of trapezium ALNC
− Area of trapezium BMNC]
1
[ Q Area of trapezium = (Sum of parallel sides)
2
× (Distance between them)]
1
1
∴ ∆ =  ( BM + AL ) ( ML ) + ( AL + CN )
2
2

1

( LN ) − ( BM + CN ) ( MN ) 
2

29
and Area of quadrilateral ABCD : The area of a
quadrilateral can be found out by dividing the
quadrilateral into two triangles.
∴ Area of quadrilateral ABCD
A(x1, y1)
D(x4, y4)
B
(x2, y2)
1
1
=  ( BM + AL ) (OL − OM ) + ( AL + CN ) (ON − (OL )
2
2

1

− ( BM + CN ) (ON − OM ) 
2

C(x3, y3)
= Area of ∆ABC + Area of ∆DAC
x y 2 1 x 3 y 3
1 x 1 y 1

+ 
 2
+ 1
= 
2  x 2 y 2 2 x 3 y 3 2 x 1 y 1
1
1
=  (y 2 + y 1 ) ( x 1 − x 2 ) + (y 1 + y 3 ) ( x 3 − x 1 )
x y 1 1 x 3 y 3
1 x 4 y 4
2
2
+
 1
+ 1
+ 
1

y
2
2
x
y
x
 2 x4 y4



1
1
3
3
− (y 2 + y 3 ) ( x 3 − x 2 ) 
2

x y 3 1 x 4 y 4
x y
1  x 1 y 1
1
|+ 
 3
+ 1
 2 2
+ 1

=| 
= [ x 1 (y 2 + y 1 − y 1 − y 3 ) + x 2 ( −y 2
y
x
2
2
2
y
x
y
2
x
x
y








2
2
3
3
4
4
1
1
2
  x 3 y 3  x 1 y 1
− y 1 + y 2 + y 3 ) + x 3 (y 1 + y 3 − y 2 − y 3 )]

= −
Q
1
  x 1 y 1  x 3 y 3
= [ x 1 (y 2 − y 3 ) + x 2 (y 3 − y 1 ) + x 3 (y 1 − y 2 )]
2
∴ Area of polygon whose vertices are
The area of triangle ABC will come out to be a positive
( x 1 , y 1 ), ( x 2 , y 2 ), ( x 3 , y 3 ), . . . , ( x n , y n ) is
quantity only when the vertices A, B, C are taken in
x y n
x 2 y 2  x 3 y 3
1 x1 y1 
anticlockwise direction and if points A, B, C are taken in
|
 n
+ ..... +
+
+
|
2 x 2 y 2  x 3 y 3  x 4 y 4
clockwise direction then the area will be negative and if
 x 1 y 1
the points A, B, C are taken arbitrary then the area will be
Or
positive or negative, the numerical value being the same
Stair Method Repeat first coordinates one time in last for
in all cases.
A
A
down arrow use positive sign and for up arrow use
Thus in general i.e.,
+
–
negative sign.
B
C B
C
1
Area of ∆ ABC = | x 1 (y 2 − y 3 ) + x 2 (y 3 − y 1 )
2
+ x 3 (y 1 − y 2 ) |
This expression can be written in determinant form as
follows
x1
1
| x2
2
x3
y1
1
y2
1|
y3
1
Corollary 1 : Area of triangle can also be found by easy
method
x y 2 1 x 3 y 3
1 x 1 y 1
|
+ 
 2
+ 1
∆ =| 
2  x 2 y 2 2 x 3 y 3 2 x 1 y 1
∴
Area of polygon =
1
2
x1
y1
x2
y2
x3
y3
xn
yn
x1
y1
1
= | {( x 1 y 2 + x 2 y 3 +… + x n y 1 )
2
− (y 1 x 2 + y 2 x 3 +… + y n x 1 )}|
30
Textbook of Coordinate Geometry
Corollary 2 : If the coordinates of the vertices of the
triangle are given in polar form i.e.,
A (r1 , θ 1 ), B (r2 , θ 2 ), C (r 3 , θ 3 ).
Y
A(r1, θ1)
Corollary 4 : Area of the triangle formed by the lines of
the form y = m 1 x + c 1 , y = m 2 x + c 2 and y = m 3 x + c 3 is
1(c − c 3 ) 2 (c 3 − c 1 ) 2 (c 1 − c 2 ) 2

∆=  2
+
+
2 (m 2 − m 3 ) (m 3 − m 1 ) (m 1 − m 2 )
Remarks
B
(r2, θ2)
C(r3, θ3)
X
O
Then, area of triangle
1
= | r1 r2 sin (θ 1 − θ 2 ) + r2 r 3 sin (θ 2 − θ 3 ) + r 3 r1 sin (θ 3 − θ 1 )|
2
1
= | ∑ r1 r2 sin (θ 1 − θ 2 ) |
2
Corollary 3 : If a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0 and
a 3 x + b 3 y + c 3 = 0 are the sides of a triangle, then the area
of the triangle is given by (without solving the vertices)
2
a1 b1 c 1
1
∆=
a2 b2 c 2
2 | C 1C 2C 3 |
a3 b3 c3
where, C 1 , C 2 , C 3 are the cofactors of c 1 , c 2 , c 3 in the
determinant
a2 b2
Here,
C1 =
= (a 2 b 3 − a 3 b 2 )
a3 b3
and
C2 =
a3
b3
a1
b1
C3 =
a1
b1
a2
b2
a1
b1
c1
a2
a3
b2
b3
c 2 = c 1C 1 + c 2C 2 + c 3C 3
c3
and
1
| { x (5 + 2) − 3 ( −2 − y ) + 4 (y − 5)}|
Area of ∆ PBC
=2
1
Area of ∆ ABC
| {6 (5 + 2) − 3 ( −2 − 3) + 4 (3 − 5)}|
2
P(x, y)
A(6, 3)
= (a 1 b 2 − a 2 b 1 )
C(4, –2)
=
| 7 x + 7y − 14 | 7 | x + y − 2 | | x + y − 2 |
=
=
| 49 |
49
7
y Example 51. Find the area of the pentagon whose
vertices are A (1, 1), B (7, 21), C (7 − 3) , D (12 , 2) and
E (0, − 3) .
∆
Area of triangle =
2 | ∆1 ∆2 ∆ 3 |
a1
b1
c1
∆ = a2
a3
b2
b3
c 2 , ∆1 =
a2
c3
a3
a1
b3
b1
∆3 =
Sol. We have
B
(–3, 5)
Or
and
y Example 50. The coordinates of A,B, C are
(6, 3),( −3, 5) and (4, − 2) respectively and P is any points
( x , y ) . Show that the ratio of the areas of the triangles
| x + y − 2|
PBC and ABC is
.
7
= (a 3 b 1 − a 1 b 3 )
2
where,
1. If area of a triangle is given then, use ± sign.
2. The points A ( x1, y1 ), B ( x 2, y2 ) and C ( x 3, y3 ) are collinear, then
area of ( ∆ ABC) = 0.
3. Four given points will be collinear, then area of the
quadrilateral is zero.
4. Area of the triangle formed by the points ( x1, y1 ), ( x 2, y2 ) and
1 x − x3 x2 − x3
|
( x 3, y3 ) is ∆ = | 1
2 y1 − y3 y2 − y3
1
5. If one vertex ( x 3, y3 ) is at (0, 0) then, ∆ = | x1 y2 − x 2 y1 |
2
a1
b1
a2
, ∆2 =
b2
a3
b2
b3
Sol. The required area
1 1 1
12 2
 
0 –3

 7 –3

7 21

 +
|
+
+
|
+
= 
2 7 21 7 –3 12 2 0 –3 1 1
1
= |(21 − 7 ) + ( −21 − 147 ) + (14 + 36) + ( −36 − 0) + (0 + 3)|
2
1
137
sq units
= | − 137 | =
2
2
Chap 01 Coordinate System and Coordinates
y Example 52. Show that the points (a , 0), (0, b ) and
1 1
(1, 1) are collinear, if + = 1
a b
h k
1
Now, Area of ∆PAB = | 3
4
2
5 −2
Sol. Let A ≡ (a , 0), B ≡ (0, b ) and C ≡ (1, 1)
or
⇒
⇒
Now, points A , B,C will be collinear, if area of ∆ABC = 0
1 a 0 0 b 1 1
or
|
+
+
|= 0
2 0 b 1 1 a 0
⇒
or
or
| (ab − 0) + (0 − b ) + (0 − a )| = 0
ab − a − b = 0
1 1
a + b = ab or + = 1
a b
we get
Solving
Since, ∆ ABC is equilateral
3
(side)2
4
3
=
( AB )2
4
∴ Area of ∆ ABC =
6h + 2k − 26 = ± 20
6h + 2k − 46 = 0 or 6h + 2k − 6 = 0
3h + k − 23 = 0 or 3h + k − 3 = 0
h − 3k − 1 = 0 and 3h + k − 23 = 0,
h = 7, k = 2
h − 3k − 1 = 0 and 3h + k − 3 = 0,
Sol. The given lines are :
7 x − 2y + 10 = 0
7 x + 2y − 10 = 0
9x + y + 2 = 0
2
7 −2 10
1
7
∴ Area of triangle ∆ =
2 −10

2 | C 1C 2C 3 | 
1
2
9
= Irrational
and
…(ii)
Rational = Irrational
which is contradiction.
Hence, x 1, y1, x 2 , y 2 , x 3 , y 3 cannot all be rational.
and
y Example 54. The coordinates of two points A and B
are (3, 4) and ( 5, − 2) respectively. Find the coordinates
of any point P if PA = PB and area of ∆ APB is 10.
Sol. Let coordinates of P be (h ,k ).
Q
PA = PB ⇒ ( PA )2 = ( PB )2
⇒
( h − 3) + ( k − 4 ) = ( h − 5) + ( k + 2)
⇒
( h − 3) 2 − ( h − 5) 2 + ( k − 4 ) 2 − ( k + 2) 2 = 0
⇒
⇒
⇒
(2h − 8) (2) + (2k − 2) ( −6) = 0
( h − 4 ) − 3 ( k − 1) = 0
h − 3k − 1 = 0
...(i)
7 2

C 1 =
= 7 − 18 = − 11,
9 1
1
9
 = − 18 − 7 = − 25
C 2 =
7
−
2

From Eqs. (i) and (ii),
2
1
y Example 55. Find the area of the triangle formed by
the straight lines 7 x − 2y + 10 = 0, 7 x + 2y − 10 = 0 and
9 x + y + 2 = 0 (without solving the vertices of the
triangle).
where,
3
{( x 1 − x 2 )2 + (y1 − y 2 )2 }
4
2
1 | = 10
h = 1, k = 0
Hence, the coordinates of P are (7, 2 ) or (1, 0 ).
Sol. Let A ( x 1, y1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) be the vertices of a
triangle ABC. If possible let x 1, y1, x 2 , y 2 , x 3 , y 3 be all
rational.
1
Now, area of ∆ ABC = | x 1 (y 2 − y 3 ) + x 2 (y 3 − y1 )
2
+ x 3 (y 1 − y 2 ) |
= Rational
…(i)
2
1
we get
y Example 53. Prove that the coordinates of the
vertices of an equilateral triangle can not all be
rational.
=
Solving
31
2
…(i)
7 −2
 = 14 + 14 = 28,
C 3 =
2
7
7 −2 10
7
2 −10 = 10C 1 − 10 C 2 + 2C 3


1
2
9
= 10 × ( −11) − 10 × ( −25) + 2 × 28 = 196
1
∴ From Eq. (i), ∆ =
× (196)2
2 | − 11 × ( −25) × 28 |
=
196 × 196
686
sq units
=
2 × 11 × 25 × 28 275
y Example 56. If ∆ 1 is the area of the triangle with
vertices (0, 0), (a tan α , b cot α ), (a sin α , b cos α ) ; ∆ 2 is
the area of the triangle with vertices
(a, b ), (a sec 2 α , b cos ec 2 α ) , (a + a sin 2 α , b + b cos 2 α )
and ∆ 3 is the area of the triangle with vertices (0, 0),
(a tan α , − b cot α ), (a sin α , b cos α ) . Show that there is
no value of α for which ∆ 1 , ∆ 2 and ∆ 3 are in GP.
32
Textbook of Coordinate Geometry
Sol. We have, ∆1 =
=
1
| (a tan α ) (b cos α ) − (a sin α ) (b cot α ) |
2
(Q one vertex is (0,0))
1
| ab || sin α − cos α |
2
... (i)
2
2
2
1 a − (a + a sin α ) a sec α − (a + a sin α )
and ∆ 2 = |
|
2 b − (b + b cos 2 α ) b cosec2 α − (b + b cos 2 α )
1
|(a tan α ) (b cos α ) − ( −b cot α ) (a sin α ) |
2
1
... (iii)
= | ab| |sin α + cos α |
2
Since, ∆1, ∆ 2 , ∆ 3 are in GP, then ∆1∆ 3 = ∆22
and
∆3 =
⇒
1
1
| ab || sin α − cos α | × | ab ||sin α + cos α |
2
2
1
2
2
[from Eqs. (i), (ii) and (iii)]
= | ab | | cos 2α |
4
| sin 2 α − cos 2 α | = | cos 2α | 2
(See remark 4)
2
2
2
1 −a sin α a (tan α − sin α )
|
= |
2 −b cos2 α b (cot 2 α − cos2 α )
− sin 2 α
1
= | ab | × |
2
− cos2 α
=
=
=
=
=
=
⇒
sin 2 α ( sec2α − 1 )
cos α ( cosec α − 1 )
2
2
|
 − sin α sin α tan α
1
|
| ab | × 
|
2
2
2
2
− cos α cos α cot α
2
2
2
1
| ab | × | − sin 2 α cos 2 α cot 2 α + sin 2 α cos 2 α tan 2 α|
2
1
| ab | × | − cos 4 α + sin 4 α |
2
1
| ab | × |sin 2 α + cos 2 α | × |sin 2 α − cos 2 α |
2
1
| ab | × |1| × | − cos 2α |
2
1
... (ii)
| ab | × | cos 2α|
2
⇒
| − cos 2α | = | cos 2α | 2
⇒
| cos 2α | = | cos 2α| 2
⇒
∴
⇒
or
and
or
| cos 2α | (1 − | cos 2α | ) = 0
(Q| cos 2α | ≠ 0)
1 − | cos 2α | = 0
| cos 2α | = 1
cos2α = ± 1 or cos2α = 1
cos2α = − 1
2α = 2nπ, 2α = (2p + 1) π
π
or
α = n π, α = p π + ; n , p ∈ I
2
For these values of α the vertices of the given triangles are
not defined. Hence ∆1, ∆ 2 and ∆ 3 cannot be in GP for any
value of α.
Exercise for Session 3
1.
The coordinates of the middle points of the sides of a triangle are (4, 2), (3, 3) and (2, 2), then coordinates of
centroid are
(a) (3, 7 / 3)
2.
(d) (3, 4)
(b) (−1, 0)
(c) (1, 1)
1
(d)  , 1
2 
If the orthocentre and centroid of a triangle are ( −3, 5) and (3, 3) then its circumcentre is
(a) (6, 2)
4.
(c) (4, 3)
The incentre of the triangle whose vertices are ( −36, 7), (20, 7) and (0, − 8) is
(a) (0, − 1)
3.
(b) (3, 3)
(b) (3, − 1)
(c) (−3, 5)
(d) (−3, 1)
An equilateral triangle has each side equal to a. If the coordinates of its vertices are ( x1, y1), ( x 2, y 2 ) and ( x 3, y 3 )
x1 y1 1
then the square of the determinant x 2 y 2 1 equals
x3 y3 1
(a) 3a4
5.
4
(b)
3a
2
3
(c) a4
4
3
(d) a4
8
The vertices of a triangle are A (0, 0), B (0, 2) and C(2 , 0) . The distance between circumcentre and orthocentre
is
(a) 2
(b)
1
2
(c) 2
(d)
1
2
Chap 01 Coordinate System and Coordinates
6.
7.
A (a, b ), B ( x1, y1) and C ( x 2, y 2 ) are the vertices of a triangle. If a, x1, x 2 are in GP with common ratio r and
b , y1, y 2 are in GP with common ratio s, then area of ∆ABC is
(a) ab (r − 1) (s − 1) (s − r )
1
(b) ab (r + 1) (s + 1) (s − r )
2
1
(c) ab (r − 1) (s − 1) (s − r )
2
(d) ab (r + 1) (s + 1) (r − s )
 1
2 
The points ( x + 1, 2), (1, x + 2), 
,
 are collinear, then x is equal to
x
+
1
x
+ 1

(a) −4
8.
(b) −8
(b) 2
(d) 8
(c) 2
(d) 1
The nine point centre of the triangle with vertices (1, 3 ), (0, 0) and (2, 0) is

3
(a)  1,

 2 
10.
(c) 4
The vertices of a triangle are (6, 0), (0, 6) and (6, 6). Then distance between its circumcentre and centroid, is
(a) 2 2
9.
33
2 1
(b)  ,

 3 3
 2 3
(c)  ,

3 2 
1
(d)  1,


3
The vertices of a triangle are (0, 0), (1, 0) and (0, 1). Then excentre opposite to (0, 0) is
(a)  1 −

1
1
, 1+

2
2
(b)  1 +

1
1
, 1+

2
2
(c)  1 +

1
1
, 1−

2
2
(d)  1 −

1
1
, 1−

2
2
11.
If α , β, γ are the real roots of the equation x 3 − 3px 2 + 3qx − 1 = 0, then find the centroid of the triangle whose
 1
 1  1
vertices are α ,  , β,  and  γ,  .
 α   β
 γ
12.
If centroid of a triangle be (1, 4) and the coordinates of its any two vertices are (4, − 8) and ( −9, 7), find the area of
the triangle.
13.
Find the centroid and incentre of the triangle whose vertices are (1, 2), (2 , 3) and (3, 4).
14.
Show that the area of the triangle with vertices ( λ , λ − 2), ( λ + 3 , λ ) and ( λ + 2 , λ + 2) is independent of λ.
15.
Prove that the points (a, b + c ), (b , c + a ) and (c, a + b ) are collinear.
16.
Prove that the points (a, b ), (c, d ) and (a − c, b − d ) are collinear, if ad = bc.
17.
 y − y2
If the points ( x1, y1), ( x 2, y 2 ) and ( x 3, y 3 ) are collinear, show that ∑  1
 = 0, i.e.
 x1x 2 
y1 − y 2 y 2 − y 3 y 3 − y1
+
+
=0
x1x 2
x 2x 3
x 3 x1
∆ ABC 2
= , find x.
∆ BCD 3
18.
The coordinates of points A, B, C and D are ( −3, 5), (4, − 2), ( x , 3x ) and (6, 3) respectively and
19.
Find the area of the hexagon whose vertices taken in order are (5, 0), (4, 2), (1, 3), ( −2 , 2), ( −3, − 1) and (0, − 4).
Session 4
Locus and Its Equation, Change of Axes the
Transformation of Axes, Removal of the Term xy from
F(x, y) = ax2 + 2hxy + by2 without Changing the Origin,
Position of a Point which lies Inside a Triangle
Locus and Its Equation
How to Find the Locus of a Point
Locus : The locus of a moving point is the path traced out
by that point under one or more given conditions.
Let ( x 1 , y 1 ) be the coordinates of the moving point say P.
Now, apply the geometrical conditions on x 1 , y 1 . This
gives a relation between x 1 and y 1 . Now replace x 1 by x
and y 1 by y in the eliminant and resulting equation would
be the equation of the locus.
Corollary 1 : If x and y are not there in the question, the
coordinates of P may also be taken as ( x , y ).
Corollary 2 : If coordinates and equation are not given in
the question, suitable choice of origin and axes may be made.
Corollary 3 : To find the locus of the point of intersection
of two straight lines, eliminate the parameter or
parameters from the given lines. If more than one
parameter, then additional condition or conditions will
also be given.
For example 1. If a point P moves in a plane such that
whose distance from a fixed point O (say) in the plane is
always constant distance a. Thus the locus of the moving
point P is clearly a circle with centre O and radius a.
P
a
O
For example 2. If a point P moves in a plane such that
whose distance from two fixed points A and B (say) are
always equal i.e. PA = PB
P
P
Q
Note
Simplify the equation by squaring both sides if square roots are
there and taking LCM to remove the denominators.
y Example 57. Find the locus of a point which moves
such that its distance from the point (0, 0) is twice its
distance from the Y-axis.
P
P
Sol. Let P ( x 1, y1 ) be the moving point whose locus is required.
A
D
B
(The point P cannot be at Q because AQ ≠ BQ )
By hypothesis | OP | = 2 | PM |
⇒
( x 12
+
y12 )
Obviously all the positions of the moving point P lies on
the right bisector of AB. Thus the locus of the moving
point P is the right bisector of AB.
(Q P lies in any quadrant)
= 2| x 1|
Y
P(x1, y1)
M
Equation of a Locus
A relation f ( x , y ) =0 between x and y which is satisfied
by each point on the locus and such that each point
satisfying the equation is on the locus is called the
equation of the locus.
X′
X
O
Y′
35
Chap 01 Coordinate System and Coordinates
Squaring both sides, then
x 12 + y12 = 4 x 12
Changing ( x 1, y1 ) to ( x , y ), then
x 2 + y 2 = c 2 − a2
⇒
which is the required locus of P.
3x 12 − y12 = 0
Changing ( x 1, y1 ) to ( x , y ), then
3x 2 − y 2 = 0
which is the required locus of P.
y Example 58. Find the locus of the moving point P
such that 2PA = 3PB, where A is (0, 0) and B is (4, − 3) .
Sol. Let P ( x 1, y1 ) be the moving point whose locus is required.
Y
y Example 60. A point moves such that the sum of its
distances from two fixed points (ae , 0) and ( −ae , 0) is
always 2a. Prove that the equation of the locus is
2
x2 y
+
= 1, where b 2 = a 2 (1 − e 2 )
a2 b 2
Sol. Let P ( x 1, y1 ) be the moving point whose locus is required
and A (ae , 0) and B ( −ae , 0) be the given fixed points.
Y
P(x1, y1)
X′
P(x1, y1)
X
A
(0, 0)
X′
X
A(ae, 0)
B(–ae, 0) O
B(4, –3)
Y′
Y′
By hypothesis,
2 PA = 3 PB or 4 ( PA )2 = 9 ( PB )2
or
⇒
or
4 { x 12 + y12 } = 9 {( x 1 − 4 )2 + (y1 + 3)2 }
⇒
4
or
5x 12
+
( x 12
+
y12 )
5y12
=9
( x 12
+
y12
− 8x 1 + 6y1 + 25)
− 72x 1 + 54y1 + 225 = 0
Let
Changing ( x 1, y1 ) to ( x , y ), then
5x 2 + 5y 2 − 72x + 54y + 225 = 0
and
which is the required locus of P.
y Example 59.A point moves so that the sum of the
squares of its distances from two fixed points A (a , 0)
and B ( −a , 0) is constant and equal to 2c 2 , find the
locus of the point.
Sol. Let P ( x 1, y1 ) be the moving point whose locus is required.
Y
O
B(–a, 0)
A(a, 0)
X
Y′
By hypothesis,
⇒
( PA ) + ( PB ) = 2c
2
2
2
( x 1 − a ) + (y1 − 0)2 + ( x 1 + a )2 + (y1 − 0)2 = 2c 2
2
⇒
2x 12 + 2y12 + 2a 2 = 2c 2
or
x 12 + y12 = c 2 − a 2
( x 1 − ae )2 + (y1 − 0)2 + ( x 1 + ae )2 + (y1 − 0)2 = 2a
( x 12 + y12 − 2aex 1 + a 2e 2 ) + ( x 12 + y12 + 2aex 1 + a 2e 2 )
= 2a
l = x 12 + y12 − 2aex 1 + a 2e 2
...(i)
(l − m method)
m = x 12 + y12 + 2aex 1 + a 2e 2
then, Eq. (i) can be written as
l + m = 2a
and
l − m = − 4aex 1
or
( l + m ) ( l − m ) = − 4aex 1
or
...(ii)
2a ( l − m ) = − 4aex 1
[from Eq. (ii)]
l − m = − 2ex 1
or
...(iii)
Adding Eqs. (ii) and (iii), then
2 l = 2a − 2ex 1 or l = a − ex 1
Squaring both sides,
l = a 2 − 2aex 1 + e 2 x 12
P(x1, y1)
X′
| PA | + | PB | = 2a
By hypothesis,
⇒
x 12 + y12 − 2aex 1 + a 2e 2 = a 2 − 2aex 1 + e 2 x 12
⇒
(1 − e 2 )x 12 + y12 = a 2 (1 − e 2 )
or
x 12
a2
+
y12
a 2 (1 − e 2 )
x 12
y12
=1
a
b2
Changing ( x 1, y1 ) to ( x , y ), then
x2 y2
+
=1
a2 b2
which is the required locus of P.
or
2
+
=1
[Qb 2 = a 2 (1 − e 2 )]
36
Textbook of Coordinate Geometry
y Example 61. Find the equation of the locus of a
point which moves so that the difference of its
distances from the points ( 3, 0) and ( −3, 0) is 4 units.
Sol. Let P ( x 1, y1 ) be the moving point whose locus is required
and A (3, 0) and B ( −3, 0) be the given fixed points.
Y
B(–3, 0)
O
Squaring both sides,
4l = 9 x 12 + 24 x 1 + 16
⇒
A(3, 0)
4 ( x 12 + y12 + 6x 1 + 9 ) = 9 x 12 + 24 x 1 + 16
x 12 y12
−
=1
4
5
Changing ( x 1, y1 ) by ( x , y ), then
x2 y2
−
=1
4
5
which is the required locus of P.
⇒
P(x1, y1)
X′
Adding Eqs. (ii) and (iii),
2 l = ( 3x 1 + 4 )
X
5x 12 − 4y12 = 20 or
y Example 62. The ends of the hypotenuse of a right
angled triangle are (6, 0) and (0, 6 ) . Find the locus of
the third vertex.
Y′
By hypothesis
| PB | − | PA | = 4 (assume | PB | > | PA |)
⇒
( x 1 + 3) 2 + ( y 1 − 0) 2 − ( x 1 − 3) 2 + ( y 1 − 0) 2 = 4
⇒
( x 12 + y12 + 6x 1 + 9 ) = 4 + ( x 12 + y12 − 6x 1 + 9 )
Sol. Let C ( x 1, y1 ) be the moving point (third vertex) whose locus
is required and A (6, 0) and B (0, 6) be the given vertices.
Y
C(x1,y1)
Squaring both sides then,
x 12 + y12 + 6x 1 + 9 = 16 + x 12 + y12 − 6x 1 + 9 + 8
B
(0, 6)
( x 12 + y12 − 6x 1 + 9 )
or
(12x 1 − 16) = 8 ( x 12 + y12 − 6x 1 + 9 )
or
(3x 1 − 4 ) = 2 ( x 12 + y12 − 6x 1 + 9 )
X′
O
Again, squaring both sides, then
9 x 12 − 24 x 1 + 16 = 4 x 12 + 4y12 − 24 x 1 + 36
5x 12
or
⇒
x 12
−
y12
−
4y12
= 20
By hypothesis
( AC )2 + ( BC )2 = ( AB )2
=1
( x 1 + 3) 2 + ( y 1 − 0) 2 − ( x 1 − 3) 2 + ( y 1 − 0) 2 = 4
⇒
( x 12 + y12 + 6x 1 + 9 ) − ( x 12 + y12 − 6x 1 + 9 ) = 4
Let
l=
and
m=
+
x 12
+
⇒
⇒
l + m = 3x 1
⇒
2x 12 + 2y12 − 12x 1 − 12y1 = 0
or
x 12 + y12 − 6x 1 − 6y1 = 0
2
2
which is the required locus of third vertex C.
Aliter 1. Slope of AC × slope of BC = − 1
 y1 − 0   y1 − r 
⇒
 = −1
 ×

 x 1 − c   x 1 − 0
x 12 + y12 − 6x 1 − 6y1 = 0
Aliter 2. Mid-point of AB is M ≡ (3, 3)
Q
... (ii)
l − m = 12x 1
( l + m ) ( l − m ) = 12x 1
( l + m ) ( 4 ) = 12x 1
( x 1 − 6) + ( y 1 − 0) + ( x 1 − 0) + ( y 1 − 6) 2 = 62 + 62
or
− 6x 1 + 9
then, Eq. (i) can be written as
l − m =4
and
⇒
...(i)
+ 6x 1 + 9
y12
(Q ∠ ACB = 90° )
⇒
2
Changing ( x 1, y1 ) by ( x , y ), then
x 2 + y 2 − 6x − 6y = 0
⇒
y12
X
Y′
4
5
Changing ( x 1, y1 ) by ( x , y ), then
x2 y2
−
=1
4
5
which is the required locus of P.
Aliter (l − m method) :
Since,
| PB | − | PA | = 4
x 12
A(6, 0)
( 3 − 0) 2 + ( 3 − 6) 2 = ( x 1 − 3) 2 ( y 1 − 3) 2
or
[from Eq. (ii)]
...(iii)
MA = MB = MC ⇒ ( MA )2 ≡ ( MC )2
x 12 + y12 − 6x 1 − 6y1 = 0
∴ Required locus is
x 2 + y 2 − 6x − 6y = 0
Chap 01 Coordinate System and Coordinates
y Example 63. Find the equation of the locus of a
point which moves so that the sum of their distances
from ( 3, 0) and ( −3, 0) is less than 9.
⇒
y − 1
 y − 1 
x=

 1 +
 2  
2 
⇒
 y − 1  y + 1
x=

 
 2   2 
Sol. Let P ( x 1, y1 ) be the moving point whose
locus is required and A (3, 0 ) and B ( −3, 0 ) are the given
points.
P(x1, y1)
B(–3, 0)
O
4x = y 2 − 1
or
y 2 = 4x + 1
y Example 65. A stick of length l rests against the
floor and a wall of a room. If the stick begins to slide
on the floor, find the locus of its middle point.
Y
X′
or
A(3, 0)
Sol. Let the cross section of the floor and wall be taken as the
coordinate axes and AB be one of the position of the stick.
Let the mid-point of AB be P ( x 1,y1 ), then coordinates of A
and B are (2x 1, 0) and (0, 2y1 ) respectively.
X
Y
Y′
B (0, 2y1)
By hypothesis,
| PA | + | PB | < 9
⇒
{( x 1 − 3)2 + (y1 − 0)2 } + {( x 1 + 3)2 + (y1 − 0)2 } < 9
⇒
( x 12 + y12 − 6x 1 + 9 ) < 9 − ( x 12 + y12 + 6x 1 + 9 )
P(x1, y1)
On squaring, we get
( x 12 + y12 − 6x 1 + 9 ) < (81 + x 12 + y12 + 6x 1 + 9 )
X′
O
− 18 ( x 12 + y12 + 6x 1 + 9 )
(Qa < b ⇒ a < b provided a >0)
2
2
⇒
−12x 1 − 81 < −18 ( x 12 + y12 + 6x 1 + 9 )
⇒
( 4 x 1 + 27 ) > 6 ( x 12 + y12 + 6x 1 + 9 )
(Q If a > b, then −a < − b)
+ 729 + 216x >
⇒
20x 12
+
36y12
36x 12
+
36y12
+ 216x 1 + 324
< 405
Changing ( x1, y1 ) by ( x, y ), then
20x 2 + 36y 2 < 405
which is the required locus of P.
y Example 64. Find the locus of a point whose
coordinate are given by x = t + t 2 , y = 2t + 1, where t is
variable.
Sol. Given,
and
From Eq. (ii),
x = t + t2
...(i)
y = 2t + 1
 y − 1
t =

 2 
...(ii)
 y − 1  y − 1
x=

 +
 2   2 
Y′
But given, | AB | = l
⇒
( AB )2 = l 2
⇒
(2x 1 − 0)2 + (0 − 2y1 )2 = l 2
⇒
4 x 12 + 4y12 = l 2
which is the required locus of P.
Aliter :
Since,
| AB | = l
Let
∠OAB = α
∴
OA = l cosα and OB = l sinα
then,
A ≡ (l cos α, 0)
and
B ≡ (0, l sin α )
Y
B
...(iii)
On eliminating t from Eqs. (i) and (iii), we get required
locus as
2
X
A(2x1, 0)
Changing ( x 1, y1 ) by ( x , y ), then
4 (x 2 + y 2 ) = l 2
On squaring, we get
16x 12
37
P(x, y)
X′
α
O
Y′
A
X
38
Textbook of Coordinate Geometry
Let P ( x , y ) be the mid-point of AB, then
2x = l cosα
2y = l sinα
Squaring and adding Eqs. (i) and (ii), then
4 (x 2 + y 2 ) = l 2
...(i)
...(ii)
which is the required locus of P.
y Example 66. Find the locus of the point of
intersection of the lines x cos α + y sin α = a and
x sin α − y cos α = b , where α is variable.
Sol. Given equations are
...(i)
x cos α + y sin α = a
and
...(ii)
x sin α − y cos α = b
Here, α is a variable, on eliminating α. Squaring and adding
Eqs. (i) and (ii), we get required locus as
( x cos α + y sin α )2 + ( x sin α − y cos α )2 = a 2 + b 2
or
( x 2 cos 2 α + y 2 sin 2 α + 2xy cos α sin α )
+ ( x 2 sin 2 α + y 2 cos 2 α − 2xy sin α cos α ) = a 2 + b 2
⇒
x 2 (cos 2 α + sin 2 α ) + y 2 (sin 2 α + cos 2 α ) = a 2 + b 2
⇒
x 2 + y 2 = a2 + b2
which is the required locus of G.
(ii) If the circumcentre be C ( x , y ) . Since in semicircle angle
90° , then AB is the diameter of the circumcircle OAB.
∴ Circumcentre is the mid-point of AB.
a+0
Then
⇒ a = 2x
x=
2
0+b
and
⇒ b = 2y
y=
2
On substituting in a 2 + b 2 = 1, we get
(2x )2 + (2y )2 = 1
1
4
which is the required locus of C.
x2 + y2 =
or
y Example 68. Two points P and Q are given, R is a
variable point on one side of the line PQ such that
∠ RPQ − ∠ RQP is a positive constant 2α. Find the
locus of the point R.
Sol. Let the X-axis along QP and the middle point of PQ is
origin and let coordinates of moving point R be ( x 1, y1 ).
Y
y Example 67. A variable line cuts X-axis at A, Y-axis at
B, where OA = a, OB = b ( O as origin) such that
a 2 + b 2 = 1.
Find the locus of
(i) centroid of ∆ OAB
(ii) circumcentre of ∆ OAB
Sol.
(i) Coordinates of A and B are (a , 0) and (0, b ) respectively.
If centroid of ∆ OAB be G ( x , y )
0+a+0
then
⇒ a = 3x
x=
3
0+0+b
and
⇒ b = 3y
y=
3
Y
R(x1, y1)
y1
X′
Q
(–a, 0)
b
X
OP = OQ = a
Let
then coordinates of P and Q are (a, 0) and ( −a, 0)
respectively.
Draw RM perpendicular on QP
MR = y1
∴
OM = x 1 and
...(i)
and in ∆RMQ,
a
O
A(a, 0)
X
On substituting in a + b = 1, we get
2
(3x )2 + (3y )2 = 1
x2 + y2 =
1
9
RM
RM
y1
=
=
QM OQ + OM a + x 1
...(ii)
But given
Y′
or
P
(a, 0)
Y′
tan φ =
2
x1 M
and let
∠ RPM = θ and ∠ RQM = φ
Now, in ∆RMP,
RM
RM
y1
tanθ =
=
=
MP OP − OM a − x 1
B(0, b)
X′
O
⇒
∴
⇒
∠ RPQ − ∠ RQP = 2α
θ − φ = 2α
tan (θ − φ ) = tan 2α
tan θ − tan φ
= tan 2α
1 + tan θ tan φ
(constant)
Chap 01 Coordinate System and Coordinates
y1
y1
−
a − x1 a + x1
= tan 2α
 y1   y1 
1+ 
.

 a − x1   a + x1 
⇒
[from Eqs. (i) and (ii)]
2x 1y1
⇒
a 2 − x 12 + y12
= tan 2α
or
a 2 − x 12 + y12 = 2x 1y1 cot 2α
or
x 12 − y12 + 2x 1y1 cot 2α = a 2
Hence, locus of the point R ( x 1, y1 ) is
x 2 − y 2 + 2xy cot 2α = a 2
Change of Axes OR the
Transformations of Axes
In coordinate geometry we have discussed the coordinates
of a point or the equation of a curve are always considered
on taking a fixed point O as the origin and two
perpendicular straight lines through O as the coordinates
axes. For convenient the coordinates of the point or the
equation of the curve changes when either the origin is
changed or the direction of axes or both are suitably.
These processes in coordinate geometry are known as the
transformations or change of axes. This process of
transformation of coordinates will be of great advantage
to solve most of the problems very easily.
(i) Change of origin OR Shifting of origin
(Translation of Axes)
To change the origin of coordinates to another point (h, k )
whereas the directions of axes remain unaltered.
Y
Y'
P(x, y) w.r.t.
old axes
P(X, Y)
Y w.r.t. new axes
y
X
O'
N
k
h
O
X'
L
x
M
X
Let O be the origin of coordinates and OX , OY be the
original coordinate axes. Let O ′ be the new origin and
(h, k ) its coordinates referred to the original axes. Draw
two lines O ′ X ′ and O ′ Y ′ through O ′ and parallel to OX
and OY respectively. Let P ( x , y ) be any point referred to
the original axes OX , OY . Again suppose that the
39
coordinates of the same point P referred to the new axes
O ′ X ′ , O ′ Y ′ are ( X , Y ) .
From O ′ draw O ′ L perpendicular to OX. from P draw PM
perpendicular to OX to meet O ′ X ′ in N. Then
OL = h, O ′ L = k, OM = x , PM = y , O ′ N = X
and
PN = Y
we have
x = OM = OL + LM = OL + O ′ N = h + X
i.e.
...(i)
x = X +h
and
y = PM = PN + NM = PN + O ′ L = Y + k
i.e.
...(ii)
y =Y + k
from Eqs. (i) and (ii),
X = x − h and Y = y − k
Thus, if origin is shifted to point (h, k ) without rotation of
axes, then new equation of curve can be obtained by
putting x + h in place of x and y + k in place of y.
Remarks
1. In this case axes are shifted parallel to themselves, then it is
also called Transformation by parallel axes.
2. Inverse translation or shifting the origin back : Some
times it is required to shift the new origin back. Then putting
x − h in place of x and y − k in place of y in any equation of
curve referred to the new origin to get the corresponding
equation referred to the old origin.
3. The above transformation is true whether the axes be
rectangular or oblique.
y Example 69. Find the equation of the curve
2x 2 + y 2 − 3x + 5y − 8 = 0 when the origin is
transferred to the point ( −1, 2) without changing the
direction of axes.
Sol. Here, we want to shift the origin to the point ( −1, 2) without changing the direction of axes. Then we replace x by
x − 1 and y by y + 2 in the equation of given curve, then
the transformed equation is
2 ( x − 1) 2 + ( y + 2) 2 − 3 ( x − 1) + 5 ( y + 2) − 8 = 0
⇒
2x 2 + y 2 − 7 x + 9y + 11 = 0
y Example 70. The equation of a curve referred to the
new axes, axes retaining their direction and origin is
(4, 5) is x 2 + y 2 = 36 . Find the equation referred to the
original axes.
Sol. Here we want to shift the ( 4, 5) to the origin without
changing the direction of axes. Then we replace x by x − 4
and y by y − 5 in the equation of given curve then the
required equation is
( x − 4 ) 2 + (y − 5 ) 2 = 36
⇒
x 2 + y 2 − 8 x − 10y + 5 = 0
40
Textbook of Coordinate Geometry
y Example 71. Shift the origin to a suitable point so
that the equation y 2 + 4 y + 8 x − 2 = 0 will not contain
term in y and the constant.
Sol. Let the origin be shifted to the point (h , k ) without
changing the direction of axes. Then we replace x by x + h
and y by y + k in the equation of the given curve then the
transformed equation is
(y + k )2 + 4 (y + k ) + 8 ( x + h ) − 2 = 0
⇒
y 2 + (2k + 4 ) y + 8x + (k 2 + 4k + 8h − 2) = 0
Since, this equation is required to be free from the term
containing y and the constant, we have
2k + 4 = 0 and k 2 + 4k + 8h − 2 = 0
∴
k = −2
h=
and
3
4
3

Hence, the point to which the origin be shifted is  , − 2 .
4

y Example 72. At what point the origin be shifted, if
the coordinates of a point ( −1, 8 ) become ( −7, 3) ?
Sol. Let the origin be shifted to the point (h , k ) without
changing the direction of axes. Then we replace x by x + h
and y by y + k and we get new co-ordinates. Here, given
old coordinates and new coordinates are ( −1, 8) and ( −7, 3)
respectively.
We have
−1 + h = − 7 and 8 + k = 3
⇒
h = − 6 and k = − 5
Hence, the origin must be shifted to ( −6, − 5).
(ii) Rotation of Axes
(Change of Directions of Axes)
To find the change in the coordinates of a point when the
directions of axes are rotated through an angle θ the
origin being fixed.
Y´
Y
θ
= OL − QN = ON cos θ − PN sin θ
= X cos θ − Y sin θ
i.e.
...(i)
x = X cos θ − Y sin θ
and
y = PM = PQ + QM = PQ + NL
= PN cos θ + ON sin θ
= Y cos θ + X sin θ
i.e.
...(ii)
y = X sin θ + Y cos θ
Now, multiplying Eqs. (i) by cos θ and Eq. (ii) by sin θ and
adding we get
...(iii)
X = x cos θ + y sin θ
Also, subtracting the product of Eq. (i) by sin θ from the
product of Eq. (ii) by cos θ , we get
...(iv)
Y = − x sin θ + y cos θ
also x 2 + y 2 = X 2 + Y 2 = OP 2 are unchanged i.e. the
distance of the point P from the origin O remains
unaffected by the rotation of axes.
Rule : When the axes are rotated through θ, replace ( x , y )
by ( x cos θ − y sin θ, x sin θ + y cos θ ).
Shifting the coordinate axes back : Some times it is
required to shift the new coordinates axes back. Then
replace ( x , y ) by
( x cos θ + y sin θ, − x sin θ + y cos θ ).
Independent Proof (Aliter)
The relations X = x cos θ + y sin θ and
Y = − x sin θ + y cos θ can be obtained independently.
Proof : Draw PM and PN perpendiculars to OX and OX ′
and also ML and MQ perpendiculars to PN and OX ′
respectively.
P
Q
Now,
OM = x , PM = y , ON = X , PN = Y
x = OM = OL − ML
Q Angle between any two lines = Angle between 


 their perpendiculars i.e. ∠XOX ′ = ∠NPM = θ 
X´
N
Y´
θ
P Old (x, y)
New (X, Y)
θ
O
Y
M
L
X
Let OX and OY be the original system of coordinate axes.
Let OX ′ and OY ′ be the new axes obtained by rotating the
original axes through an angle θ. Let P be a point in the
plane whose coordinates are ( x , y ) and ( X , Y ) referred to
old and new axes respectively. Draw PM and PN
perpendiculars to OX and OX ′ and also NL and NQ
perpendiculars to OX and PM. We have
θ
Q
θ
X´
N
θ
L
θ
O
We have,
and
M
OM = x , PM = y , ON = X
PN = Y
X
41
Chap 01 Coordinate System and Coordinates
Also, angle between any two lines
= Angle between their perpendiculars lines
i.e.
∠ XOX ′ = ∠ MPN = ∠ PMQ = θ
∴
X = ON = OQ + QN
i.e.
and
= OQ + ML
= OM cos θ + PM sin θ
(QOQ = OM cos θ and ML = PM sin θ)
= x cos θ + y sin θ
X = x cos θ + y sin θ
Y = PN = PL − NL = PL − QM
= PM cos θ − OM sin θ
(Q PL = PM cos θ and QM = OM sin θ )
= y cos θ − x sin θ
Y = − x sin θ + y cos θ
i.e.
Remark :
The results Eqs. (i), (ii), (iii) and (iv) can be conveniently
remembered by the following methods.
(i) Light heavy method : Let x , y be light and X ,Y be
heavy then heavy X ,Y down and x , y up then
x
y
X
cosθ
sinθ
Y
− sinθ
cosθ
third row is obtained by differentiating second row with
respect to θ .
For remembrance C
(Civil Services)
S
↓
↓
cos θ
sin θ
↓
↓
−sin θ
cos θ
C → cos θ 

 i.e.

S → sin θ 

 (cos θ ) ′ = − sin θ 


 (sin θ ) ′ = cos θ 
(a) Finding x and y in terms of X and Y
x = Sum of the products of the elements in the left
most column with the corresponding elements of the
first column
i.e.
x = X cos θ − Y sin θ
and y = Sum of the products of the elements in the
left most column with the corresponding elements of
the second column.
i.e.
y = X sin θ + Y cos θ
Hence, x = X cos θ − Y sin θ 

y = X sin θ + Y cos θ
(b) Finding X and Y in terms of x and y
X = Sum of the products of the elements of top-row
with the corresponding elements of first row.
i.e.
X = x cos θ + y sin θ
and Y = Sum of the products of the elements of
top-row with the corresponding elements of second
row.
i.e.
Y = − x sin θ + y cos θ
Hence, X = x cos θ + y sin θ 

Y = − x sin θ + y cos θ
(ii) Matrix method :
 X   cos θ sin θ   x 
x 
Y  =  − sin θ cos θ  y  = A y  (say)
  
 
 
and
 x  cos θ − sin θ   X 
X 
y  = sin θ cos θ  Y  = A ′ Y 
  
 
 
where, A ′ is the transpose matrix of A.
(iii) Complex number method :
Let
z = x + iy
and
Z = X + iY , where i = −1
then
z = Ze iθ
…(i)
i.e.
( x + iy ) = ( X + iY ) (cos θ + i sin θ )
On comparing real and imaginary parts, we get
x = X cos θ − Y sin θ

y = X sin θ + Y cos θ
Z = ze −iθ
Again from Eq. (i),
i.e.
( X + iY ) = ( x + iy ) (cos θ − i sin θ )
On comparing real and imaginary parts, we get
X = x cos θ + y sin θ 

Y = − x sin θ + y cos θ
y Example 73. If the axes are turned through 45°, find
the transformed form of the equation
3x 2 + 3y 2 + 2xy = 2.
1
2
Replacing ( x , y ) by ( x cos θ − y sin θ, x sin θ + y cos θ )
 x − y x + y
i.e.
,


 2
2 
Sol. Here, θ = 45 ° so sin θ = cos θ =
Then, 3x 2 + 3y 2 + 2xy = 2 becomes
2
2
 x − y  x + y
 x + y
 x − y
3
 =2
 
 +2
 +3
 2   2 
 2 
 2 
42
Textbook of Coordinate Geometry
⇒
3 (2x 2 + 2y 2 ) + 2 ( x 2 − y 2 ) = 4
⇒
8x + 4y = 4
or
2x 2 + y 2 = 1
2
2
which is free from the term containing xy.
y Example 74. Prove that if the axes be turned
π
the equation x 2 − y 2 = a 2 is transformed
through
4
to the form xy = λ . Find the value of λ .
π
1
so sin θ = cos θ =
4
2
Replacing ( x , y ) by ( x cos θ − y sin θ, x sin θ + y cos θ )
 x − y x + y
i.e.
,


 2
2 
Sol. Here, θ =
then, x 2 − y 2 = a 2 becomes
2
2
 x + y
 x − y
2
 =a
 −

 2 
 2 
⇒
 x − y x + y  x − y x + y
2
+
−

 
 =a
 2
2   2
2 
⇒
 2x   −2y 
2
 =a
  
 2  2 
or
xy = −
a2
2
a2
Comparing it with xy = λ, then we get λ = − .
2
y Example 75. Through what angle should the axes be
rotated so that the equation 9 x 2 − 2 3 xy + 7 y 2 = 10
may be changed to 3x 2 + 5y 2 = 5 ?
Sol. Let angle be θ then replacing ( x , y ) by
( x cos θ − y sin θ, x sin θ + y cos θ )
then,
9 x 2 − 2 3xy + 7y 2 = 10 becomes
9 ( x cos θ − y sin θ )2 − 2 3
( x cos θ − y sin θ ) ( x sin θ + y cos θ )
+ 7 ( x sin θ + y cos θ )2 = 10
⇒
x 2 (9 cos 2 θ − 2 3 sin θ cos θ + 7 sin 2 θ )
+ 2xy ( −9 sin θ cos θ − 3 cos 2θ + 7 sin θ cos θ )
+ y 2 (9 cos 2 θ + 2 3 sin θ cos θ + 7 cos 2 θ ) = 10
On comparing with 3x 2 + 5y 2 = 5 (coefficient of xy)
we get − 9 sin θ cos θ − 3 cos 2θ + 7 sin θ cos θ = 0
or
sin 2θ = − 3 cos 2θ
or
tan 2θ = − 3 = tan (180° − 60° )
or
∴
2 θ = 120°
θ = 60°
y Example 76. If ( x , y ) and ( X , Y ) be the coordinates
of the same point referred to two sets of rectangular
axes with the same origin and if ux + vy , when u and v
are independent of X and Y become VX + UY , show
that u 2 + v 2 = U 2 + V 2
Sol. Let the axes rotate an angle θ and if ( x , y ) be the point with
respect to old axes and ( X , Y ) be the co-ordinates with
respect to new axes, then
x + iy = ( X + iY ) e iθ = ( X + iY ) (cos θ + i sin θ )
On comparing real and imaginary parts, we get
 x = X cos θ − Y sin θ

y = X sin θ + Y cos θ
Then, ux + vy = u ( X cos θ − Y sin θ ) + v ( X sin θ + Y cos θ )
= (u cos θ + v sin θ ) X + ( −u sin θ + v cos θ ) Y
but given new curve VX + UY
then, VX + UY = (u cos θ + v sin θ )
X + ( −u sin θ + v cos θ )Y
On comparing the coefficients of X andY , we get
...(i)
V = u cos θ + v sin θ
and
...(ii)
U = − u sin θ + v cos θ
Squaring and adding Eqs. (i) and (ii), we get
V 2 + U 2 = (u cos θ + v sin θ )2 + ( − u sin θ + v cos θ )2
= u2 + v 2
Hence,
u2 + v 2 = U 2 + V 2
Aliter 1 (By matrix method) :
x  cos θ − sin θ X  X cos θ − Y sin θ
…(i)
=
y  = sin θ
cos θ Y  X sin θ + Y cos θ
  
x 
X cos θ + Y sin θ
Q ux + vy = [u v ]   = [u v ] 

y
 
X sin θ + Y cos θ
[from (i)]
= uX cos θ − uY sin θ + vX sin θ + vY cos θ
= X (u cos θ + v sin θ ) + Y ( − u sin θ + v cos θ )
but given new curve VX + UY
Then, VX + UY = X (u cos θ + v sin θ ) + Y
( − u sin θ + v cos θ )
On comparing the coefficients of X andY , we get
...(ii)
V = u cos θ + v sin θ
...(iii)
U = − u sin θ + v cos θ
Squaring and adding Eqs. (ii) and (iii), we get
u2 + v 2 = U 2 + V 2
Aliter 2 (Best approach) :
ux + vy = Re ((u − iv ) ( x + iy ))
= Re ((u − iv ) ( X + iY ) e iθ )
and
VX + UY = Re ((V − iU ) ( X + iY ) )
From Eqs. (i) and (ii), we get
V − iU = (u − iv )e iθ
Taking modulus both sides, then
...(i)
...(ii)
Chap 01 Coordinate System and Coordinates
| V − iU | = | u − iv | | e iθ |
⇒
(V + U ) = u + v
or
u + v =U + V
2
2
2
2
2
2
2 .
2

 x + y 
−5  −3 + 
  + 20
 2 

1
43

 x − y 

 −2 + 
 2 


 x + y 
− 22 −3 + 
  − 14 = 0
 2 

2
(iii) Double Transformation
x 2 − 14 xy − 7y 2 − 2 = 0
⇒
(Origin Shifted and Axes Rotated)
If origin is shifted to the point. O ′ (h, k ) and at the same
time the directions of axes are rotated through an angle θ
in the anticlockwise sense such that new coordinates of
P ( x , y ) become ( X , Y ).
Y
Y´
P (x, y)
(X, Y)
θ
X´
θ
O´
Clearly, h ≠0
Rotating the axes through an angle θ, we have
x = X cos θ − Y sin θ
and
y = X sin θ + Y cos θ
f ( x , y ) = ax 2 + 2hxy + by 2
Q
X
O
Removal of the Term xy from
f ( x, y) = ax 2 + 2hxy + by 2
without Changing the Origin
Then, we get
...(i)
x = h + X cos θ − Y sin θ
and
...(ii)
y = k + X sin θ + Y cos θ
In practice we have to replace x by h + x cos θ − y sin θ
and y by k + x sin θ + y cos θ.
Again, if we want to shift the coordinate axes back to their
original positions, then we obtained X and Y by solving
Eqs. (i) and (ii), then
X = ( x − h ) cos θ + (y − k ) sin θ
and
Y = − ( x − h ) sin θ + (y − k ) cos θ
y Example 77. What does the equation
2x 2 + 4 xy − 5y 2 + 20x − 22y − 14 = 0 becomes when
referred to rectangular axes through the point
( −2 , − 3) , the new axes being inclined at an angle of
45° with the old ?
Sol. Let O ′ ( −2, − 3) be the new origin and axes are rotated
about O ′ through an angle 45° in anticlockwise direction
then replacing x and y by
− 2 + x cos 45° − y sin 45°
and
−3 + x sin 45° + y cos 45°
x
y
−

 x + y
i.e. −2 + 
 and −3 + 
 respectively in the
 2 
 2 
given curve, then the new equation of curve will be
After rotation, new equation is
F ( X , Y ) = (a cos 2 θ + 2h cos θ sin θ + b sin2 θ ) X 2
+ 2 { (b − a ) cos θ sin θ + h (cos 2 θ − sin2 θ ) } XY
+ (a sin2 θ − 2h cos θ sin θ + b cos 2 θ ) Y 2
Now, coefficient of XY = 0
a −b
Then, we get
cot 2θ =
2h
Remark
Usually, we use the formula, tan2θ =
rotation, θ.
a− b
However, if a = b, we use cot 2θ =
as in this case tan2θ is not
2h
defined.
y Example 78. Given the equation
4 x 2 + 2 3xy + 2y 2 = 1, through what angle should the
axes be rotated so that the term in xy be wanting from
the transformed equation.
Sol. Comparing the given equation with ax 2 + 2hxy + by 2 , we
get a = 4, h = 3 b = 2 . If axes are to be rotated at θ, then
tan 2θ =
∴
2h
2 3
π
=
= 3 = tan
2
3
a−b
π
π
π 4π
,π +
⇒ 2θ = ,
3
3
3 3
π 2π
θ= ,
6 3
2θ =
2


 x + y 
 x − y 
 x − y  
2 − 2 + 

   −3 + 
  + 4  −2 + 
 2 



2 
2 


2h
for finding the angle of
a− b
44
Textbook of Coordinate Geometry
Position of a Point which Lies
Inside a Triangle
If any point say ( P ) lies within the triangle ABC,
then
∆1 + ∆2 + ∆ 3 = ∆
A
P
B
where,
Also,
C
∆ = Area of triangle ABC,
∆ 1 = Area of ∆PBC,
∆ 2 = Area of ∆PCA,
∆ 3 = Area of ∆PAB
∆ 1 ≠ 0, ∆ 2 ≠ 0, ∆ 3 ≠ 0, ∆ ≠ 0
(Each individual area must be non-zero)
y Example 79. Find λ if ( λ , λ + 1) is an interior point of
∆ ABC where, A ≡ (0, 3) ;B ≡ ( −2 , 0) and C ≡ (6, 1) .
Sol. The point P ( λ , λ + 1) will be inside the triangle ABC, then
Area of ∆ PBC + Area of ∆ PCA + Area of ∆ PAB = Area of
∆ ABC
 λ λ + 1 1
 λ λ + 1 1
 λ λ + 1 1
1
1
1
3 1|
0 1| + 
⇒ |−2
1 1| + | 0
| 6

 2
 2
2 
0 1
1 1
3 1
–2
0
 6
 0 3 1
1
= | −2 0 1|

2
 6 1 1
⇒
| 7 λ + 6 | + | 8λ − 12 | + | λ + 4 | = 22
For λ < − 4 :
++++++++
3/2
7λ + 6 – – – – – + + + + + + + + + + +
–6/7
++++++++++++++
λ + 4– –
–4
8λ – 12 – – – – – – – –
Then, − (7 λ + 6) − (8λ − 12) − ( λ + 4 ) = 22
5
⇒
−16λ = 20 ∴ λ = −
4
which is impossible.
6
For −4 ≤ λ < − :
7
Then, − (7 λ + 6) − (8λ − 12) + ( λ + 4 ) = 22
6
⇒
−14 λ = 12 ∴ = −
7
which is impossible.
6
3
For − ≤ λ < :
7
2
Then, (7 λ + 6) − (8λ − 12) + λ + 4 = 22 ⇒ 22 = 22
6
Q at λ = − , area of ∆ PBC = 0
7
−6
∴
λ≠
7
 6 3
λ ∈ − , 
∴
 7 2
3
For λ ≥ :
2
Then, 7 λ + 6 + 8λ − 12 + λ + 4 = 22 ⇒ 16λ = 24
3
λ=
∴
2
3
Q at λ = , area of ∆PCA = 0
2
3
∴
λ≠
2
 6 3
Hence, value of λ ∈  − ,  .
 7 2
Chap 01 Coordinate System and Coordinates
45
Exercise for Session 4
1.
The equation of the locus of points equidistant from ( −1, − 1) and (4, 2) is
(a) 3x − 5y − 7 = 0
2.
(d) x 2 + y 2 = a 2
(b) 2x 2 − y 2 = 8
(c) x 2 − y 2 = 4
(d) 2x 2 + 3y 2 = 5
(b) x 2 + y 2 = 2
(c) xy = 3
(d) x 2 + 2y 2 = 3
(b) x + 2y = 2
(c) 2y − x = 2
(d) 2y − 3x = 5
The equation 4xy − 3x 2 = a 2 become when the axes are turned through an angle tan−1 2 is
(a) x 2 + 4y 2 = a 2
7.
(c) xy = a 2
If a point moves such that twice its distance from the axis of x exceeds its distance from the axis of y by 2, then
its locus is
(a) x − 2y = 2
6.
(b) 2x 2 − y 2 = 2a 2
If the coordinates of a variable point P be (cos θ + sin θ, sin θ − cos θ ), where θ is the parameter, then the locus
of P is
(a) x 2 − y 2 = 4
5.
(d) x − 3y + 5 = 0
1
1

If the coordinates of a variable point P be t + , t −  , where t is the variable quantity, then the locus of P is

t
t
(a) xy = 8
4.
(c) 4x + 3y + 2 = 0
The equation of the locus of a point which moves so that its distance from the point (ak , 0) is k times its
a 
distance from the point  , 0 , (k ≠ 1) is
k 
(a) x 2 − y 2 = a 2
3.
(b) 5x + 3y − 9 = 0
(b) x 2 − 4y 2 = a 2
(c) 4x 2 + y 2 = a 2
(d) 4x 2 − y 2 = a 2
Transform the equation x 2 − 3xy + 11x − 12y + 36 = 0 to parallel axes through the point ( −4, 1) becomes
ax 2 + bxy + 1 = 0 then b 2 − a =
(a)
1
4
(b)
1
16
(c)
1
64
(d)
1
256
8.
Find the equation of the locus of all points equidistant from the point (2 , 4) and the Y-axis.
9.
Find the equation of the locus of the points twice as far from ( −a , 0) as from (a , 0).
10.
OA and OB are two perpendicular straight lines. A straight line AB is drawn in such a manner that OA + OB = 8.
Find the locus of the mid point of AB.
11.
The ends of a rod of length l move on two mutually perpendicular lines. Find the locus of the point on the rod
which divides it in the ratio 1 : 2.
12.
The coordinates of three points O, A, B are (0, 0), (0, 4) and (6, 0) respectively. A point P moves so that the area
of ∆POA is always twice the area of ∆POB. Find the equation to both parts of the locus of P.
13.
14.
15.
16.
 ab

What does the equation (a − b ) ( x 2 + y 2 ) − 2abx = 0 become, if the origin be moved to the point 
,0 ?
a − b 
The equation x 2 + 2xy + 4 = 0 is transformed to the parallel axes through the point (6, λ). For what value of λ its
new form passes through the new origin ?
1°
Show that if the axes be turned through 7 ; the equation 3x 2 + ( 3 − 1) xy − y 2 = 0 become free of xy in its
2
new form.
Find the angle through which the axes may be turned so that the equation Ax + By + C = 0 may reduce to the
form x = constant, and determine the value of this constant.
17.
Transform 12x 2 + 7xy − 12y 2 − 17x − 31y − 7 = 0 to rectangular axes through the point (1, − 1) inclined at an
 4
angle tan−1  to the original axes.
 3
Shortcuts and Important Results to Remember
1 If D, E, F are the mid-points of the sides BC, CA, AB of
∆ABC, the
A=E + F−D
B=F+ D−E
and
C=D+E−F
2 Orthocentre, nine point centre, centroid, circumcentre of a
triangle are collinear. Centroid divides the line joining
orthocentre and circumcentre in the ration
2 :1 (Internally) and nine point centre is the mid-point of
orthocentre and circumcentre.
3 The circumcentre of a right angled triangle is the
mid-point of the hypotenuse.
4 In an equilateral triangle orthocentre, nine point centre,
centroid, circumcentre, incentre coincide.
5 The distance between the orthocentre and circumcentre
in an equilateral triangle is zero.
6 The orthocentre of a triangle having vertices (α , β ), (β, α )
and (α , α ) is (α , α ).
7 Orthocentre of the triangle formed by the points

1
 1   1  1 
, − αβγ
α ,  ; β,  ;  γ,  is  −
 α   β  γ   αβγ

i.e. all points and orthocentre lie on xy = 1.
8 Points in a triangle : Centroid (G ) , Incentre (I ), Excentres
(I1, I2 , I3 ), Orthocentre (O ), Circumcentre (C ) are given by
 m1 x1 + m2 x2 + m3 x3 m1 y1 + m2 y2 + m3 y3 
,

 , where

m1 + m2 + m3
m1 + m2 + m3 
G
I
I1
I2
I3
O
C
m1
1
sin A
− sin A
sin A
sin A
tan A
sin 2A
m2
1
sin B
sin B
− sin B
sin B
tan B
sin 2B
m3
1
sin C
sin C
sin C
− sin C
tan C
sin 2B
and vertices
A ≡ ( x1, y1 ), B ≡ ( x2 , y2 ), C ≡ ( x3 , y3 ) and A, B, C are the
angles of ∆ABC.
9 If the circumcentre and centroid of a triangle are
respectively (α , β ), (γ, δ ), then orthocentre will be
(3γ − 2α , 3δ − 2β).
10 If ABCD is a parallelogram, then D = A − B + C.
11 If D, E, F are the mid-points of the sides BC, CA, AB of
∆ABC, then the centroid of ∆ABC = centroid of ∆DEF. If
area of ∆ABC = ∆, then area of ∆AFE = area of ∆BDF =
∆
area of ∆CED = area of ∆DEF = and area of
4
parallelogram CEFD = area of parallelogram BDEF = area
∆
of parallelogram AEDF =
2
12 Orthocentre of the right angle triangled ABC, right angled
at A is A .
13 Circumcentre of the right angled triangle ABC, right
B+C
angled at A is
.
2
14 X-axis divides the line segment joining ( x1, y1 ), ( x2 , y2 ) in
the ratio − y1 : y2 and Y-axis divides the same line
segment in the ratio− x1 : x2 .
15 Area of the triangle formed by ( x1, y1), ( x2 y2 ), ( x3 , y3 ) is
1 x1 − x3 x2 − x3
|
|.
2 y1 − y3 y2 − y3
16 The area of the triangle formed by y = m1 x + c1,
y = m2 x + c 2 , y = m3 x + c 3 is
1 (c1 − c 2 )2
Σ
.
2 | m1 − m2 |
17 Area of the quadrilateral formed by
( x1, y1 ) ( x2 , y2 ) ( x3 , y3 ) , ( x4 , y4 ) is
1 x1 − x3 x2 − x4
|
|
2 y1 − y3 y2 − y4
18 If ( x1, y1 ), ( x2 , y2 ) are the ends of the hypotenuse of a right
angled isosceles triangle, then the third vertex is given by
 x1 + x2 ± ( y1 − y2 ) y1 + y2 m ( x1 − x2 )
,




2
2
19 Given the two vertices ( x1, y1 ) and ( x2 , y2 ) of an equilateral
triangle, then its third vertex is given by
 x1 + x2 ± 3 ( y1 − y2 ) y1 + y2 m 3 ( x1 − x2 )
,




2
2
20 Circumcentre of the triangle formed by the points
( x1, y1 ), ( x2, y2 ) and ( x3 , y3 ) is same as that of triangle
formed by the points (0, 0 ),
( x2 − x1, y2 − y1 ), ( x3 − x1, y3 − y1 ).
JEE Type Solved Examples :
Single Option Correct Type Questions
Ex. 3. Orthocentre of triangle with vertices (0, 0 ), (3, 4)
and (4, 0) is
This section contains 5 multiple choice examples. Each
example has four choice (a), (b), (c) and (d) out of which
ONLY ONE is correct.
l
Ex. 1. Locus of centroid of the triangle whose vertices are
(a cos t , a sin t ), (b sin t , - b cos t ) and (1, 0 ), where t is a
parameter, is
l
(a) (3 x - 1)2 + (3y )2 = a 2 - b 2
(c) (3 x + 1)2 + (3y )2 = a 2 + b 2
(d) (3 x + 1)2 + 3y 2 = a 2 - b 2
Sol. (b) Let A º (a cost, a sin t ), B º (b sin t, - b cost ) and C º (1, 0 )
a cost + b sin t + 1
Þ3 x - 1 = a cost + b sin t
3
a sin t - b cost
and y =
Þ3y = a sin t - b cost
3
On squaring and adding Eqs. (i) and (ii), we get
(3 x - 1 ) 2 + (3y ) 2 = a 2 + b 2
æ 3ö
(c) ç3, ÷
è 4ø
(d) (3, 9)
®
æ a cost + b sin t + 1 a sin t - b cost ö
\Centroid º ç
,
÷
ø
è
3
3
x=
(b) (3, 12)
Sol. (c) Denote the points are ( x1, y1 ), ( x 2, y 2 ) and ( x 3, y 3 ) from the
matrix
é x - x 3 y1 - y 3 ù
P =ê 1
ë x 2 - x 3 y 2 - y 3 úû
é - 4 0ù
=ê
ë - 1 4 úû
(b) (3 x - 1)2 + (3y )2 = a 2 + b 2
Let
æ 5ö
(a) ç3, ÷
è 4ø
\
…(i)
…(ii)
\Circumcentre of the triangle
æ x + x 2 + l(y1 - y 2 ) y1 + y 2 - l ( x1 - x 2 ö
,
ºç 1
÷
ø
è
2
2
æ 3 - 4 l 4 + 3 l ö æ 13 ö
,
ºç
÷ º ç2, ÷
è 2
2 ø è 8ø
which is locus of centroid.
Ex. 2. The incentre of triangle with vertices (1, 3 ), (0, 0 )
and ( 2, 0 ) is
l
æ 3ö
æ2 3 ö
æ 1 ö
æ2 1 ö
(a) ç1,
÷ (b) ç ,
÷ (d) ç1,
÷
÷ (c) ç ,
è 3ø
è
ø
3 3
è 2 ø
è3 2 ø
Sol. (d)
®
R 1× R 2
|P |
( -4 ) ( -1 ) + 0
1
=
=- 16
4
l=
i.e.
æ 13 ö
C º ç2, ÷
è 8ø
æ7 4ö
and centroid G º ç , ÷
è3 3 ø
\Orthocentre
4
13 ö
æ 7
H º ç3 ´ - 2 ´ 2, 3 ´ - 2 ´ ÷
è 3
3
8ø
Y
B (1, Ö3)
æ 3ö
º ç3, ÷
è 4ø
Aliter : Let orthocentre
H º ( a, b )
O
Let
and
X
æ 1 ö
Incentre º ç1,
÷
è
3ø
B (3, 4)
Y
O º ( 0, 0 ), A º (2, 0 )
B º (1, 3 )
Q
OA = 2 = OB = AB
Þ DOAB is an equilateral.
\ Incentre = Centroid
æ0+2+1 0+ 0+ 3ö
ºç
,
÷
3
3
è
ø
\
A (2, 0)
F
E
H
O
\ Slope of BH ´ slope of OA = - 1
æ b - 4 ö æ 0ö
Þ
ç
÷ ´ ç ÷ = -1
è a -3ø è 4ø
D A(4, 0)
X
48
Textbook of Coordinate Geometry
\
a -3 = 0
Þ
a =3
and slope of AH ´ slope of OB = - 1
æ b - 0 ö æ 4ö
Þ
ç
÷ ´ ç ÷ = -1
è a - 4ø è 3 ø
From Eq. (i) ,
b=
1 1 1
1
= | x1 y1| | r r 1 | = 0
2
r2 r2 1
…(i)
(QC1 and C 2 are identical)
Þ Points A, B, C are collinear.
Ex. 5. Let A be the image of ( 2, - 1) with respect to Y -axis.
Without transforming the origin, coordinate axis are turned
at an angle 45° in the clockwise direction. Then, the
coordiates of A in the new system are
l
3
4
æ 3ö
Hence, orthocentre is ç3, ÷ .
è 4ø
Ex. 4. If x 1 , x 2 , x 3 as well as y 1 , y 2 , y 3 are in GP, with the
same common ratio, then the points ( x 1 , y 1 ), ( x 2 , y 2 ) and
(x3 , y 3 )
l
(a) lie on a straight line
(b) lie on an ellipse
(c) lie on a circle
(d) are vertices of a triangle
3 ö
æ 1
(a) ç ,÷
è
2
2ø
1 ö
æ 3
(b) ç ,÷
è
2
2ø
æ 1 3 ö
(c) ç ,
÷
è 2 2ø
æ 3 1 ö
(d) ç ,
÷
è 2 2ø
Sol. (a) Since, the image of (h, k ) w.r.t. Y -axis is ( -h, k ).
\Coordinate of A are ( -2, - 1 ).
If ( X , Y ) are the coordinates of A w.r.t. the new coordinate axes
obtained by turning the axes through an angle 45° in the
clockwise direction, then
X = - 2 cos( -45° ) - sin( -45° )
2
1
1
=+
=2
2
2
and
Y = 2 sin( -45° ) - cos( -45° )
2
1
3
==2
2
2
3 ö
æ 1
\ Required coordinates are ç - , ÷.
è 2
2ø
Sol. (a) Let common ratio of GP is r, then x 2 = x1r , x 3 = x1r 2, y 2 = y1r
and y 3 = y1r 2.
Let A º ( x1, y1 ), B º ( x 2, y 2 ) and C º ( x 3, y 3 )
x y1 1
1 1
\ Area of DABC = | x 2 y 2 1 |
2 x y 1
3
3
x1
y1 1
1
= | x1r y1r 1 |
2
x1r 2 y1r 2 1
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 3 multiple choice examples. Each
example has four choices (a), (b), (c) and (d). Out of which
MORE THAN ONE may be correct.
l
Ex. 6. Let S1, S 2 , ¼, be squares such that for each n ³1,
Q Area of Sn < 1 Þan2 < 1
100
<1
Þ
2n - 1
Þ
2n - 1 > 100 > 2 6
Þ
Þ
\ n = 8, 9, 10, ¼
the length of a side of Sn equals the length of a diagonal of
Sn + 1 . If the length of a side S1 is 10 cm, then for which of the
following value of n is the area of Sn less than 1 sq cm?
(a) 7
(b) 8
(c) 9
(d) 10
Sol. (b,c,d)
If a be the side of the square, then diagonal d = a 2 by
hypothesis
an = 2 an + 1
Þ
Þ
an - 1
an - 2
an
a
=
=
= ¼= 1 n
2 ( 2 )2 ( 2 )3
( 2)
10
a1
=
an =
( 2 )n - 1 (2 ) (n - 1)/2
an + 1 =
n -1 >6
n >7
Ex 7. If each of the vertices of a triangle has integral
coordinates, then the triangles may be
l
(a) right angled
(c) isosceles
(b) equilateral
(d) scalene
Sol. (a,c,d) Let A º ( x1, y1 ), B º ( x 2, y 2 ) and C º ( x 3, y 3 ) be the vertices
of triangle ABC. Given x1, y1, x 2, y 2, x 3, y 3be all integers.
1
Now, area of DABC = x1(y 2 - y 3 ) + x 2(y 3 - y1 ) + x 3(y1 - y 2 )
2
…(i)
= Rational
Chap 01 Coordinate System and Coordinates
If DABC is equilateral then,
3
Area of DABC =
(side ) 2
4
3
=
{( x1 - x 2 ) 2 + (y1 - y 2 ) 2 }
4
…(ii)
= Irrational
It is clear from Eqs. (i) and (ii), DABC can not be equilateral.
(a) (1, 6)
æ5 ö
(c) ç , 6 ÷
è6 ø
1ö
æ
(d) ç -7, ÷
è
8ø
Sol. (b,c,d) Let vertex of the DABC be A( x, y )
\
Þ
AB = AC
( AB ) 2 = ( AC ) 2
Þ
( x - 1 ) 2 + (y - 3 ) 2 = ( x + 2 ) 2 + (y - 7 ) 2
Þ
Ex. 8. ABC is an isosceles triangle. If the coordinates of
the base are B(1, 3 ) and C ( -2, 7 ). The coordinates of vertex A
can be
l
æ 1 ö
(b) ç - , 5÷
è 2 ø
49
6 x - 8y + 43 = 0
…(i)
æ 1 ö æ5 ö
Here, use observe that the coordinates ç - , 5 ÷ , ç , 6 ÷ and
è 2 ø è6 ø
1ö
æ
ç -7, ÷ satisfy the Eq. (i).
è
8ø
JEE Type Solved Examples :
Paragraph Based Questions
n
This section contains one solved paragraph based 3
multiple choice questions. Each of these questions has
four choices (a), (b), (c) and (d) out of which ONLY ONE is
correct.
Paragraph
(Q. Nos. 9 to 11)
æ 1 ö æ 1ö æ 1ö
If A ça , ÷, B çb , ÷, C ç g , ÷ be the vertices of a DABC, where
è a ø è bø è g ø
a , b are the roots of x 2 - 6ax + 2 = 0; b , g are the roots of
x 2 - 6bx + 3 = 0 and g , a are the roots of x 2 - 6cx + 6 = 0;
a, b, c being positive.
9. The value of a + b + c is
(a) 1
(c) 3
(b) 2
(d) 5
10. The coordinates of centroid of DABC is
æ 11ö
(a) ç1, ÷
è 9ø
æ 1 11 ö
(b) ç , ÷
è 3 18 ø
æ 11 ö
(c) ç 2, ÷
è 18 ø
æ 2 11 ö
(d) ç , ÷
è 3 19 ø
11. The coordinates of orthocentre of DABC is
ö
æ 1
(a) ç - , - 2÷
ø
è 2
ö
æ 1
(b) ç - , - 3 ÷
ø
è 3
æ 1
ö
(c) ç - , - 5÷
è 5
ø
æ 1
ö
(d) ç - , - 6 ÷
è 6
ø
Sol. Q a, b are the roots of x 2 - 6ax + 2 = 0
\
a + b = 6a
and
ab = 2
Again, b, g are the roots of x 2 - 6bx + 3 = 0
…(i)
…(ii)
\
b + g = 6b
and
bg = 3
Again, g, a are the roots of x 2 - 6cx + 6 = 0
…(iii)
…(iv)
\
g + a = 6c
and
g a =6
from Eqs. (ii), (iv) and (vi), we get
ab × bg × ga = 2 × 3 × 6
Þ
abg = 6
\
a = 2, b = 1, g = 3
…(v)
…(vi)
9. (b) Adding Eqs. (i) , (iii) and (v), we get
2( a + b + g ) = 6 (a + b + c )
1
or
a + b + c = (2 + 1 + 3 ) = 2
3
æ Sa S ab ö
10. (c) Centroid of DABC º ç
,
÷
è 3 3 abg ø
æ 2 + 1 + 3 2 ×1 + 1 ×3 + 3 ×2 ö
ºç
,
÷
è
ø
3
3 ×2 ×1 ×3
æ 11 ö
º ç2, ÷
è 18 ø
æ 1
ö
11. (d) Orthocentre of DABC º ç , - abg ÷
è abg
ø
æ 1
ö
º ç - , - 6÷
è 6
ø
50
Textbook of Coordinate Geometry
JEE Type Solved Examples :
Single Integer Answer Type Questions
This section contains one solved example. The answer to
this example is a single digit integer ranging from 0 to 9
(both inclusive).
n
æ
1 ö
Ex.12. If the points ( -2, 0 ), ç -1, ÷ and (cos q, sin q ) are
è
3ø
collinear, then the number of values of q Î[0, 2 p ] is
l
Sol. (1) Since, the given points are collinear, then
-2
0 1
1
1 =0
-1
3
cos q sin q 1
Þ
cos q ö
æ 1
ö
æ
-2ç
- sin q÷ - 0 + 1 ç - sin q ÷ =0
è 3
ø
è
3 ø
Þ
3 sin q - cos q = 2
3
1
sin q - cos q = 1
2
2
pö
æ
sin ç q - ÷ = 1
è
6ø
Þ
or
p
p
= 2np +
6
2
p p 2p
for n = 0,
q= + =
Î[ 0, 2 p ]
6 2 3
Number of values q is 1.
or
q-
JEE Type Solved Examples :
Matching Type Questions
n
This section contains one solved example. Which has
four statements (A, B, C and D) given in Column I and
four statements (p, q, r and s) in Column II. Any given
statements in Column I can have correct matching with
one or more statement(s) given in Column II.
l
Ex. 13. Match the following
Column I
Column II
A. The points (l + 1, 1), (2l + 1, 3) and
(2l + 2, 2l ) are collinear then number of
values of l is
(p) a prime
number
B. Area of DABC is 20 sq units, where A , B
and C are (4 , 6), (10, 14) and (x , y)
respectively. AC is perpendicular to BC,
then number of positions of C is
(q) an odd
number
C. In a DABC coordinates of orthocentre,
centroid and vertex A are respectively
(2, 2), (2, 1) and (0, 2). The x-coordinate of
vertex B is
(r) a composite
number
D. A man starts from P(- 3, 4 ) and reaches
the point Q (0, 1) touching the X-axis at
R(l , 0) such that PR + RQ is minimum,
then 10 | l | is
(s) a perfect
number
Sol. (A) ® (p); (B) ® (r); (C) ® (p, q); (D) ® (r, s)
(A)Q Points are collinear
l+1 1 1
\
2l + 1 3 1 = 0
2l + 2 2l 1
Þ
( l + 1 )(3 - 2 l ) - 1(2 l + 1 - 2 l - 2 )
+ 1 ( 4 l2 + 2 l - 6 l - 6 ) = 0
or
2 l2 - 3 l - 2 = 0
or
(2 l + 1 ) ( l - 2 ) = 0
\
l = 2, -
1
2
Number of values of l is 2.
(B) Q Area of triangles ABC is 20 sq units.
D
(4, 6) A
O
C
B (10, 14)
E
Q C can not be at D and E
\ Four positions are possible two above AB and two below AB.
Chap 01 Coordinate System and Coordinates
(C)
P
B(a, g)
Y
51
Y
H(2, 2)
(0, 2) A
Q(0, 1)
X¢
R O
A
X
G(2, 1)
X¢
X
O
Y¢
C(a, b)
C
Y¢
Q
ÐPRA = ÐQRO
From similar triangles PAR and QOR
AR PA
l+3 4
or
=
=
RO QO
0-l 1
a+ a+ 0
=2
3
\
a =3
Þ x-coordinates of vertex B = 3
(D) For PR + RQ to be minimum, it should be the path of light
Q
or
l=-
3
5
\ 10 | l | = 6.
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions (Ex. Nos. 14 and 15) are Assertion-Reason
type examples. Each of these examples contains two
statements.
Statement I (Assertion) and Statement II (Reason)
Each of these examples also has four alternative choices,
(a), (b), (c) and (d) only one out of which is the correct
answer. You have to select the correct choice as given
below :
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is not a
correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Ex. 14. Statement I The area of the triangle formed by
the points A(100, 102 ), B (102, 105 ), C(104, 107 ) is same as the
area formed by A ¢ (0, 0 ), B ¢ ( 2, 3 ), C ¢ ( 4, 5 ).
l
Statement II The area of the triangle is constant with
respect to translation :
Sol. (a) Area of triangle is unaltered by shifting origin to any point.
If origin is shifted to (100, 102), then A, B, C becomes A¢( 0, 0 ),
B ¢(2, 3 ), C ¢( 4, 5 ) respectively. Hence, both statements are true
and Statement II is correct explanation for Statement I.
Ex. 15. Statement I If centroid and circumcentre of a
triangle are known its orthocentre can be found
Statement II Centroid, orthocentre and circumcentre of a
triangle are collinear.
l
Sol. (b)Q Centroid divides orthocentre and circumcentre in the
ratio 2 : 1 (internally).
\We can find easily orthocentre.
Þ Statement I is true, and centroid, orthocentre and
circumcentre are collinear Statement II is true but Statement II
is not correct explanation for Statement I.
52
Textbook of Coordinate Geometry
Subjective Type Examples
n
In this section, there are 7 subjective solved examples.
Sol. Since, m1 and m2 are the roots of the equation
x 2 + ( 3 + 2) x + ( 3 - 1) = 0
Ex. 16. The four points A (a, 0 ), B (b, 0 ), C ( g, 0 ) and
D ( d, 0 ) are such that a, b are the roots of equation
ax 2 + 2hx + b = 0 and g, d are those of equation
a ¢ x 2 + 2h ¢ x + b ¢ = 0. Show that the sum of the ratios in
which C and D divide AB is zero, if ab ¢ + a ¢ b = 2hh ¢ .
l
Sol. Since, a, b are the roots of ax 2 + 2hx + b = 0
2h
b
and ab =
a+b=\
a
a
and g, d are the roots of a ¢ x 2 + 2h ¢ x + b ¢ = 0
2h ¢
b¢
and gd =
then,
g + d=a¢
a¢
then
m1 + m2 = - ( 3 + 2 ), m1m2 = ( 3 - 1 )
\
m1 - m2 = (m1 + m2 ) 2 - 4m1m2
= (3 + 4 + 4 3 - 4 3 + 4 ) = 11
and coordinates of the vertices of the given triangle are ( 0, 0 ),
(c / m1, c ) and (c / m2, c ).
…(i)
Hence, the required area of triangle =
...(ii)
1 æ 1
1 ö 1 | m - m1|
= c2 ç - ÷ = c2 2
2 è m1 m2 ø 2
| m1m2|
Let C divides AB in the ratio l :1
then
(α, 0)
A
(γ, 0)
λ
(β, 0)
B
C
1
11
= c2
2 ( 3 - 1)
1
l ×b + 1 × a
l+1
g-a
l=
b-g
g=
\
=
\
but given
Þ
Þ
Þ
or
l
(α, 0)
A
µ
(δ, 0)
D
1
(β, 0)
B
m ×b + 1 × a
d=
m+1
d-a
m=
b-d
l+m=0
g -a d-a
+
=0
b-g b-d
( a + b ) ( g + d) - 2 ab - 2 gd = 0
æ 2h ö æ 2h ¢ ö 2b 2b ¢
=0
ç- ÷ ç÷- è a øè a¢ø a
a¢
[from Eqs. (i) and (ii)]
ab ¢ + a ¢b = 2hh ¢
Ex. 17. If m1 and m 2 are the roots of the equation
x 2 + ( 3 + 2 ) x + ( 3 - 1) = 0
Show that the area of the triangle formed by the lines
æ 33 + 11 ö 2
y = m1 x , y = m 2 x and y = c is ç
÷c .
4
ø
è
1 2
11 ( 3 + 1 )
c ×
2
( 3 - 1) ( 3 + 1)
æ 33 + 11 ö 2
=ç
÷c
4
è
ø
and let D divides AB in the ratio m :1
then
1 c
c
´c ´c
2 m1
m2
Ex. 18. If x coordinates of two points B and C are the
roots of equation x 2 + 4 x + 3 = 0 and their y coordinates are
l
the roots of equation x 2 - x - 6 = 0. If x coordinate of B is
less than x coordinate of C and y coordinate of B is greater
than the y coordinate of C and coordinates of a third point A
be (3, - 5 ) , find the length of the bisector of the interior angle
at A.
Sol. Q
and
x 2 + 4 x + 3 = 0 Þ x = - 1, - 3
x 2 - x - 6 = 0 Þ x = - 2, 3
Also, given that x and y coordinates of B are respectively less
than and greater than the corresponding coordinates of C.
\
B º ( -3, 3 ) and C º ( -1, - 2 )
Now,
AB = (3 + 3 ) 2 + ( -5 - 3 ) 2 = 10
and
AC = (3 + 1 ) 2 + ( -5 + 2 ) 2 = 5
AB 2
=
AC 1
Let AD be the bisector of Ð BAC, then
BD AB 2
=
=
DC AC 1
\
Chap 01 Coordinate System and Coordinates
\
Y
B (–3,3)
Þ
X´
X
O
D
Þ
A (3,–5)
Thus, D divides BC internally in the ratio 2 : 1
æ 2 ( -1 ) + 1 ( -3 ) 2 ( -2 ) + 1 (3 ) ö
Dºç
,
\
÷
è
ø
2+1
2+1
Thus,
3r 2 = 4a 2 - 4a + 4
5ö
1ö
æ
æ
AD = ç3 + ÷ + ç - 5 + ÷
è
ø
è
3
3ø
=
2
196 196 14 2
units
+
=
9
9
3
Ex. 19. The distance between two parallel lines is unity.
A point P lies between the lines at a distance a from one of
them. Find the length of a side of an equilateral triangle
PQR, vertex Q of which lies on one of the parallel lines and
vertex R lies on the other line.
and
then,
Given,
In D PQL,
\
PQ = QR = RP = r
Ð PQL = q
Ð XQR = q + 60°
PL = a and RN = 1 unit
PL a
sin q =
=
QP r
…(i)
a = r sin q
Y
R (r, θ + 60°)
r
r
1
60°
X′
(0, 0) Q
θ
P (r, θ)
N
L
Y′
and in DQRN, sin( q + 60° ) =
a
r
RN 1
=
QR r
Ex. 20. In a D ABC , A º (a, b) , B º (1, 2 ), C º ( 2 , 3 ) and
point A lies on the line y = 2 x + 3 where a, b Î I. If the area
of D ABC be such that [ D ] = 2 , where [.] denotes the greatest
integer function, find all possible coordinates of A.
l
l
Sol. Let
[from Eq. (i)]
= 4 (a 2 - a + 1 )
2
\
r=
(a 2 - a + 1 )
3
Hence, length of side of an equilateral triangle
2
(a 2 - a + 1 ) units.
=
3
æ 5 1ö
D º ç- , - ÷
è 3 3ø
2
Now,
ìï 1 a
3
a 2 üï
´ 1 - 2 ý =1
rí ´ +
2
r ïþ
ïî 2 r
Þ
or
Y´
ì1
ü
3
r í sin q +
cos qý = 1
2
2
î
þ
Þ
Þ
×
×
r sin( q + 60° ) = 1
r {sin q cos60° + cos q sin 60° } = 1
3
a
+
(r 2 - a 2 ) = 1
2
2
3
a
(r 2 - a 2 ) = 1 2
2
2
a
3 2
(r - a 2 ) = 1 +
-a
4
4
3r 2 - 3a 2 = 4 + a 2 - 4a
Þ
(–1,–2)C
53
X
Sol. Q( a, b ) lies on y = 2 x + 3
then
b = 2a + 3
Thus, the coordinates of A are ( a, 2 a + 3 )
1 ì a 2 a + 3½ ½1 2½ ½2
3 ½ü
+
+
D = í½
2 ½ ½2 3½ ½a 2 a + 3½ýþ
2 î½1
…(i)
1
= | 2 a - (2 a + 3 ) + 3 - 4 + 4 a + 6 - 3 a|
2
1
= | a + 2|
2
but
[D] = 2
é1
ù
Þ
êë 2 | a + 2 | úû = 2
| a + 2|
Þ
2£
<3
2
Þ
4 £| a + 2 | < 6
Þ
4 £a + 2 <6
and
-6 < a + 2 £ - 4
Þ
2 £ a < 4 and -8 < a £ - 6
Q
a ÎI
\
a = 2 , 3 , - 7, - 6
Hence, possible coordinates of A are (2, 7 ), (3, 9 ), ( -7, - 11 ) and
( -6, - 9 ).
54
Textbook of Coordinate Geometry
Ex. 21. Let S be the square of unit area. Consider any
quadrilateral which has one vertex on each side of S. If a ,b,c
and d denote the lengths of the sides of the quadrilateral,
prove that 2 £a 2 + b 2 + c 2 + d 2 £ 4.
l
Sol. Given S be the area of square with vertices
O ( 0, 0 ), A (1, 0 ), B (1, 1 ), C ( 0, 1 ).
Let PQRM be the quadrilateral with vertices
P ( p, 0 ), Q (1, q ), R (r , 1 ) and M( 0, m )
Sol. Since, the circumcentre of the triangle is at the origin O, we
have OA = OB = OC = R, where R is the circumradius of the
circumcircle.
\
(OA ) 2 = R 2 Þa 2 + a 2 tan 2 a = R 2 Þ a = R cos a
and sides MP = a , PQ = b, QR = c, RM = d
Y
Therefore, the coordinates of A are ( R cos a, R sin a ).
R(r, 1)
(0, 1)C
d
c
a
b
B(1, 1)
O
Similarly, the coordinates of B are ( R cosb, R sin b ) and those of
C are ( R cos g, R sin g ).
Q (1, q)
(0, m)M
X′
Ex. 22. The circumcentre of a triangle with vertices
(a , a tan a ), B (b, b tan b) and C (c , c tan g ) lies at the origin,
where 0 < a, b, g < p / 2 and a + b + g = p. Show that its
orthocentre lies on the line
æbö
ægö
æbö
ægö
æa ö
æa ö
4 cos ç ÷ cos ç ÷ cos ç ÷ x - 4 sin ç ÷ sin ç ÷ sin ç ÷ y =y
è2ø
è2ø
è 2ø
è 2ø
è 2ø
è 2ø
l
P(p, 0)
Thus the coordinates of centroid G of D ABC are
R
æR
ö
ç (cos a + cosb + cos g ), (sin a + sin b + sin g ) ÷
è3
ø
3
X
A(1, 0)
Now, if P (h, k ) is the orthocentre of D ABC, then
R
1 ×h + 2 × 0
(cos a + cosb + cos g ) =
3
1+2
Y′
2
Then,
2
a =p +m
2
O
b 2 = (1 - p ) 2 + q 2
1
c 2 = (1 - q ) 2 + (1 - r ) 2
G
2
d 2 = r 2 + (1 - m ) 2
2
2
2
2
2
2
2
\ a + b + c + d = p + (1 - p ) + q + (1 - q )
P
2
+ r 2 + (1 - r ) 2 + m 2 + (1 - m ) 2
= 2 [ p 2 + q 2 + r 2 + m 2 - p - q - r - m + 2]
2
éæ
1ö
=2 êçp - ÷ +
2ø
êë è
(QPG : GO = 2 : 1 )
2
2
2
ù
1ö
1ö
æ
æ 1ö
æ
çq - ÷ + çr - ÷ + çm - ÷ + 1 ú ³2
è
è 2ø
è
2ø
2ø
úû
Þ
a 2 + b 2 + c 2 + d 2 ³2
also, since
\
0 £ x £1
x 2 £1
\
a 2 £ 1, b 2 £ 1, c 2 £ 1, d 2 £ 1
Þ
a2 + b2 + c2 + d 2 £4
From Eqs. (i) and (ii), we get
2 £a2 + b2 + c2 + d 2 £4
Þ
R (cos a + cosb + cos g ) = h
and
R
1 ×k + 2 × 0
(sin a + sin b + sin g ) =
3
1+2
Þ
R (sin a + sin b + sin g ) = k
...(i)
...(ii)
Dividing (ii) by (i), then
…(i)
Þ
sin a + sin b + sin g k
=
cos a + cosb + cos g h
Þ
4 cos( a / 2 ) cos(b / 2 ) cos( g / 2 ) k
=
1 + 4 sin( a / 2 ) sin(b / 2 ) sin( g / 2) h
(Because, a + b + y = p by identity)
…(ii)
Hence, the orthocentre P (h, k ) lies on the line
æ aö
æbö
ægö
æ aö
æbö
ægö
4 cos ç ÷ cos ç ÷ cos ç ÷ x - 4 sin ç ÷ sin ç ÷ sin ç ÷ y = y
è2ø
è2ø
è2ø
è2ø
è2ø
è2ø
#L
Coordinate System and Coordinates Exercise 1 :
Single Option Correct Type Questions
n
This section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c), (d) out of which
ONLY ONE is correct.
1. Vertices of a variable triangle are (3, 4), (5 cos θ, 5 sin θ)
and (5 sin θ, − 5 cos θ), where θ ∈ R. Locus of its
orthocentre is
(a) x 2 + y 2 + 6 x
(b) x 2 + y 2 − 6 x
(c) x 2 + y 2 + 6 x
(d) x 2 + y 2 − 6 x
+ 8y − 25 = 0
+ 8y − 25 = 0
− 8y − 25 = 0
− 8y − 25 = 0
2. If a rod AB of length 2 units slides on coordinate axes in
the first quadrant. An equilateral triangle ABC is
completed with C on the side away from O. Then, locus
of C is
(a) x 2 + y 2 − xy + 1 = 0
(b) x 2 + y 2 − xy 3 + 1 = 0
(c) x 2 + y 2 + xy 3 − 1 = 0
(d) x 2 + y 2 − xy 3 − 1 = 0
3. The sides of a triangle are 3x + 4y, 4 x + 3y and 5x + 5y
units, where x > 0, y > 0. The triangle is
(a) right angled
(c) obtuse angled
(b) acute angled
(d) isosceles
4. Let P and Q be the points on the line joining A( −2, 5)
and B (3, 1) such that AP = PQ = QB. Then, the mid-point
of PQ is
 1 
1 
(a)  , 3
(b)  − , 4
 4 
2 
(c) ( 2, 3 )
(d) ( − 1, 4 )
5. A triangle ABC right angled at A has points A and B as
(2, 3) and (0, − 1) respectively. If BC = 5 units, then the
point C is
(a) ( 4, 2)
(b) ( −4, 2)
(c) ( −4, 4 )
(d) ( 4, − 4 )
6. The locus of a point P which divides the line joining
(1, 0) and (2 cos θ, 2 sin θ) internally in the ratio 2 : 3
for all θ is
(a) a straight line
(b) a circle
(c) a pair of straight lines (d) a parabola
7. The points with the coordinates (2a, 3a ), (3b, 2b ) and (c , c )
are collinear
(a) for no value of a, b , c
c
(c) if a, , b are in HP
5
(b) for all values of a, b , c
2c
(d) if a, , b are in HP
5
8. The vertices of a triangle are (0, 3), (−3, 0) and (3, 0). The
coordinates of its orthocentre are
(a) (0, − 2)
(b) (0, 2)
(c) (0, 3 )
(d) (0, − 3 )
9. ABC is an equilateral triangle such that the vertices B
and C lie on two parallel lines at a distance 6. If A lies
between the parallel lines at a distance 4 from one of
them, then the length of a side of the equilateral triangle
is
88
(a) 8
(b)
3
4 7
(c)
(d) None of these
3
10. A, B, C are respectively the points (1, 2), ( 4, 2), ( 4, 5). If
T 1 , T 2 are the points of trisection of the line segment AC
and S 1 , S 2 are the points of trisection of the line segment
BC, the area of the quadrilateral T 1 S 1 S 2 T 2 is
3
5
(c) 2
(d)
(a) 1
(b)
2
2
11. (i) The points ( −1, 0), ( 4, − 2) and (cos 2 θ, sin 2 θ) are
collinear
 1 − tan 2 θ 2 tan θ 
(ii) The points ( −1, 0), ( 4, − 2) and 
,

 1 + tan 2 θ 1 + tan 2 θ
are collinear
(a)
(b)
(c)
(d)
both statements are equivalent
statemetn (i) has more solution than statement (ii) for θ
statement (ii) has more solution than statement (i) for θ
None of the above
12. If α 1 , α 2 , α 3 , β 1 , β 2 , β 3 are the values of n for which
n −1
n −1
r =0
r =0
∑ x 2r is divisible by ∑ x r , then the triangle having
vertices (α 1 , β 1 ), (α 2 , β 2 ) and (α 3 , β 3 ) cannot be
(a) an isosceles triangle
(b) a right angled isosceles triangle
(c) a right angled triangle
(d) an equilateral triangle
3
13. A triangle ABC with vertices A( −1, 0), B  −2,  and

4
7

C  −3, −  has its orthocentre at H. Then, the

6
orthocentre of triangle BCH will be
(a) ( −3, − 2)
(c) ( −1, 2)
(b) (1, 3 )
(d) None of these
56
Textbook of Coordinate Geometry
4
14. If ∑ ( x i2 + y i2 ) ≤ 2x 1 x 3 + 2x 2 x 4 + 2y 2y 3 + 2y 1y 4 , the
i =1
points ( x 1 , y 1 ), ( x 2 , y 2 ), ( x 3 , y 3 ), ( x 4 , y 4 ) are
(a) the vertices of a rectangle
(b) collinear
(c) the vertices of a trapezium
(d) None of the above
#L
15. Without change of axes the origin is shifted to (h, k ),
then from the equation x 2 + y 2 − 4 x + 6y − 7 = 0,
then term containing linear powers are missing, then
point (h, k ) is
(b) ( −3, 2)
(d) ( −2, − 3 )
(a) (3, 2)
(c) ( 2, − 3 )
Coordinate System and Coordinates Exercise 2 :
More than One Correct Option Type Questions
n
This section contains 7 multiple choice questions. Each
questions has four choices (a), (b), (c), (d) out of which
MORE THAN ONE may be correct.
16. If ( −6, − 4 ), (3, 5), ( −2, 1) are the vertices of a
parallelogram, then remaining vertex can be
(a) (0, − 1)
(c) ( − 11, − 8 )
(b) ( − 1, 0)
(d) (7, 10)
17. If the point P( x , y ) be equidistant from the points
A(a + b, a − b ) and B (a − b, a + b ) then
(a) ax = by
(b) bx = ay
(c) x 2 − y 2 = 2(ax + by )
(d) P can be (a, b )
(a) parallelogram
(c) rhombus
(b) rectangle
(d) square
20. The medians AD and BE of the triangle with vertices
A(0, b ), B(0, 0) and C (a, 0) are mutually perpendicular if
(a) b = a 2
(b) a = b 2
(c) b = − a 2
(d) a = − b 2
21. The points A( x , y ), B(y, z ) and C (z, x ) represents the
(a) x = y
(c) z = x
rational numbers, then which of the following points of
the triangle will always have rational coordinates
#L
the vertices of a
vertices of a right angled triangle, if
18. If the coordinates of the vertices of a triangle are
(a) centroid
(c) circumcentre
19. The points A( −4, − 1), B( −2, − 4 ), C ( 4, 0) and D(2, 3) are
(b) incentre
(d) orthocentre
(b) y = z
(d) x = y = z
22. Let the base of a triangle lie along the line x = a and be
of length 2a. The area of this triangles is a 2 , if the vertex
lies on the line
(a) x = − a
a
(c) x =
2
(b) x = 0
(d) x = 2a
Coordinate System and Coordinates Exercise 3 :
Paragraph Based Questions
n
This section contains 2 solved paragraphs based upon
each of the paragraph, 3 multiple choice questions have
to be answered. Each of these question has four choices
(a), (b), (c) and (d) out of which ONLY ONE is correct.
Paragraph I
(Q. Nos. 23 to 25)
ABC is a triangle right angled at A , AB = 2AC. A ≡ (1, 2),
B ≡ ( −3, 1). ACD is an equilateral triangle. The vertices of two
triangles are in anticlockwise sense. BCEF is a square with
vertices in clockwise sense.
23. If area of ∆ACF is S, then the value of 8S is
(a) 42
(c) 62
(b) 51
(d) 102
24. Length of DE is
(a) 17
(4 − 3 )
(b)
17
2
(8 + 3 )
(c)
17
2
(4 + 3 )
(d) 15 ( 4 + 3 )
25. The y-coordinate of the centroid of the square BCEF is
1
4
5
(c) −
4
(a) −
3
4
7
(d) −
4
(b) −
Chap 01 Coordinate System and Coordinates
Paragraph II
26. Length of perpendicular from M to OA is equal to
(Q. Nos. 26 to 28)
(a) 3
 1
Let O( 0, 0) , A( 2, 0) and B 1,  be the vertices of a triangle.

3
Let R be the region consisting of all those points P inside ∆OAB
satisfying.
d ( P , OA ) ≤ min {d ( P , OB ), d ( P , AB )}, where, d denotes the
distance from the point P to the corresponding line. Let M be the
peak of region R.
#L
57
(b)
1
3
(c) 3
(d) 2 − 3
27. The perimeter of region R is equal to
(a) 4 − 3
(b) 4 + 3
(c) 4 + 3 3
(d) 2 + 4 ( 2 − 3 )
28. The area of region R is equal to
(a) 2 − 3
(b) 2 + 3
(c) 2 3
(d) 4 + 3
Coordinate System and Coordinates Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 5 questions. The answer to each
example is a single digit integer ranging from 0 to 9 (both
inclusive).
29. If the area of the triangle formed by the points
(2a, b ), (a + b, 2b + a ) and (2b, 2a ) be ∆ 1 and the area of the
triangle whose vertices are (a + b, a − b ), (3b − a, b + 3a ) and
(3a − b, 3b − a ) be ∆ 2 , then the value of ∆ 2 / ∆ 1 is
30. The diameter of the nine point circle of the triangle with
vertices (3, 4 ), (5 cos θ, 5 sin θ) and (5 sin θ, − 5 cos θ), where
θ ∈ R, is
#L
31. The ends of the base of an isosceles triangle are
(2 2, 0) and (0, 2 ). One side is of length 2 2. If ∆ be
the area of triangle, then the value of [∆ ] is (where
[⋅] denotes the greatest integer function)
32. If ( x , y ) is the incentre of the triangle formed by the
points (3, 4 ), ( 4, 3) and (1, 2), then the value of x 2 is
33. Let P and Q be points on the line joining A( −2, 5)
and B(3, 1) such that AP = PQ = QB. If mid-point of
b
PQ is (a, b ), then the value of is
a
Coordinate System and Coordinates Exercise 5 :
Matching Type Questions
n
The section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and four
statements (p, q, r and s) in Column II. Any given statement in Column I can have correct matching with one or more
statement(s) given in Column II.
34. Consider the triangle with vertices A(0, 0), B (5, 12) and C(16, 12).
Column I
Column II
A. If(λ , µ ) are the coordinates of centroid of
triangle ABC, then (λ + µ ) is divisible by
(p)
B. If (λ , µ) are the coordinates of circumcentre of
triangle ABC, then 2λ is divisible by
(q)
5
C. If (λ , µ ) are the coordinates of incentre of
triangle ABC, then µ is divisible by
(r)
7
D. If (λ , µ ) are the coordinates of excentre
opposite to vertex B, then λ + µ is divisible by
(s)
3
35. The vertices of a triangle are A( − 10, 8), B (14, 8) and
C( −10, 26). Let G, I, H , O be the centroid, incentre,
orthocentre, circumcentre respectively of ∆ABC.
Column I
9
Column II
A. The inradius r is
(p) a prime number
B. The circumradius R is
(q) an even number
C. The area of ∆IGH is
(r) a composite number
D. The area of ∆OGI is
(s) a perfect number
58
#L
Textbook of Coordinate Geometry
Coordinate System and Coordinates Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos. 36 to 39) are Assertion-Reason
type questions. Each of these questions contains two
statements :
Statement I (Assertion) and
Statement II (Reason)
Each of these question also has four alternative choices
(a), (b), (c) and (d), only one out of which is the correct
answer.
You have to select the correct choice as given below
(a) Statement I is true, statement II is true; statement II is a
correct explanation for statement I
(b) Statement I is true, statement II is true; statement II is not a
correct explanation for statement I
(c) Statement I is true, statement II is false
(d) Statement I is false, statement II is true
36. The vertices of a triangle an A(1, 2), B ( −1, 3) and C(3, 4 ).
Let D, E, F divide BC , CA, AB respectively in the same
ratio.
Statement I The centroid of triangle DEF is (1, 3).
Statement II The triangle ABC and DEF have the same
centroid.
37. Statement I Let the vertices of a ∆ABC be A( −5, − 2),
B(7, 6) and C(5, − 4 ). Then, the coordinates of the
circumcentre are (1, 2).
Statement II In a right angled triangle, the mid point
of the hypotenuse is the circumcentre of the triangle.
38. A line segment AB is divided internally and externally in
the same ratio at P and Q respectively and M is the
mid-point of AB.
A
M
P
B
Q
Statement I MP, MB, MQ are in G.P
Statement II AP, AB and AQ are in H.P.
39. Statement I Transformation of the equation
x 2 − 3xy + 11x − 12y + 36 = 0 to parallel axes through the
1
point ( − 4, 1) becomes ax 2 + bxy + 1 = 0, then b 2 − a = .
64
Statement II If the axes turned through an angle θ,
then the equation f ( x , y ) = 0 is transformed by replacing
( x , y ) by (( x cos θ − y sin θ), ( x sin θ + y cos θ)).
Coordinate System and Coordinates Exercise 7 :
Subjective Type Questions
n
In this section, there are 7 subjective questions.
40. If A ( x 1 , y 1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) are the vertices of a
∆ ABC and ( x , y ) be a point on the internal bisector of
angle A, then prove that
x y 1  x y 1
bx 1 y 1 1 + c  x 1 y 1 1 = 0

 

x 2 y 2 1  x 3 y 3 1
where, AC = b and AB = c .
41. If a, b, c be the pth, qth and rth terms respectively of a
HP, show that the points (bc , p ), (ca , q ) and (ab , r ) are
collinear.
42. A line L intersects three sides BC, CA and AB of a
triangle in P, Q, R respectively, show that
BP CQ AR
⋅
⋅
= −1
PC QA RB
 a 3 a 2 − 3  b 3 b 2 − 3
,
,
 and
,
a − 1 a − 1  b − 1 b − 1 
43. If the points, 
 c 3 c 2 − 3
 are collinear for three distinct values a, b, c

c − 1 c − 1 
and a ≠ 1, b ≠ 1 and c ≠ 1, then show that
abc − (bc + ca + ab ) + 3 (a + b + c ) = 0
44. Show that the area of the triangle whose sides are
∆2
, where C 1 , C 2
2 | C 1C 2C 3 |
and C 3 are the cofactors of c 1 , c 2 and c 3 respectively in
the determinant
a 1 b1 c 1
ar x + br y + c r = 0, r = 1, 2, 3 is
∆ = a2
b2
c2
a3
b3
c3
Chap 01 Coordinate System and Coordinates
45. If A 1 , A 2 , A 3 , ..., A n are n points in a plane whose
coordinates are ( x 1 , y 1 ), ( x 2 , y 2 ), ( x 3 , y 3 ) , ..., ( x n , y n )
respectively. A 1 A 2 is bisected in the point G 1 ; G 1 A 3 is
divided at G 2 in the ratio 1 : 2 ; G 2 A 4 is divided at G 3 in
the ratio 1 : 3; G 3 A 5 at G 4 in the 1 : 4 and so on until all
the points are exhausted. Show that the coordinates of
the final point so obtained are
x 1 + x 2 + ..... + x n
y 1 + y 2 + ..... + y n
and
n
n
#L
59
46. If by change of axes without change of origin, the
expression ax 2 + 2hxy + by 2 becomes
a1 x 12 + 2h 1 x 1 y 1 + b1 y 12 , prove that
(i) a + b = a1 + b1
(ii) ab − h 2 = a1 b1 − h 12
(iii) (a − b ) 2 + 4h 2 = (a1 − b1 ) 2 + 4h 12
Coordinate System and Coordinates Exercise 8 :
Questions Asked in Previous 13 Year’s Exams
n
This section contains questions asked in IIT-JEE, AIEEE,
JEE Main & JEE Advanced from year 2005 to 2017.
47. If a vertex of a triangle is (1, 1) and the mid-points of
two side through this vertex are ( − 1, 2) and (3, 2), then
the centroid of the triangle is
[AIEEE 2005, 3M]
 1 7
 7
(b) 1, 
(a)  , 
3 3
 3
 1 7
(c)  − , 
 3 3
7

(d)  − 1, 

3
48. Let O(0, 0), P(3, 4 ), Q(6, 0) be the vertices of the triangle
OPQ. The point R inside the triangle OPQ is such that the
triangles OPR, PQR, OQR are of equal area. The
coordinates of R are
[IIT-JEE 2007, 3M]
4 
 2
(b) 3, 
(a)  , 3
3 
 3
 4
(c) 3, 
 3
 4 2
(d)  , 
 3 3
49. Let A(h, k ), B(1, 1) and C(2, 1) be the vertices of a right
angled triangle with AC as its hypotenuse. If the area of
the triangle is 1, then the set of values which ‘k’ can take
is given by
[AIEEE 2007, 3M]
(a) {1, 3 }
(c) ( − 1, 3}
(b) {0, 2}
(d) { − 3, − 2}
50. Three distinct points A, B and C are given in the
2-dimensional coordinates plane such that the ratio of
the distance of any one of them from the point (1, 0) to
1
the distance from the point ( −1, 0) is equal to . Then, the
3
circumcentre of the triangle ABC is at the point
[AIEEE 2009, 4M]
5 
(a)  , 0
4 
5 
(c)  , 0
3 
5 
(b)  , 0
2 
(d) (0, 0)
51. The x-coordinate of the incentre of the triangle that has
the coordinates of mid-points of its sides are (0, 1), (1, 1)
and (1, 0) is
[JEE Main 2013, 4M]
(a) 2 + 2
(b) 2 − 2
(c) 1 + 2
(d) 1 − 2
52. The number of points, having both coordinates are
integers, that lie in the interior of the triangle with
vertices (0, 0), (0, 41) and (41, 0) is
[JEE Main 2015, 4M]
(a) 820
(b) 780
(c) 901
(d) 861
53. Let k be an integer such that the triangle with vertices
(k , − 3k ), (5, k ) and ( −k , 2) has area 28 sq units. Then, the
orthocentre of this triangle is at the point
[JEE Main 2017, 4M]
 1
(a)  2, 
 2
1

(b)  2, − 

2
 3
(c) 1, 
 4
3

(d) 1, − 

4
Answers
Exercise for Session 4
Exercise for Session 1
1. (d)
4. (b)
2. (b)
5. (b)
8. (−6, 0)
7. (0, ± 4)
9. θ = α
3. (b)
6. (±5, 0)
10. x2 + y2 = 2 ax
Exercise for Session 2
1. (d)
2. (c)
3. (b)
4. (b)
5. (d)
6. (a)
7. (c)
8. (c)
9. 8a
13.  ,
3 5
5 2
 and
 2 2
2
63
15. (5, 1) and (−3, 3)
17. a = b = 2 −
65
1+ 3 7 − 5 3  1 − 3 7 + 5 3 
18. 
,
,
 or 

2
2
 2
  2

14.
3
Exercise for Session 3
1. (a)
2. (b)
8. (c)
7. (a)
12. 333 sq units
2
3 45
18. −
,
67 17
3. (a)
9. (d)
4. (c)
10. (b)
5. (a)
11. ( p, q)
13. Centroid = incentre ≡ (2 , 3)
19. 34 sq units
(b)
2. (d)
3. (c)
4. (b)
5. (c)
(c)
8. y2 − 8 y − 4x + 20 = 0
3x2 + 3 y2 − 10 ax + 3a2 = 0 10. x + y = 4
12. x2 − 9 y2 = 0
9x2 + 36 y2 = 4l 2
1
2
2
2
2 4
13. (a − b) (x + y ) = a b
14. −3
3
C
−1  B 
16. tan   , −
17. xy = 0
 A
(A 2 + B 2 )
1.
7.
9.
11.
6. (c)
6. (b)
Chapter Exercises
(d)
2. (d)
3. (c)
4. (a)
5. (a)
6. (b)
(d)
8. (c)
9. (c)
10. (b)
11. (b)
12. (d)
(d)
14. (a)
15. (c)
16. (b,c,d) 17. (b,d) 18. (a,c,d)
(a,b) 20. (b,d)
21. (a,b,c) 22. (b,d) 23. (b)
24. (b)
(d)
26. (d)
27. (d)
28. (a)
29. (4)
30. (5)
(3)
32. (9)
33. (6)
(A ) → ( p, q); (B ) → ( p, r ); (C ) → ( p, s); (D ) → ( p)
(A ) → (q, r , s); (B ) → ( r ); (C ) → (q, r , s); (D ) → ( p)
(a)
37. (a)
38. (a)
39. (b)
47. (b)
48. (c)
49. (c)
50. (a)
51. (b)
52. (b)
53. (a)
1.
7.
13.
19.
25.
31.
34.
35.
36.
3. Let a = 3x + 4y , b = 4x + 3y , c = 5x + 5y
Solutions
∴
a2 + b2 − c2
2ab
(3 x + 4y ) 2 + ( 4 x + 3y ) 2 − (5 x + 5y ) 2
=
2(3 x + 4y ) ( 4 x + 3y )
xy
=−
<0
(3 x + 4y ) ( 4 x + 3y )
cosC =
1. Since, the distance of all points from O( 0, 0) = 5
∴ O( 0, 0 ) be the circumcentre.
Centroid of triangle (G )
 3 + 5 cosθ + 5 sin θ ) 4 + 5 sin θ − 5 cosθ 
,
≡



3
3
Let H(α , β ) be the orthocentre of the triangle, then
α = 3 + 5 cosθ + 5 sin θ
⇒
(α − 3 ) = 5 cosθ + 5 sin θ
and
β = 4 + 5 sin θ − 5 cosθ
⇒
(β − 4 ) = 5 sin θ − 5 cosθ
On squaring and adding Eqs. (i) and (ii), we get
(α − 3 ) 2 + (β − 4 ) 2 = 25 + 25
⇒
α 2 + β 2 − 6α − 8β − 25 = 0
Hence, the locus of H(α , β ) is
x 2 + y 2 − 6 x − 8y − 25 = 0
2.
⇒
C > 90 °
Hence, triangle is obtuse angled.
4.
A
…(ii)
B
Q
AP = PQ = QB
Q
…(i)
P
1 
∴ mid-point of PQ = mid-point of AB =  , 3
2 
5. z A = 2 + 3i ; z B = − i and AB = ( 4 + 16) = 20 = 2 5
Y
A (2, 3)
C
Y
C(a, b)
5
X¢
B
2
2
X
O
(0, –1) B
60°
X¢
q
O
A
D
Y¢
X
∴
Y¢
Let
and
Let
∴
AB = AC = 2
∠BAO = θ
C ≡ (α , β )
α = 2 cosθ + 2 cos(180 ° − (60 ° + θ ))
= 2 cosθ − 2 cos(60 ° + θ )

1
3
= 2 cosθ − 2 cosθ × − sin θ ×

2
2 

= cosθ + 3 sin θ
and
∴
⇒
or
∴
Eliminating θ from Eqs. (i) and (ii), we get
α 2 + β 2 − αβ 3 − 1 = 0
Hence, the locus of H(α , β ) is
x 2 + y 2 − xy 3 − 1 = 0
zC − z A AC iπ / 2 i
=
e
=
z B − z A AB
2
i
zC − (2 + 3i ) = ( −i − 2 − 3i )
2
zC = 2 + 3i − i + 2 = 4 + 2i
C ≡ ( 4, 2 )
6. Let P ≡ (h, k ) and A ≡ (1, 0), B ≡ (2 cosθ, 2 sin θ )
…(i)
β = 2 sin(180 ° − (60 ° + θ ))
= 2 sin(60 ° + θ )

1
3
= 2 sin θ × + cosθ ×

2
2 

= sin θ + 3 cosθ
( AC ) = ( BC ) 2 − ( AC ) 2 = 5
…(ii)
PA : PB = 2 : 3
4 cosθ + 3
∴
h=
⇒ 5h − 3 = 4 cosθ
5
4 sinθ + 0
and
k=
⇒ 5k = 4 sinθ
5
OA squaring and adding Eqs. (i) and (ii), we get
(5h − 3 ) 2 + (5k ) 2 = 16
∴ Locus of C is (5 x − 3 ) 2 + (5y ) 2 = 16
or
25( x 2 + y 2 ) − 30 x − 7 = 0
which is a circle.
Q
…(i)
…(ii)
62
Textbook of Coordinate Geometry
2a 3a 1
1
7.
3b 2b 1 = 0
2
c c 1
T1 divides AC in the ratio 1 : 2 internally, then
 1 × 4 + 2 × 1 1 × 5 + 2 × 2
T1 ≡ 
,



1+2
1+2
or
T1 ≡ (2, 3 )
and T2 is the mid-point of T1 and C, then
 2 + 4 3 + 5
T2 ≡ 
,

 2
2 
⇒
or
2a(2b − c ) − 3a(3b − c ) + 1(3bc − 2bc ) = 0
− 5ab + ac + bc = 0
5ab
or
c=
a+ b
2c
2ab
or
=
5 a+b
2c
Hence, a, , b are in HP.
5
or
T2 ≡ (3, 4 )
S1 divides BC in the ratio 1 : 2 (internally), then
B
Y
8.
45° 45°
45°
C (–3, 0)
45°
X
A (3, 0)
O
Y¢
It is clear from the figure orthocentre ≡ ( 0, 3 )
C
(Q BS1 = S1S 2 = S 2C )
or
S1 ≡ ( 4, 3 )
and S 2 is the mid-point of S1 and C, then
 4 + 4 3 + 5
S2 ≡ 
,

 2
2 
or
S 2 = ( 4, 4 )
Area of quadrilateral T1S1S 2T2
2
3
4
3
=1 4
2
4
3
4
2
3
9. Let, A( 0, 0)
Coordinate of B are (2 cot θ, 2 ) and coordinate of C are
{ 4 cot (60 ° − θ ), − 4 }
y=x tanq
Y
y=2
2
6
S2
 1 × 4 + 2 × 4 1 × 5 + 2 × 2
,
S1 ≡ 



1+2
1+2
B (0, 3)
X¢
S1
A
(0,0)
q
60°–q
B
X
C
y=–4
1
(6 + 16 + 16 + 9 ) − (12 + 12 + 12 + 8 )
2
3
= sq units.
2
y=x tan (q – 60°)
∴
⇒
⇒
⇒
AB = AC
( AB ) 2 = ( AC ) 2
4 cot 2 θ + 4 = 16 cot 2 (60 ° − θ ) + 16
4cosec 2θ = 16cosec 2 (60 ° − θ )
sin(60 ° − θ ) = 2 sin θ
 3

1
cosθ − sin θ = 2 sin θ

2
 2

tanθ =
∴
11. Statement (i) :
Q Points ( −1, 0 ), ( 4, − 2 ) and (cos2θ, sin 2θ ) are collinear.
− 1 − cos2θ 0 − sin 2θ
=0
∴
4 − cos2θ − 2 − sin 2θ
⇒
3
5
⇒
or
or
∴ The required length
= AB = ( 4 cot 2 θ + 4 )
=2
10.
A
28 4 7
=
3
3
T1
Q A ≡ (1, 2 ), B ≡ ( 4, 2 ), C ≡ ( 4, 5 )
=
or
T2
C
(Q AT1 = T1T2 = T2C )
or
and
2 + sin 2 θ + 2 cos2θ + sin 2θ cos2θ
+ 4 sin 2θ − sin 2θ cos2θ = 0
5 sin 2θ + 2(1 + cos2θ ) = 0
10 sin θ cosθ + 4 cos2 θ = 0
2 cosθ (5 sin θ + 2 cosθ ) = 0
2
cosθ = 0 and tanθ = −
5
π
θ = mπ +
2
 −2 
θ = nπ + tan −1   ; m, n ∈ I
 5
Chap 01 Coordinate System and Coordinates
15. ( x + h ) 2 + (y + k ) 2 − 4( x + h ) + 6(y + k ) − 7 = 0
Statement (ii) :
 1 − tan θ 2 tan θ 
,
Q Points ( −1, 0 ), ( 4, − 2 ) and 
 are collinear.
 1 + tan 2 θ 1 + tan 2 θ 
2
1 − tan 2 θ
1 + tan 2 θ
1 − tan 2 θ
4−
1 + tan 2 θ
−1 −
∴
2 tan θ
1 + tan 2 θ
=0
2 tan 2 θ
−2 −
2
1 + tan θ
0−
⇒ 10 tan 3 θ + 4 tan 2 θ + 10 tan θ + 4 = 0
or
(2 tan 2 θ + 2 ) (5 tan θ + 2 ) = 0
or
tan 2 θ ≠ − 1
2
∴
tanθ = −
5
 −2 
⇒ θ = pπ + tan −1   , p ∈ I
 5
⇒ x 2 + y 2 + 2 x(h − 2 ) + 2y (k + 3 ) + (h 2 + k 2 − 4h + 6k − 7 ) = 0
According to question
h − 2 = 0, k + 3 = 0
∴
h = 2, k = − 3
hence,
(h, k ) ≡ (2, − 3 ).
16. If the remaining vertex is (h, k ), then
Case I
Y
∑x
X¢
∑x
(h , k )
(–6, –4)
Y¢
= integer
h − 2 = 3 − 6, k + 1 = 5 − 4
∴ Fourth vertex is ( −1, 0 ).
r
r=0
⇒
1 ⋅ (1 − x 2n )
(1 − x 2 )
= integer
1 ⋅ (1 − xn )
(1 − x )
⇒
1 + xn
= integer
1+x
Case II
(h, k)
13. The orthocentre of ∆BCH is the vertex A( −1, 0).
14. Given, x12 + y12 + x 22 + y 22 + x 32 + y 32 + x 42 + y 42
≤ 2 x1 x 3 + 2 x 2 x 4 + 2y 2y 3 + 2y1y 4
⇒ ( x1 − x 3 ) 2 + ( x 2 − x 4 ) 2 + (y 2 − y 3 ) 2 + (y1 − y 4 ) 2 ≤ 0
or ( x1 − x 3 ) 2 + ( x 2 − x 4 ) 2 + (y 2 − y 3 ) 2 + (y1 − y 4 ) 2 = 0
or
x1 = x 3 , x 2 = x 4 , y 2 = y 3 , y1 = y 4
x1 + x 2 x 3 + x 4
or
=
2
2
y1 + y 2 y 3 + y 4
and
=
2
2
or h = − 1, k = 0
Y
(–2, 1)
X¢
n must be 1, 3, 5, 7, 9, 11, ……
vertices are (1, 7), (3, 9), (5, 11)
∴
Here, ( AB ) 2 = 8, ( BC ) 2 = 8, (CA ) 2 = 32
triangle cannot be an equilateral.
∴
D(x4, y4)
X
O
2r
r=0
n −1
(3, 5)
(–2, 1)
n −1
12. Q
63
(3, 5)
X
O
(–6, –4)
Y¢
h + 3 = − 2 − 6, k + 5 = 1 − 4 or h = − 11, k = − 8
∴ Fourth vertex is ( −11, − 8 ).
Case III
Y
(–2, 1)
(3, 5)
O
X¢
(h , k )
X
(–6, –4)
Y¢
B(x2, y2)
h − 6 = − 2 + 3, k − 4 = 1 + 5
or
h = 7, k = 10
∴ Fourth vertex is (7, 10).
17. (PA )2 = (PB )2
A(x1, y1)
C(x3, y3)
Hence, AB and CD bisect each other
Therefore, ACBD is a parallelogram. Also,
( AB ) 2 = ( x1 − x 2 ) 2 + (y1 − y 2 ) 2 = ( x 3 − x 4 ) 2 + (y 3 − y 4 ) 2 = (CD ) 2
Thus, ACBD is a parallelogram and AB = CD.
Hence, it is a rectangle.
⇒ ( x − a − b ) 2 + (y − a + b ) 2 = ( x − a + b ) 2 + (y − a − b ) 2
⇒ ( x − a + b ) 2 − ( x − a − b ) 2 = (y − a + b ) 2 − (y − a − b ) 2
⇒
(2 x − 2a )(2b ) = (2y − 2a ) (2b )
or
x =y
and P is the mid-point of AB i.e., (a, a )
If P be (a, b ), then bx = ay
(Qa = b )
64
Textbook of Coordinate Geometry
Now, in right angled triangle ABG,
( BA ) 2 = ( AG ) 2 + ( BG ) 2
1
1
⇒
b 2 = (a 2 + 4b 2 ) + (a 2 + b 2 )
9
9
⇒
a 2 = 2b 2
a = ±b 2
∴
18. Let vertices of triangle ABC are
A ≡ ( x1 , y1 ), B ≡ ( x 2 , y 2 ) and C ≡ ( x 3 , y 3 )
where, x1 , x 2 , x 3 , y1 , y 2 , y 3 ∈ Q
a = BC = ( x 2 − x 3 ) + (y 2 − y 3 )
∴
2
2
= rational or irrational
similarly,
b = CA, c = AB
Then, incentre
 ax + bx 2 + cx 3 ay1 + by 2 + cy 3 
≡ 1
,

 a+b+c
a+b+c 
21. Q
( AB ) 2 = ( x − y ) 2 + (y − z ) 2
( BC ) 2 = (y − z ) 2 + (z − x ) 2
and (CA ) 2 = (z − x ) 2 + ( x − y ) 2
Case I If x = y , then
( AB ) 2 + (CA ) 2 = ( BC ) 2
Case II If y = z , then
( AB ) 2 + ( BC ) 2 = (CA ) 2
Case III If z = x, then
( BC ) 2 + (CA ) 2 = ( AB ) 2
∴Incentre has rational or irrational coordinates but centroid,
circumcentre and orthocentre have always rational coordinates.
19. Since, A ≡ (−4 − 1), B ≡ (−2, − 4), C ≡ (4, 0), D ≡ (2, 3)
D
C
22.
A
C (a, 2a+l)
B
∴
( AB ) = ( −4 + 2 ) + ( −1 + 4 ) = 13
( BC ) 2 = ( −2 − 4 ) 2 + ( −4 − 0 ) 2 = 52
(CD ) 2 = ( 4 − 2 ) 2 + ( 0 − 3 ) 2 = 13
and
( DA ) 2 = (2 + 4 ) 2 + (3 + 1 ) 2 = 52
Clearly,
AB = CD and BC = AD
∴ABCD is a parallelogram.
Again,
( AC ) 2 = ( −4 − 4 ) 2 + ( −1 − 0 ) 2 = 65
and
( BD ) 2 = ( −2 − 2 ) 2 + ( −4 − 3 ) 2 = 65
∴
AC = BD
Hence, ABCD is a rectangle.
2
20.
2
2
(a, l)
⇒
⇒
⇒
Q
⇒
and
⇒
A
(given)
Sol. (Q. Nos. 23 to 25)
E a, b
2 2
90°
B(x, y)
A ≡ (a, λ )
B ≡ (a, 2a + λ )
C ≡ ( x, y )
∆ABC = a 2
1
× 2a × | x − a | = a 2
2
| x − a| = a
x −a = ± a
x = 0, x = 2a
Let
then,
and
Area of
⇒
b
x–a
x=a
Y
A (0, b)
X¢
[from Eqs. (i) and (ii)]
AB = (1 + 3 ) 2 + (2 − 1 ) 2 = 17
1
1
AC = AB =
17
2
2
G
E
(0, 0)
B
D a, 0
2
C (a, 0)
(a 2 + b 2 )
2
2
BE = ×
3
3
2
1
( BG ) 2 = (a 2 + b 2 )
9
X
F
BG =
…(i)
C
a2

2
2
AG = AD = ×  + b 2 
 4

3
3
1
( AG ) = (a 2 + 4b 2 )
9
2
60°
D
60°
60°
…(ii)
A
B
Chap 01 Coordinate System and Coordinates
For coordinates of C :
zc − z A
AC i π / 2 i
=
=
e
2
z B − z A AB
i
⇒ zc − (1 + 2i ) = ( −3 + i − 1 − 2i )
2
i
1
= ( −4 − i ) = − 2i +
2
2
3
or
zc =
2
3 
C ≡  , 0
∴
2 
n
B 1, 1
Ö3
F
15°
X¢
Given,
and
1
i 3
+ 3 −i +
4
4

3

5
z D =  + 3 + i 1 +


4
4

=
5
3
D ≡  + 3, 1 +

4
4
23.
∴
1
0
1| =
7
1
−4 −
2
8S = 51
2
R = IO + OA + AI
= r cosec 15 ° + 2 + r cosec 15 °
= 2r cosec15° + 2
2 2
= 2(2 − 3 ) ⋅
+2
( 3 − 1)
= ( 3 − 1)2 ⋅
⇒

4
2
2
25. y-coordinate of the centroid of the square
BCEF =
= 2 + 4 (2 − 3 )
1
2
1
= × 2 × (2 − 3 ) = (2 − 3 )
2
28. Area of region R = × OA × r
2
7
2=−7
2
4
2 2
+2
( 3 − 1)
= 2 2( 3 − 1) + 2
51
8
2
 11
17
3
3

=  + 3 +  +
(8 + 3 )
 =
4

2
4
2
0−
O ≡ ( 0, 0 ) A ≡ (2, 0 )
 1 
B ≡ 1,


3
27. Perimeter of region
2

5
1
3 9
24. DE =  + 3 −  + 1 +
+ 
4
X
=2 − 3
 1 9
E ≡ ,− 
 2 2
2
A (2, 0)
Let
P ≡ ( x, y )
If I be the incentre of ∆OAB.
If in radius = r, then
ID = IE = IF = r, if P at I ,
then
d ( P , OA ) = d ( P , OB ) = d ( P , AB ) = r ,
But
d ( P , OA ) ≤ min. {d ( P , OB ), d ( P , AB )}
which is possible only P lies in the ∆OIA
∴ M be the peak of region R.
∴
M=I
7
zF = − 4 − i
2
7

F ≡  −4, −  , then

2
1
1 3
S= |
2 2
D
15°
= 1 × (2 − 3 )
9 
⇒ z F − ( −3 + i ) = − i  − i 
2 
∴
r
r
26. Length of perpendicular from M to OA = ID = r = OD tan15°
For coordinates of F :
zF − zB
= e −iπ / 2 = − i
zC − z B
⇒
O
E
Y¢
1 i 3 1

⇒ z D − (1 + 2i ) =  +
  − 2i 

2  2
2
∴
Sol. (Q. Nos. 26 to 28)
Y
For coordinates of D :
zD − zA
= eiπ / 3
zC − z A
or
65
29. We know that the area of the triangle formed by joining the
mid-points of any triangle is one fourth of that triangle, then
∆ 2 = 4 ∆1
∆2
∴
=4
∆1
30. Since, the distance of all points from O(0, 0) = 5
∴Circumradius ( R ) = 5
Hence, diameter of nine point circle = R = 5
66
Textbook of Coordinate Geometry
31. Let B ≡ (2 2, 0), C ≡ (0, 2 )
(B)
C (16, 12)
BC = (8 + 2 ) = 10
∴
A
20
11
O (l, 4)
2Ö2
2Ö2
A (0, 0)
B (5, 12)
13
(OA ) 2 = (OB ) 2 = (OC ) 2
⇒ λ + µ 2 = ( λ − 5 ) 2 + (µ − 12 ) 2 = ( λ − 16 ) 2 + (µ − 12 ) 2
⇒ ( λ − 5 ) 2 + (µ − 12 ) 2 = ( λ − 16 ) 2 + (µ − 12 ) 2
or
( λ − 5 ) = − ( λ − 16 )
21
∴
λ=
⇒ 2 λ = 21
2
(C) Incentre of ∆ABC is
 11 × 0 + 13 × 16 + 20 × 5 11 × 0 + 13 × 12 + 20 × 12
,




11 + 13 + 20
11 + 13 + 20
2
B
D
AD = ( AB ) − ( BD )
2
and
C
2
10
11

= 8 −  =

4
2
1
⋅ BC ⋅ AD
2
1
55
11
= ⋅ 10 ⋅
=
2
2
2
∆=
∴
⇒
≡ (7, 9 )
Here,
µ =9
(D) Excentre opposite to vertex B is
 11 × 0 + 13 × 16 − 20 × 5 11 × 0 + 13 × 12 − 20 × 12
,




11 + 13 − 20
11 + 13 − 20
[∆] = 3
32. Let A ≡ (3, 4), B ≡ (4, 3) and C ≡ (1, 2)
a = | BC | = (9 + 1 ) = 10
∴
b = | CA | = ( 4 + 4 ) = 2 2
Here,
∴
c = | AB | = (1 + 1 ) = 2
⇒
=
∴
33.
A
10 × 3 + 2 2 × 4 + 2 × 1
10 + 2 2 + 2
x=
35.
R
18 (b)
Q
1 
≡  , 3 = (a, b )
2 
1
b
a = , b =3 ⇒ =6
2
a
34. (A) Centroid of ∆ABC is
 0 + 5 + 16 0 + 12 + 12
,

 ≡ (7, 8 )


3
3
Here,
∴
λ = 7, µ = 8
λ + µ = 15
30 (a)
B
Q
AP = PQ = QB
Let R is the mid-point of PQ such that PR = RQ.
Now,
AR = AP + PR = QB + RQ = RB
⇒
AR = RB
∴ R bisects the line segment AB, then
 − 2 + 3 5 + 1
R≡
,

 2
2 
∴
C (–10, 26)
3 ( 10 + 3 2 )
=3
( 10 + 3 2 )
x2 = 9
P
≡ (27, − 21 )
λ = 27, µ = − 21
λ + µ =6
A
(–10, 8)
24 (c)
B
(14, 8)
− 10 8 1
1
∆ = | 14
8 1 | = 216
2
− 10 26 1
a + b + c 30 + 18 + 24
=
= 36
2
2
 − 10 + 14 − 10 8 + 8 + 26
,
G≡



3
3
s=
(given)
≡ ( −2, 14 )
 30 × − 10 + 24 × − 10 + 18 × 14 30 × 8 + 24 × 26 + 18 × 8
I ≡
,



30 + 24 + 18
30 + 24 + 18
≡ ( −4, 14 )
H = A ≡ ( −10, 8 )
O = mid-point of BC ≡ (2, 17 )
Chap 01 Coordinate System and Coordinates
∆ 216
=
=6
s
36
BC 30
(B) R =
=
= 15
2
2
(A) r =
− 4 14 1
1
(C) Area of ∆IGH = | − 2 14 1 | = 6
2
− 10 8 1
2 17 1
1
(D) Area of ∆OGI = | − 2 14 1 | = 3
2
− 4 14 1
comparing with ax 2 + bxy + 1 = 0.
1
3
a = ,b = −
∴
8
8
∴
b2 − a
9 1 1
=
− =
64 8 64
⇒ Statement I is true.
Hence, both statements are true and statement II is not correct
explanation of statement I.
40. Let AD be the internal bisector of angle A.
A (x1, y1)
36. The centroid of triangle ABC is G(1, 3). If D divided BC in the
ratio λ : 1, then
A/2 A/2
P
 3λ − 1 4λ + 3
 λ + 3 3λ + 4
D≡
,
,
, E ≡ 

λ +1 λ +1
λ + 1 λ + 1 
and
 −λ + 1 3λ + 2
F ≡
,

 λ +1 λ +1
AB = ( − 5 − 7 ) 2 + ( −2 − 6 ) 2 = 4 13
BC = (7 − 5 ) 2 + (6 + 4 ) 2 = 2 26
and
CA = (5 + 5 ) 2 + ( −4 + 2 ) 2 = 2 26
Q ( BC ) 2 + (CA ) 2 = ( AB ) 2 and | BC | = | CA |
⇒ ∆ABC is right angled isosceles triangle and right angle at C.
∴Circumcentre is mid-point of A, B ≡ (1, 2 ). Both statements
are true and statement II is correct explanation of statement I.
38. Let A ≡ (0, 0), M ≡ (h, k ), B ≡ (2h, 2k ) and let same ratio λ : 1
 2h λ 2kλ 
 2h λ 2kλ 
P ≡
,
,
, Q = 

 λ + 1 λ + 1
 λ − 1 λ − 1
2λ
(λ + 1)
2λ
AQ =
(λ − 1)
AP − AB λ − 1
=
=
AB − AQ λ + 1
AP =
Q
(h 2 + k 2 ), AB = 2 (h 2 + k 2 ),
∴
BD AB c
=
=
DC AC b
 cx + bx 2 cy 3 + by 2 
D≡ 3
,

 c+b
c+b 
Let P ≡ ( x, y ). According to question, P ( x, y ) lies on AD,
therefore P , A, D are collinear

x
y
1
 x
y1
1
1
= 0

∴
cx 3 + bx 2 cy 3 + by 2 1

 c+b
c+b


x
y
1 

1 
or
x1
y1
1 
= 0
(c + b )
cx 3 + bx 2 cy 3 + by 2 c + b
y 1
y 1  x
x
 x1 y1 1 +  x1 y1 1 = 0
or

 

cx 3 cy 3 c bx 2 by 2 b
 using the addition

 property of determinants
2
AP
AQ
39. Replacing x by (x − 4) and y by y + 1 is
x − 3 xy + 11 x − 12y + 36 = 0, then
( x − 4 ) 2 − 3( x − 4 ) (y + 1 ) + 11( x − 4 ) − 12(y + 1 ) + 36 = 0
⇒
x 2 − 3 xy + 8 = 0
x 2 3 xy
or
−
+1=0
8
8
C (x3, y3)
D
(h + k )
2
∴ AP , AB and AQ are in HP.
⇒ AM + MP , AM + MB and AM + MQ are in HP
∴ MP , MB and MQ are in GP.
Both statements are true and statement II is correct
explanation of statement I.
2
C (x2, y2)
Then,
∴The centroid of triangle DEF is (1, 3).
Both statements are true and statement II is correct
explanation of statement I.
37. Q
67
or
Hence,
x
c
x1
y
y1
x 3 y 3
x y
b
x1 y1
x 2 y 2
1  x y

1
 + b x1 y1
1 x 2 y 2
1  x y

1
 + c x1 y1
1 x 3 y 3
1
1
= 0
1
1
1
= 0
1
41. Let first term and common difference of corresponding AP are
A and D.
Since,
∴
⇒
Similarly,
pth term of HP = a
1
pth term of AP =
a
1
A + (p − 1) D =
a
1
A + (q − 1 ) D =
b
...(i)
...(ii)
68
Textbook of Coordinate Geometry
A + (r − 1 ) D =
and
1
c
...(iii)
Subtracting Eq. (ii) from Eq. (i),
1 1
−
a b
and subtracting Eq. (iii) from Eq. (ii),
1 1
(q − r ) D = −
b c
Dividing Eq. (iv) by Eq. (v),
1 1
−
p −q a b
(b − a )
bc
=
=
×
1
1
q −r
ba
(c − b )
−
b c
p − q bc − ac
⇒
=
q − r ac − ab
(p − q ) D =
Let A (bc , p ), B (ca , q ), C (ab , r )
q−p
Slope of AB =
ca − bc
r −q
and
Slope of BC =
ab − ca
q−p
=
ca − bc
…(iv)
(h, k)
A
Q
X´
...(vi)
42. Suppose the equation of the line L is ax + by + c = 0. Let the
coordinates of A, B, C be respectively ( x1 , y1 ) ; ( x 2 , y 2 ) ; ( x 3 , y 3 ).
Suppose P divides BC in the ratio λ : 1 , then coordinates of P
 λx + x 2 λy 3 + y 2 
are  3
,
 since P lies on L.
 λ+1
λ+1 
 λa − a 
 µh + a µk 
P ≡
, 0 ; Q ≡ 
,

 λ+1 
 µ + 1 µ + 1
and
 − aν + h
k 
R≡
,

 ν+1
ν + 1
Q
λa − a
λ+1
1 µh + a
2 µ+1
−aν + h
ν+1
C
⇒ a ( λx 3 + x 2 ) + b ( λy 3 + y 2 ) + c ( λ + 1 ) = 0
⇒
λ (ax 3 + by 3 + c ) + (ax 2 + by 2 + c ) = 0
λ BP
(ax 2 + by 2 + c )
∴
=
=−
1 CP
(ax 3 + by 3 + c )
CQ
(ax 3 + by 3 + c )
Similarly,
=−
QA
(ax1 + by1 + c )
AR
(ax1 + by1 + c )
and
=−
RB
(ax 2 + by 2 + c )
BP CQ AR
Hence,
⋅
⋅
= −1
PC QA RB
X
Now,
Since, points P ,Q, R are collinear.
 λx + x 2 
 λy + y 2 
a 3
 +b 3
 +c=0
 λ+1 
 λ+1 
C
(a, 0)
Case I : Two sides divided internally and one side divided
externally.
Hence, λ, µ are positive and ν is negative
P
⇒
P
Let
BC = 2a
∴
OB = OC = a
Then,
B ≡ ( −a, 0 )
and
C ≡ (a , 0 )
Let
A ≡ (h, k )
Let P , Q, R divides BC , CA and AB in the ratio λ : 1, µ : 1 and
ν:1
BP
CQ
AR
i.e.
= λ,
= µ,
=ν
PC
QA
RB
∴
B
O
B
(–a, 0)
Y´
Since,
Slope of AB = slope of BC
Hence, A, B, C are collinear.
R
R
Y
...(v)
[from Eq. (vi)]
A
Aliter :
Let middle point O of BC as the origin, BC as the X-axis and
yy ′ as Y-axis.
0
1
µk
1 =0
µ+1
k
1
ν+1
or
0 λ+1
λa − a
µh + a µk µ + 1 = 0
−aν + h k ν + 1
or
λa − a 0 λ + 1
k µh + a µ µ + 1 = 0
− aν + h 1 ν + 1
Applying R2 → R2 − µR3 , then
λa − a 0 λ + 1
k a + aµv 0 1 − µv = 0
−av + h 1 v + 1
or
−k
λa − a λ + 1
=0
a + aµv 1 − µv
Chap 01 Coordinate System and Coordinates
or
−ka
λ −1 λ + 1
=0
1 + µv 1 − µv
or − ka {( λ − 1 ) (1 − µv ) − ( λ + 1 ) (1 + µν )} = 0
or
− ka { − 2 λµν − 2 }
⇒
2ka ( λµν + 1 ) = 0
or
λµν = − 1
BP CQ AR
or
⋅
⋅
= −1
PC QA RB
(Q ka ≠ 0 )
Case II : All the three sides will be divided externally
R
Y
A
X´
B
O
C
P
X
Y´
Q
Here, λ, µ and ν are negative.
Now, in case I formula for internal division will be used then
we can show that
i.e.
λµν = − 1
BP CQ AR
⋅
⋅
= −1
PC QA RB
43. Since, given points are collinear, then
a3
a −1
1 b3
2 b −1
c3
c −1
or
where, ∆c be the determinant of cofactors of ∆.
Let,
1
b3 b2 − 3 b − 1 = 0
(a − 1 )(b − 1 )(c − 1 ) 3
c c2 − 3 c − 1
c2 − 3 c − 1
a 3 –3 a
a 3 a 2 a a 3 a 2 –1
3
2
3
2
b b b + b b –1 + b 3 –3 b
c 3 –3 c
c 3 c 2 –1
c3 c2 c
a 3 –3 –1
+ b 3 –3 –1 = 0
c 3 –3 –1
or
then,
1 a a2 1 a2 a3
1 a a3
2
3
2
−abc 1 b b + 1 b b −3 1 b b 3 + 0 = 0
1 c c2 1 c2 c3
1 c c3
A1
∆c = A2
A3
B1 C1
B2 C 2
B3 C 3
A1 = b2c 3 − b3c 2 , B1 = a 3c 2 − a 2c 3 , C1 = a 2b3 − a 3b2
A2 = b3c1 − b1c 3 , B2 = c 3a1 − c1a 3 , C 2 = a 3b1 − a1b3
A3 = b1c 2 − b2c1 , B3 = a 2c1 − a1c 2 , C 3 = a1b2 − a 2b1
Given lines are
a3 a2 − 3 a − 1
b3 b2 − 3 b − 1 = 0
c3
⇒
or − abc (a − b ) (b − c ) (c − a )
+ (ab + bc + ca ) (a − b ) (b − c ) (c − a )
− 3 (a + b + c ) (a − b ) (b − c ) (c − a ) = 0
⇒ − (a − b ) (b − c ) (c − a )
{abc − (ab + bc + ca ) + 3 (a + b + c )} = 0
∴ abc − (ab + bc + ca ) + 3 (a + b + c ) = 0
Q a,b, c are distinct



∴a ≠ b, b ≠ c, c ≠ a

∴(a − b ) (b − c ) (c − a ) ≠ 0 


a, b, c are distinct.
Aliter : Suppose the given points lie on the line
lx + my + n = 0
 t3 
 t 2 − 3
then,
l 
 +m
 +n=0
 t − 1
 t −1
where, t = a, b, c
...(i)
⇒
lt 3 + mt 2 + nt − (3m + n ) = 0
i.e. a, b, c are the roots of Eq. (i)
m
then,
a+b+c=−
l
n
ab + bc + ca =
l
(3m + n )
abc =
l
Now, abc − (bc + ca + ab ) + 3 (a + b + c )
(3m + n ) n 3m
=
− −
=0
l
l
l
Hence, abc − (bc + ca + ab ) + 3 (a + b + c ) = 0
44. Q ∆c = ∆2
a2 − 3
1
a −1
b2 − 3
1 =0
b −1
c2 − 3
1
c −1
a3 a2 − 3 a − 1
⇒
69
a1 x + b1y + c1 = 0
...(i)
a 2 x + b2y + c 2 = 0
...(ii)
…(iii)
a 3 x + b3y + c 3 = 0
Let P be the point of intersection of Eqs. (ii) and (iii), then
1
x
y
=
=
b2c 3 − b3c 2 c 2a 3 − c 3a 2 a 2b3 − a 3b2
x
y
1
i.e.
=
=
A1 B1 C1
∴
A B 
P ≡  1 , 1
 C1 C1 
Similarly, if Q and R be the points of intersections of Eqs. (iii)
and (i) and Eqs. (i) and (ii) respectively, then
A B 
A B 
Q ≡  2 , 2, R ≡  3 , 3
 C2 C2 
 C3 C3 
70
Textbook of Coordinate Geometry
A1
C1
1 A
∴ Area of ∆PQR = | 2
2 C2
A3
C3
B1
C1
B2
C2
B3
C3
= x12 (a cos2 α + h sin 2α + b sin 2 α )
+ x1y1 (2h cos2α − a sin 2α + b sin 2α )
+ y12 (a sin 2 α − h sin 2α + b cos2 α ) ...(i)
But, from question the expression ax 2 + 2hxy + by 2 transforms
into
…(ii)
a1 x12 + 2h1 x1y1 + b1y12
Therefore, the expressions (i) and (ii) are the same. Hence,
equating the coefficients, we get
1
1 |=
1
A1
1
A2
2 | C1C 2C 3 |
A3
B1 C1
B2 C 2
B3 C 3
2
a1 b1 c1
1
∆2
=
a 2 b2 c 2 =
2 | C1C 2C 3 |
2 | C1C 2C 3 |
a 3 b3 c 3
…(iv)
 x + x 2 y1 + y 2 
G1 ≡  1
,

 2
2 
or
a1 + b1 = a + b
Again,
a1 − b1 = a (cos α − sin α )
3
2
A3
1
G3
⇒
(a1 − b1 ) = (a − b ) cos2α + 2h sin 2α
(a1 − b1 ) 2 = (a − b ) 2 cos2 2α + 4h 2 sin 2 2α
…(vii)
+ 4h (a − b ) sin 2α cos2α
Again, from Eq. (iv)
2h1 = 2h cos2α − (a − b ) sin 2α
∴
4h12 = 4h 2 cos2 2α + (a − b ) 2 sin 2 2α
…(viii)
− 4h (a − b ) sin 2α cos2α
On adding Eqs. (vii) and (viii), we get
(a1 − b1 ) 2 + 4h12 = (a − b ) 2 + 4h 2
Again,
4a1b1 − 4h12 = (a1 + b1 ) 2 − (a1 − b1 ) 2 − 4h 2
= (a + b ) 2 − (a − b ) 2 − 4h 2 [from Eq. (vi)]
2
Hence,
a1b1 − h1 = ab − h 2
A2
G1
Since, G2 divided G1 A3 internally in the ratio 1 :2
  x1 + x 2 

y + y2
 + 1.y 3 
 + 1. x 3 2 ⋅  1
 2 ⋅ 



2
2

,
G2 ≡ 
∴
2+1
2+1






 x + x 2 + x 3 y1 + y 2 + y 3 
,
G2 ≡  1



3
3
47. Vertex of triangle is (1, 1) and mid-point of sides through this
vertex are ( −1,2 ) and (3, 2 )
Again G3 divides G2 A4 internally in the ratio 1 :3
  x1 + x 2 + x 3 

y + y2 + y3
 + 1. y 4 
 + 1. x 4 3 ⋅  1
 3 ⋅ 



3
3

,
∴
G3 ≡ 
3+1
3+1






A (1, 1)
(–1, 2)
 x + x 2 + x 3 + x 4 y1 + y 2 + y 3 + y 4 
,
G3 ≡  1



4
4
(3, 2)
B
Proceeding in this way, we can show that
 x + x 2 + ...+ xn y1 + y 2 + K + yn 
Gn − 1 ≡  1
,



n
n
or we can say that the coordinates of the final point are
x1 + x 2 + ...+ xn
y + y 2 + ...+ yn
and 1
n
n
2
+ 2h sin 2α + b (sin 2 α − cos2 α )
1
G2
...(vi)
2
A5
4
1
G4
…(v)
a1 + b1 = a (cos2 α + sin 2 α ) + b (sin 2 α + cos2 α )
A1
i.e.
2h1 = 2h cos2α − a sin 2α + b sin 2α
A1 ≡ ( x1 , y1 ) and A2 ≡ ( x 2 , y 2 )
A4
i.e.
…(iii)
b1 = a sin 2 α − h sin 2α + b cos2 α
On adding Eqs. (iii) and (v), we get
45. Since, G1 is the middle point of A1 A2 where
∴
a1 = a cos2 α + h sin 2α + b sin 2 α
C
⇒ Vertex B and C come out to be ( −3, 3 ) and (5, 3 )
1 −3 + 5 1 + 3 + 5
 7
,
⇒ 1, 
∴ Centroid is
 3
3
3
48.
Y
P (3, 4)
46. Let the axes be rotated through an angle α, since ( x , y )
be the coordinates with respect to old axes and ( x 1 , y 1 )
be the coordinates with respect to new axes, then
x = x1 cosα − y1 sin α
and
y = x1 sin α + y1 cosα
Now, the expression ax 2 + 2hxy + by 2 becomes
a ( x1 cosα − y1 sin α ) 2 + 2h ( x1 cosα − y1 sin α )
( x1 sin α + y1 cosα ) + b( x1 sin α + y1 cosα ) 2
R
X¢
O
Q(6, 0)
Y¢
X
Chap 01 Coordinate System and Coordinates
Q Area of ( ∆OPR ) = Area of ( ∆PQR ) = Area of ( ∆OQR )
∴ By geometry R should be the centroid of∆OPQ
 0 + 3 + 6 0 + 4 + 0  4
∴
,
R≡
 ≡ 3, 
  3

3
3
⇒ x-coordinate of incentre
2 × 0 + 20 2 × 0 + 2 × 2
=
2+2+2 2
2
=
=2 − 2
2+ 2
49. Given, the vertices of a right angled triangle are A(1, k ), B (1, 1)
and C(2, 1 ) and area of ∆ABC = 1 sq unit
71
52.
C
(0, 41)
Y
Y
B
A (1, k)
B (1, 1)
X¢
C (2, 1)
X
O
X¢
O
A(41, 0)
X
Y¢
Y¢
We know that, area of right angled triangle
1
1
= × BC × AB = (1 ) (k − 1 )
2
2
1
= k − 1 ⇒ ± (k − 1 ) = 2
2
⇒
k = − 1, 3
∴
K = { −1, 3 }
50. Let P ≡ (1, 0) and Q ≡ (− 1, 0)
Given that,
AP BP CP 1
=
=
=
AQ BQ CQ 3
⇒
Let
3AP = AQ or 9 ( AP ) 2 = ( AQ ) 2
A ≡ ( x, y ), then
a(( x − 1 ) 2 + (y − 0 ) 2 ) = ( x + 1 ) 2 + (y − 0 ) 2
⇒ 8 x 2 + 8y 2 − 20 x + 8 = 0
5
x2 + y 2 − x + 1 = 0
⇒
2
Circumcentre of ∆ABC = Centre of circle Eq. (i)
5 
=  , 0
4 
53.
…(i)
⇒
⇒
or
∴
and
(Q BC = a, CA = b, AB = c )
Y
C (0, 2)
k = 2, k ≠ −
→
(1, 1)
(1, 0)
A (2, 0)
Now, x-coordinate of incentre is given as
ax1 + bx 2 + cx 3
a+b+c
X
(Q D < 0)
23
(Qk ∈ I )
5
∴Vertices an A ≡ (2, − 6 ), B ≡ (5, 2 ), C ≡ ( −2, 2 ).
Denote the points are ( x1 , y1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) from the
 x − x 3 y1 − y 3  4 −8
matrix P =  1

= 
x 2 − x 3 y 2 − y 3  7 0 
∴
B
(0, 0)
+ 13k + 10 = ± 56
+ 13k − 46 = 0
+ 13k + 66 = 0
+ 13k − 46 = 0
+ 13k + 66 ≠ 0
23
k = 2, −
5
∴
a = 2, b = 2 2 , c = 2
(0, 1)
5k 2
5k 2
5k 2
5k 2
5k 2
⇒
51. From the figure, we have
x1 = 0, x 2 = 0, x 3 = 2
Total number of integral points inside the square
OABC = 40 × 40 = 1600.
Number of integral points on AC = Number of integral points
on OB = 40 (namely (1, 1), (2, 2), (3, 3 ), … , (40, 40)
Hence, number of integral points inside the
1600 − 40
∆OAC =
= 780
2
Aliter : x > 0, y > 0 x + y < 41 or 0 < x + y < 41
∴ Number of integral points inside the ∆OAC = 40C 2 = 780.
k −3k 1
1
| 5
k 1 | = 28
2
−k
2 1
λ=
→
R 1 ⋅ R 2 28 1
= =
56 2
|P |
1
1


 7 + × − 8 − 4 − × − 3
2
2
∴Circumcentre of triangle is 
,

2
2




 3 5
 5 2
or  , −  and centroid is  , − 
 2 4
 3 3
then, orthocentre
3 2
5
5
 1
=  × 3 − 2 × , − × 3 + 2 ×  or 2,  ,
3
 2
2 3
4
CHAPTER
02
The Straight Lines
Learning Part
Session 1
●
●
●
●
●
Definition
Slope or Gradient of a Line
Lines Parallel to Coordinate Axes
Different Forms of the Equation of a Straight Line
The Distance Form or Symmetric Form or Parametric
Form of a Line
●
●
●
●
Angle of Inclination of a Line
Angle Between Two Lines
Intercepts of a Line on Axes
Reduction of General Equation to Standard Form
Session 2
●
●
●
Position of Two Points Relative to a Given Line
Equations of Lines Parallel and Perpendicular to
a Given Line
Distance Between Two Parallel Lines
●
●
Position of a Point which Lies Inside a Triangle
Distance of a Point From a Line
●
Area of Parallelogram
●
Concurrent Lines
How to Find Circumcentre and Orthocentre by Slopes
Session 3
●
●
Points of Intersection of Two Lines
Family of Lines
●
Session 4
●
●
●
Equations of Straight Lines Passing Through a Given
Point and Making a Given Angle with a Given Line
Equation of the Bisectors
Equation of that Bisector of the Angle Between
Two Lines Which Contains a Given Point
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A Line Equally Inclined with Two Lines
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Bisector of the Angle Containing The Origin
How to Distinguish the Acute (Internal) and Obtuse
(External) Angle Bisectors
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Session 5
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The Foot of Perpendicular Drawn from the
Point ( x1 , y 1 ) to the Line ax + by + c = 0
Image or Reflection of a Point ( x1 , y 1 ) in Different Cases
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Image or Reflection of a Point ( x1 , y 1 ) about a Line
Mirror
Use of Image or Reflection
Session 6
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● Refraction of Light
Reflection of Light
Condition of Collineirty If Three Given Points are in Cyclic Order
Practice Part
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JEE Type Examples
Chapter Exercises
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Exercises with the
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Session 1
Definition, Angle of Inclination of a Line, Slope or
Gradient of a Line, Angle Between Two Lines,
Lines Parallel to Coordinate Axes, Intercepts of a
Line on Axes, Different Forms of the Equation of
A Straight Line, Reduction of General Equation to
Standard Form, The Distance Form or Symmetric
Form or Parametric Form of a Line
X¢
Definition
A straight line defined as the curve which is such that the
line segment joining any two points on it lies wholly on it.
Theorem : Show that the general equation of the first
degree in x , y represents a straight line.
Proof : The general equation of the first degree is
...(i)
ax + by + c = 0
Let P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) be the coordinates of any two
points on the curve given by Eq. (i), then
...(ii)
ax 1 + by 1 + c = 0
...(iii)
ax 2 + by 2 + c = 0
Multiplying Eq. (iii) by λ and adding to Eq. (ii), we have
a ( x 1 + λx 2 ) + b (y 1 + λy 2 ) + c (1 + λ ) = 0
or
 x + λx 2 
 y 1 + λy 2 
a 1
 +b 
 + c = 0 ( λ ≠ − 1)
 1+ λ 
 1+ λ 
This relation shows that the point
 x 1 + λx 2 y 1 + λy 2 
,

 lies on Eq. (i).
1+ λ 
 1+ λ
But from previous chapter we know that this point divides
the join of P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) is the ratio λ : 1.
Since, λ can have any value, so each point on the line PQ
lies on Eq. (i) i.e. the line wholly lies on Eq. (i). Hence, by
the definition of the straight line as given above we
conclude that Eq. (i) represents a straight line.
Hence, the general equation of first degree in x , y viz
ax + by + c = 0 represents a straight line.
Remarks
1. The number of arbitrary constants in the equation of a
straight line is two (we observe three constants a, b and c in the
equation ax + by + c = 0 of a straight line. The given equation
a
b
of line can be rewritten as   x +   y + 1= 0 or
 c
 c
a
b
px + qy + 1 = 0 where p = and q = .
c
c
Thus, we have only two arbitrary constants p and q in the
equation of a straight line.
Hence, to completely determine the equation of a straight line,
we require two conditions to determine the two unknowns in
general.
2. A straight line is briefly written as a ‘line.’
3. The equation of a straight line is the relation between x and y,
which is satisfied by the coordinates of each and every point
on the line.
Angle of Inclination of a Line
The angle of inclination of a line is the measure of the
angle between the X-axis and the line measured in the
anticlockwise direction.
Y
B
150°
X'
30°
O
A
Y'
Here, angle of inclination of line AB = 150 °.
X
Chap 02 The Straight Lines
Y
1. When two lines are parallel, they have the same inclination.
2. The inclination of a line which is parallel to X-axis or coinciding
with X-axis is 0°.
3. The angle of inclination of the line lies between 0° and 180° i.e.
π
0 < θ ≤ π and θ ≠ .
2
Y
135°
O
Remarks
X′
45°
45°
X
O
Y′
Slope or Gradient of a Line
X
45°
45°
X′
75
Y′
Theorem : If P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) are two points on a
line l, then the slope m of the line l is given by
y − y1
m= 2
,x1 ≠ x2
x2 − x1
If inclination of a line is (θ ≠ 90 ° ), then tan θ is called the
slope or gradient of the line. It is usually denoted by m.
θ is positive or negative according as it is measured in
anticlockwise or clockwise direction.
If x 1 = x 2 , then m is not defined. In that case the line is
perpendicular to X-axis.
Y
B
Y
(x2,y2) Q
π+ θ
θ
X′
O
X
A
Y′
i.e. Slope of AB = m of AB =m ( AB )
= tan θ or tan [ − ( π − θ )]
= tan ( π + θ )
= slope of BA = m of BA
=m( BA )
∴
m( AB ) = m( BA )
Hence, we do not take into consideration the direction of a
line segment while talking of its slope.
Remarks
1. Slope of a line is not the angle but is the tangent of the
inclination of the line.
2. If a line is parallel to X-axis, then its slope = tan0 ° = 0.
3. Slope of a line parallel to Y-axis or perpendicular to X-axis is not
defined. Whenever we say that the slope of a line is not defined.
4. If a line is equally inclined to the axes, then it will make an angle
of 45° or 135° with the positive direction of X-axis. Slope in this
case will be tan 45° or tan135°. i.e. ± 1
.
Y
45°
45°
45°
Y′
L
x1
M
X
x2
Y′
Proof : Given P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) are two points on
a line l, let line l makes an angle θ with positive direction
of X-axis. Draw PL, QM perpendiculars on X-axis and
PN ⊥ QM
Then,
PN = LM = OM − OL = x 2 − x 1
and
QN = QM − NM = QM − PL
= y2 − y1
Also,
∠QAM = ∠QPN = θ
Now, in ∆QPN
QN y 2 − y 1
Difference of ordinates
tan θ =
=
=
PN x 2 − x 1 Difference of abscissaes
m=
y2 − y1
x2 − x1
If x 1 = x 2 , then tan θ = ∞ or θ =
X
X′
y2
π
i.e. m is not defined or
2
the line is perpendicular to X-axis.
45°
O
O
N
y1
θ
A
X′
or
Y
X′
θ
(x1,y1) P
π– θ
45°
Remarks
135°
X
O
Y′
1. When the two lines are parallel, then their slopes are equal
i.e. m1 = m2.
2. If three points A, B, C are collinear, then slope of AB = slope of
BC = slope of AC.
76
Textbook of Coordinate Geometry
y Example 1. Find the inclination of the line whose
1
.
slope is −
3
y Example 6. For what value of k the points
(k , 2 − 2k ), ( − k + 1, 2k ) and ( − 4 − k , 6 − 2k ) are
collinear ?
Sol. Let α be the inclination of a line then its slope = tanα
Sol. Let A ≡ (k , 2 − 2k ), B ≡ ( −k + 1, 2k ) and C ≡ ( − 4 − k , 6 − 2k )
are collinear, then
Slope of AB = Slope of AC
2k − (2 − 2k ) 6 − 2k − (2 − 2k )
⇒
=
− 4 −k −k
−k +1−k
∴
tan α = −
1
3
= − tan 30 °
= tan (180 ° − 30 ° ) = tan 150 °
α = 150 °
⇒
⇒
y Example 2. Find the slope of the line through the
points (4, − 6 ), ( −2 , − 5) .
Sol. Slope of the line m =
1
− 5 − ( − 6)
=−
6
−2 − ( 4 )
∴
y Example 3. Determine λ , so that 2 is the slope of
the line through (2 , 5) and ( λ , 3) .
Sol. Slope of the line joining (2, 5) and ( λ , 3)
=
⇒
⇒
∴
⇒
⇒
⇒
3−5
−2
=2
=
λ −2 λ −2
(given)
− 2 = 2λ − 4
2λ = 2
λ =1
y Example 4. Show that the line joining the points
(2 , − 3) and ( − 5 , 1) is parallel to the line joining (7, − 1)
and (0, 3) .
4k − 2
4
=
− 2k + 1 − 4 − 2k
( 4k − 2) ( − 4 − 2k ) = 4 ( − 2k + 1)
(2k − 1) ( − 2 − k ) − ( − 2k + 1) = 0
(2k − 1) ( − 2 − k + 1) = 0
1
k ≠ , ∴ k = −1
2
Angle Between Two Lines
Theorem : The acute angle θ between the lines having
slopes m 1 and m 2 is given by
m − m 2 

θ = tan −1  1
1 + m 1m 2
Proof : Let l 1 and l 2 be two non-perpendicular lines,
neither of which is parallel to the Y-axis.
Y
Sol. Slope of the line joining the points (2, − 3) and ( − 5, 1) is
and slope of the line joining the points (7, − 1) and (0, 3) is
3 − ( − 1) − 4
=
m2 =
0−7
7
y Example 5. Find whether the points
( − a, − b ), [ − (s + 1) a, − (s + 1) b ] and [(t − 1) a, (t − 1) b ]
are collinear ?
Sol. Let A ≡ ( − a, − b ), B ≡ [ − (s + 1) a, − (s + 1) b ] and
C ≡ ((t − 1) a, (t − 1) b )
− ( s + 1) b + b b
Then, slope of AB =
=
− ( s + 1) a + a a
and
( t − 1) b + ( s + 1) b b
slope of BC =
=
( t − 1) a + ( s + 1) a a
Hence, given points are collinear.
π−θ
l2
l1
θ
1 − ( − 3)
4
m1 =
=−
− 5 −2
7
Here,
m1 = m 2
Hence, lines are parallel.
1


k ≠ (QDenominator ≠ 0)


2
X′
θ1
O
θ2
X
Y′
Let m 1 and m 2 be the slopes of two given lines l 1 and l 2
respectively. Let θ 1 and θ 2 be the inclinations of these
lines.
∴
m 1 = tan θ 1 and m 2 = tan θ 2
π

Let θ and π − θ be the angles between the lines  θ ≠  .

2
Then,
θ 2 = θ + θ 1 or θ = θ 2 − θ 1
⇒
tan θ = tan (θ 2 − θ 1 )
⇒
 tan θ 2 − tan θ 1   m 2 − m 1 
tan θ = 
=

 1 + tan θ 2 tan θ 1   1 + m 1m 2 
 m − m1 
Also, tan( π − θ ) = − tan θ = −  2

 1 + m 2m 1 
...(i)
...(ii)
77
Chap 02 The Straight Lines
From Eqs. (i) and (ii) the angle between two lines of slopes
m 1 and m 2 is given by
 m − m2 
tan θ = ±  1

 1 + m 1m 2 
  m − m2  
θ = tan −1  ±  1

  1 + m 1m 2  
⇒
Hence, the acute angle between the lines l 1 and l 2 is given
by
m − m 2 
.
θ = tan −1 1
1 + m 1m 2
Corollary 1 : If two lines, whose slopes are m 1 and m 2 are
parallel,
iff
θ = 0 ° (or π) ⇔ tan θ = 0
⇔ m1 = m2
Thus, when two lines are parallel, their slopes are equal.
Corollary 2 : If two lines, whose slopes are m 1 and m 2 are
perpendicular,
π
π
iff
θ =  or −  ⇔ cot θ = 0
2 
2
⇔ m 1m 2 = − 1
Thus, when two lines are perpendicular, the product of
their slopes is −1. The slope of each is the negative
reciprocal of the slope of other i.e. if m is the slope of a
1
line, then the slope of a line perpendicular to it is − .
m
y Example 7. Find the angle between the lines joining
the points (0, 0), (2 , 3) and (2 , − 2), ( 3 , 5).
Sol. Let the given points be A ≡ (0, 0), B ≡ (2, 3), C ≡ (2, − 2) and
D ≡ (3, 5). Let m1 and m 2 be the slopes of the lines AB and
CD respectively.
3−0 3
5 − ( − 2)
∴
m1 =
= and m 2 =
=7
2−0 2
3−2
y Example 8. The angle between two lines is
1
slope of one of them is . Find the slope of the other
2
line.
Sol. If θ be the acute angle between the lines with slopes m1 and
m 2 , then
m − m 2 

tanθ = 1
1 + m1m 2
∴
− 11 11
=  =
 23  23
∴
 11
θ = tan −1  
 23
θ=
Let
π
tan
4
then
⇒
π
4
and m1 =
1
2
1 −m 
2

= 2

 1
1 + ⋅m2

 2
1 − 2m 2

1 =
2 + m2 
⇒
1 − 2m 2
= ±1
2 + m2
Taking positive sign then,
1 − 2m 2 = 2 + m 2
1
∴
m2 = −
3
and taking negative sign then,
1 − 2m 2 = − 2 − m 2
∴
m2 = 3
Hence, the slope of the other line is either −
1
or 3.
3
y Example 9. Without using pythagoras theorem, show
that the points A ( − 1, 3), B (0, 5) and C ( 3 , 1) are the
vertices of a right angled triangle.
Sol. In ∆ ABC, we have
Let θ be the acute angle between the lines
 3

 2 −7 
m
m
−


1
2
=
tanθ =


1 + m1m 2 1 +  3 ⋅ 7
 
  2 
π
and the
4
and
Slope of side AB =
5−3
= 2 = m1
0 − ( − 1)
(say)
Slope of side BC =
1−5
4
= − = m2
3−0
3
(say)
Slope of side CA =
3−1
1
= − = m3
2
−1−3
(say)
Clearly,
 1
m1m 3 = 2 ×  −  = − 1
 2
∴ AB and CA are perpendicular to each other i.e.
∠ BAC = 90°
Hence, the given points are the vertices of a right angled
triangle.
78
Textbook of Coordinate Geometry
y Example 10. A line passes through the points
A (2 , − 3) and B (6, 3) . Find the slopes of the lines
which are
(i) parallel to AB
Y
N
(ii) perpendicular to AB
a
3 − ( − 3) 6 3
= =
6−2
4 2
X′
(ii) The slope of a line perpendicular to AB is
1
1
2
=− =−
3
3
m
2
(say)
C (x3,y3)
P(x,y)
N
X′
a
–a
O
X
Y′
Here, |a| = –a
Let
P ( x , y ) be any point on the line l, then
x = a is the required equation.
(Here, | a | = a )
Remarks
1. In particular equation of Y-axis is x = 0
(Qa=0 )
2. A line is parallel to Y-axis, at a distance from it and is on the
negative side of Y-axis, then its equation is x = − a .
(ii) Equation of a line parallel to X-axis : Let l be a
straight line parallel to X-axis and at a distance b from it, b
being the directed distance of the line from the X-axis.
Therefore, the line lies above the X-axis, if b > 0 and if
b < 0, then the line would lie below the X-axis.
B (x2,y2)
Y
Here, m1 and m 2 are rational numbers
(Q x 1, x 2 , x 3 , y1, y 2 , y 3 are integers)
m − m2
tan ( ∠BAC ) = 1
∴
1 + m1m 2
But
Y
x= |a|= –a
Sol. Let ABC be a triangle whose vertices are A ( x 1, y1 ), B ( x 2 , y 2 )
and C ( x 3 , y 3 ). Assume x 1, x 2 , x 3 , y1, y 2 , y 3 all are integers.
Let ∠BAC = 60°
y − y1
Slope of AC = 3
(say)
= m1
x 3 − x1
60°
l
a<0
y Example 11. Show that the triangle which has one of
the angles as 60°, can not have all vertices with
integral coordinates.
(x1,y1) A
X
a
Y′
Here, |a| = a
3
m1 = m =
2
y − y1
Slope of AB = 2
= m2
x 2 − x1
x= |a|= a
O
(i) Let m1 be the slope of a line parallel to AB, then
and
P(x,y)
a >0
Sol. Let m be the slope of AB. Then m =
−
l
= Rational (Qm1 and m 2 are rational)
tan ( ∠BAC ) = tan 60° = 3 = Irrational
b
(i) Equation of a line parallel to Y-axis : Let l be a
straight line parallel to Y-axis and at a distance a from it, a
being the directed distance of the line from the Y-axis .
Therefore, the line lies on the right of Y-axis if a > 0 and if
a < 0, then the line would lies on the left of Y-axis.
y=|b|=b
P(x,y)
l
b
X′
M
O
Q Rational number ≠ Irrational number
Which is contradiction so our assumption that the vertices
are integers is wrong. Hence, the triangle having one angle
of 60° can not have all vertices with integral coordinates.
Lines Parallel to Coordinate Axes
b>0
Y′
Y
Here, |b|= b
b<0
M
X¢
X
O
X
b
–b
y=|b|= –b P(x,y)
Y¢
Here, |b|= –b
l
79
Chap 02 The Straight Lines
Let P ( x , y ) be any point on the line l, then y = b is the
required equation (Here, | b | = b ).
Sol. Since, the given (both) lines are parallel to Y-axis and the
required line is equidistant from these lines, so it is also
parallel to Y-axis. Let equation of any line parallel to Y-axis is
x =a
 7 15
− + 
 2 2 8
Here, a =
= = 2 units
2
4
Hence, its equation is x = 2 .
Remarks
1. In particular equation of X-axis is y = 0
(Qb = 0 )
2. A line parallel to X-axis at a distance b from it and is on the
negative side of X-axis, then its equation is y = − b.
Sol. (i) Equation of straight line parallel to Y-axis at a distance a
units to the right is x = a .
∴ Required equation is x = 3
(ii) Equation of straight line parallel to Y-axis at a distance a
units to the left is x = − a.
∴ Required equation is x = − 2.
y Example 13. Find the equation of the straight line
parallel to X-axis and at a distance
(i) 5 units above the X-axis
(ii) 9 units below the X-axis.
Sol. (i) Equation of a straight line parallel to X -axis at a distance
b units above the X -axis is y = b.
∴ Required equation is y = 5
(ii) Equation of a straight line parallel to X -axis at a
distance b units below the X -axis is y = − b.
∴ Required equation is y = − 9
y Example 14. Find the equation of the straight line
which passes through the point (2 , − 3) and is
(i) parallel to the X-axis
(ii) perpendicular to the X-axis
Intercepts of a Line on Axes
If a line cuts X-axis at A (a , 0 ) and the Y-axis at B (0, b )
then OA and OB are known as the intercepts of the line on
X-axis and Y-axis respectively. | a | is called the length of
intercept of the line on X-axis. Intercept of a line on X-axis
may be positive or negative and | b| is called the length of
intercept of the line on Y-axis. Intercept of a line on Y-axis
may be positive or negative.
Y
B(0,b)
y intercept
y Example 12. Find the equation of the straight line
parallel to Y-axis and at a distance (i) 3 units to the
right (ii) 2 units to the left.
X′
O
Y′
If a line parallel to Y-axis, then its intercept on Y-axis is not
defined and if a line parallel to X-axis, then its intercept on
X-axis is not defined.
Intercepts in II quadrant
x =a
....(i)
Y
B (0,b)
b
X′
a
A
(–a,0)
b
X
O
X′
A(–a,0)
a
O
b
B (0,–b)
Y′
Intercept on X-axis = –a, length of
intercept on X-axis = |a|
Intercept on Y-axis = –b, length of
intercept on Y-axis = |b|
a
O
X
Y′
Intercept on X-axis = a, length of
intercept on X-axis = |a|
Intercept on Y-axis = b, length of
intercept on Y-axis = |b|
Intercepts in III quadrant
Y
...(ii)
A(a,0)
X′
Y′
Intercept on X-axis = –a, length of
intercept on X-axis = |a|
Intercept on Y-axis = b, length of
intercept on Y-axis = |b|
Since, it passes through the point (2, − 3) putting x = 2 in
Eq. (ii)
Then,
2=a ⇒ a=2
Hence, required equation of the line x = 2 .
y Example 15. Find the equation of a line which is
7
15
equidistant from the lines x = − and x = .
2
2
Intercepts in I quadrant
Y
B (0,b)
Since, it passes through the point (2, − 3) .
Putting y = − 3 in Eq. (i), then
b= −3
Hence, required equation of the line is y = − 3 .
Eq. (i) Let equation of any line perpendicular to X -axis
= Equation of any line parallel to Y-axis is
X
A(a,0)
Remark
Sol. (i) Let equation of any line parallel to X -axis is
y =b
x intercept
X
Intercepts in IV quadrant
Y
X′
a
O
A(a,0)
X
b
B (0,–b)
Y′
Intercept on X-axis = a, length of
intercept on X-axis = |a|
Intercept on Y-axis = –b, length of
intercept on Y-axis = |b|
80
Textbook of Coordinate Geometry
Different Forms of the Equation of a
Straight Line
(i) Slope-Intercept Form
Theorem : The equation of the straight line whose slope
is m and which cuts an intercept c on the Y-axis is
y = mx + c
Y
B
P (x,y)
θ
Q
θ
X′
R
M
c
O
L
X
A
Y′
Proof : Let AB be a line whose slope is m and which cuts
an intercept c on Y-axis. Let P ( x , y ) be any point on the
line. Draw PL ⊥ to X-axis and QM ⊥ to PL.
Then, from figure,
∠ PRL = ∠ PQM = θ, OQ = c
∴
PM = PL − ML = PL − OQ = y − c
and
QM = OL = x
PM
Now in ∆PQM, tan θ =
QM
y −c
m=
⇒ y = mx + c
⇒
x
which is the required equation of the line.
Remarks
1. If the line passes through the origin, then c = 0 (Q 0 = m.0 + c
⇒ c = 0) and hence equation of the line will become y = mx.
2. Equation of any line may be taken as y = mx + c.
3. If the line is parallel to X-axis, then θ = 0°
i.e. m = tan0 ° = 0. Hence, equation of the line parallel to X-axis
is y = c.
y Example 17. What are the inclination to the X-axis
and intercept on Y -axis of the line
3y = 3 x + 6 ?
Sol. The given equation can be written as
x
y=
+2
3
Now, comparing Eq. (i) with y = mx + c , then we get
1
m=
3
Let θ be the inclination to the X -axis, then
tan θ = tan 30°
∴
θ = 30° and c = 2.
y Example 18. Find the equation of the straight line
cutting off an intercept of 3 units on negative direction
 3
of Y -axis and inclined at an angle tan −1   to the
 5
axis of x.
Sol. Here,
 3
c = − 3 and θ = tan −1  
 5
3
=m
5
Hence, the equation of the line
y = mx + c
3
i.e.
y = x −3
5
or
3x − 5y − 15 = 0
or
tanθ =
y Example 19. Find the equation to the straight line
cutting off an intercept of 5 units on negative
direction of Y -axis and being equally inclined to the
axes.
Sol. Here, c = − 5
m = tan 45° or tan135°
i.e.
m = ±1
Y
135°
y Example 16. If the straight line y = mx + c passes
through the points (2 , 4 ) and ( −3 , 6 ) , find the values of
m and c .
45°
X′
O
45°
2
24
On solving Eqs. (i) and (ii), we get m = − , c =
5
5
45°
45° 45°
Sol. Since, (2, 4 ) lies on y = mx + c
∴
4 = 2m + c
Again, ( −3, 6) lies on y = mx + c
∴
6 = − 3m + c
(0,–5)
...(i)
Y′
...(ii)
...(i)
Hence, required equation is
y = ( ± 1) x − 5
or
y = ± x −5
X
81
Chap 02 The Straight Lines
y Example 20. Find the equations of the bisectors of
the angle between the coordinate axes.
Sol. Let L1 and L 2 be the straight lines bisecting the co-ordinate
axes.
Both L1 and L 2 pass through origin
∴ Equation of line through origin is y = mx
for L1, m = tan 45° = 1
Y
L1
13
5°
L2
X′
45°
X
O
∴ Equation of line L1 is y = x
i.e.
x −y =0
For L 2 , m = tan 135° = − 1
∴ Equation of line L 2 is y = − x
i.e.
x +y =0
Hence, equations of the bisectors of the angle between the
coordinate axes are x ± y = 0.
Theorem : The equation of the straight line which
passes through the point ( x 1 , y 1 ) and has the slope ‘m’ is
y − y 1 = m (x − x 1 )
Proof : Let AB be a straight line whose slope is m and
which pass through the point Q ( x 1 , y 1 ). Let the line AB
cuts X-axis at R and ∠BRX = θ, then
tan θ =m
B
Y
P(x,y)
N
θ
A
R
O
M
L
X
Y′
Let P ( x , y ) be any point on the line AB. Draws PL and QM
perpendiculars from P and Q on X-axis respectively. Also
draw QN perpendicular from Q on PL, then from figure
∠PRL = ∠PQN = θ, OL = x , OM = x 1 , PL = y , QM = y 1
Then, QN = ML = OL − OM = x − x 1
and
PN = PL − NL = PL − QM = y − y 1
m=
y − y1
x − x1
or
y − y 1 = m (x − x 1 )
which is the required equation of the line.
Aliter : Let the equation of the required line be
...(i)
y = mx + c
where, m is the slope of the line.
Since line Eq. (i) passes through the point ( x 1 , y 1 ), therefore
...(ii)
y − y 1 = m (x − x 1 )
which is the required equation of the line.
Corollary : If the line passes through the origin, then
putting x 1 = 0 and y 1 = 0 in y − y 1 = m ( x − x 1 ).
It becomes y = mx , which is the equation of the line
passing through the origin and having slope m.
Remark
(ii) The Point–Slope Form of a Line
X′
∴
y 1 = mx 1 + c
Subtracting (ii) form (i), we get
Y′
y 1)
x 1,
θ
Q(
Now, in triangle PQN ,
PN y − y 1
tan θ =
=
QN x − x 1
The equation y − y1 = m ( x − x1 ) is called point-slope form or
one point form of the equation.
y Example 21. Find the equation of a line which
makes an angle of 135° with the positive direction
of X-axis and passes through the point ( 3 , 5) .
Sol. The slope of the line = m = tan135° = − 1
Here x 1 = 3, y1 = 5 .
∴ The required equation of the line is
y − 5 = − 1 ( x − 3)
or
x +y −8=0
y Example 22. Find the equation of the straight line
bisecting the segment joining the points ( 5 , 3) and (4 , 4 )
and making an angle of 45° with the positive direction
of X-axis.
Sol. Here, m = slope of the line = tan 45° = 1.
Let A be the mid-point of (5, 3) and ( 4 , 4 ) . Then, the
coordinates of A are
9 7
5 + 4 3 + 4
,
 i.e.  ,  .

 2 2
 2
2 
Hence, the required equation of the line is
9
7

y − = 1 x − 

2
2
or
x −y −1=0
82
Textbook of Coordinate Geometry
y Example 23. Find the equation of the right bisector
of the line joining (1, 1) and ( 3 , 5) .
Sol. Let m be the slope of the line joining (1, 1) and (3, 5).
m=
Then,
5−1 4
= =2
3−1 2
⇒
∴ Slope (M) of right bisector of the join of (1, 1)
1
and
( 3, 5) = −
m
1
∴
M=−
2
 1 + 3 1 + 5
Mid-point of the join of (1, 1) and (3, 5) is 
,

 2
2 
i.e. (2, 3) .
Hence, equation of the right bisector passing through (2, 3)
1
and having slope
M = − is
2
1
y − 3 = − ( x − 2)
2
or
x + 2y − 8 = 0
(iii) The Two-Point Form of a Line
Theorem : The equation of a line passing through two
given points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by
 y − y1 
y − y1 =  2
 (x − x 1 )
 x2 − x1 
Proof :
Let AB be a line which passes through two points
Q ( x 1 , y 1 ) and R ( x 2 , y 2 ). Let P ( x , y ) be any point on
the line AB.
Y
)
,y 2
R(x 2
X′
O
A
N
Now, triangles PHQ and QKR are similar, then
PH QH
=
QK RK
B
y 1)
P(x,y)
x 1,
(
Q
H
K
M
L
X
Y′
Draws PL, QM and RN are perpendiculars from P , Q and R
on X-axis respectively. Also draws QH and RK are
perpendiculars on PL and QM respectively. Then from
figure
ON = x 2 , OM = x 1 , OL = x , RN = y 2 ,
QM = y 1 and PL = y
then,
RK = NM = OM − ON = x 1 − x 2
QH = ML = OL − OM = x − x 1
QK = QM − KM = QM − RN = y 1 − y 2
and
PH = PL − HL = PL − QM = y − y 1
or
y − y1
x − x1
=
y1 − y2 x1 − x2
 y − y1 
y − y1 =  2
 (x − x 1 )
 x2 − x1 
which is the required equation of the line.
Aliter I : Let the equation of the required line be
...(i)
y = mx + c
where m is the slope of the line.
Since, line Eq. (i) passes through the points ( x 1 , y 1 ) and
( x 2 , y 2 ) therefore
...(ii)
y 1 = mx 1 + c
and
...(iii)
y 2 = mx 2 + c
Now, subtracting Eqs. (ii) from (i), we get
...(iv)
y − y 1 = m (x − x 1 )
and subtracting Eqs. (ii) from (iii), we get
...(v)
y 2 − y 1 = m (x 2 − x 1 )
Dividing Eqs. (iv) by (v) then, we get
y − y1
x − x1
=
y2 − y1 x2 − x1
or
 y − y1 
y − y1 =  2
 (x − x 1 )
 x2 − x1 
which is the required equation of the line.
Aliter II : Since points, P ( x , y ), Q ( x 1 , y 1 ) and R ( x 2 , y 2 )
are collinear then area of ∆ PQR = 0
x
y 1
x
y 1
1
i.e.
| x 1 y 1 1 | = 0 or x 1 y 1 1 = 0
2
x2 y2 1
x2 y2 1
which is the required equation of the line.
y Example 24. Find the equation to the straight line
a
a


joining the points  at 1 ,  and  at 2 ,  .
t1 
t2 


a

Sol. The equation of the line joining the points at 1,  and

t1
a

at 2 ,  is

t2
a 
 a
−
a  t 2 t1 
 ( x − at 1 )
y− =
t 1  at 2 − at 1 




Chap 02 The Straight Lines
⇒
y−
a
a (t 1 − t 2 )
( x − at 1 )
=−
t1
at 1t 2 (t 1 − t 2 )
or
y−
1
a
( x − at 1 )
=−
t1
t 1t 2
or
t 1t 2y − at 2 = − x + at 1
or
x + t 1t 2y = a (t 1 + t 2 )
which is the required equation of the line.
y Example 25. Let ABC be a triangle with A ( −1, − 5) ,
B (0, 0) and C (2 , 2) and let D be the middle point of BC.
Find the equation of the perpendicular drawn from B to
AD.
Sol. Q D is the middle point of BC.
 0 + 2 0 + 2
∴ Coordinates of D are 
,

 2
2 
C(2,2)
X′
X
M
y Example 27. Find the equations of the medians of
a triangle, the coordinates of whose vertices are
( −1, 6 ) , ( −3 , − 9 ) and ( 5 , − 8 ) .
Sol. Let A ( −1, 6), B ( −3, − 9 ) and C (5, − 8) be the vertices of
∆ ABC. Let D , E and F be the mid-points of the sides BC , CA
and AB respectively.
 − 3 + 5 − 9 − 8
Coordinates of D ≡ 
,

 2
2 
 5 − 1 − 8 + 6
Coordinates of E ≡ 
,

 2
2 
D
(0,0)B
Hence, equation of altitude AD which passes through (0, 4 )
1
and having slope is
5
1
y − 4 = ( x − 10)
5
or
x − 5y + 10 = 0
17 

i.e. 1, − 

2
Y
i.e. (2, − 1)
Y
(–1,6)A
(–1,–5)A
Y′
X′
i.e. D(1, 1)
F
C(5,–8)
y Example 26. The vertices of a triangle are
A (10, 4 ), B ( −4, 9 ) and C ( −2 , − 1). Find the equation of
the altitude through A.
Sol. Q Slope of BC =
Y′
 −1 − 3 6 − 9 
and coordinates of F ≡ 
,

 2
2 
i.e. ( −2, − 3 / 2)
∴ Equation of the median AD = Equation of line through
17 

( −1, 6) and 1, −  is

2
y −6=
⇒
A(10,4)
D
X′
O
(–2,–1) C
Y′
−
17
−6
2
( x + 1)
1+1
29
( x + 1) or 29 x + 4y + 5 = 0
4
Equation of median BE is
−1 + 9
( x + 3) or 8x − 5y − 21 = 0
y +9 =
2+3
1
1
=
−5 5
Y
(–4,9) B
D
(–3,–9)B
−1 − 9 −10
=
= −5
2
−2 + 4
∴ Slope of altitude AD = −
X
O E
1+5
Slope of median AD =
=3
1+1
1
Slope of BM which is perpendicular to AD = − .
∴
3
Hence, equation of the line BM is
1
y − 0 = − ( x − 0) ⇒ x + 3y = 0
3
which is the required equation of the line.
X
83
y −6= −
and equation of median CF is
3
− +8
( x − 5)
y +8= 2
−2 − 5
or
13x + 14y + 47 = 0
84
Textbook of Coordinate Geometry
y Example 28. Find the ratio in which the line segment
joining the points (2 , 3) and (4 , 5) is divided by the line
joining (6 , 8 ) and ( −3 , − 2) .
Sol. The equation of line passing through (6, 8) and ( −3, − 2) is
y −8=
−2 − 8
( x − 6)
−3 − 6
Clearly P lies on Eq. (i), then
 3 + 5λ 
2 + 4λ 
10 
 + 12 = 0
 −9 
1+ λ 
 1+ λ 
20 + 40λ − 27 − 45λ + 12 + 12λ = 0
5
or
7 λ + 5 = 0 or λ = −
7
5
∴ The required ratio = λ : 1 = − : 1 = − 5 : 7
7
Hence, the required ratio is 5 : 7 (externally).
or
(iv) The Intercept Form of a Line :
Theorem : The equation of the straight line which cuts
off intercepts of lengths of a and b on X-axis and Y-axis
respectively, is
x y
+ =1
a b
Proof : Let QR be a line which cuts off intercepts OA = a
and OB = b on the X-axis and Y-axis respectively, where
a ≠ 0. The line is non vertical, because b is finite. Let
P ( x , y ) be a general point on the line.
Draws PL and PM perpendiculars on X-axis and Y-axis
respectively. Then PL = y and OL = x also join OP. Clearly,
Y
B(0,b)
M
x
P(x,y)
y
X′
A(a,0)
O
L
⇒
or
⇒
9y − 72 = 10x − 60
or
...(i)
10x − 9y + 12 = 0
Let the required ratio be λ : 1.
Now, the coordinates of the point P which divide the line
segment joining the points (2, 3) and ( 4, 5) in the ratio λ : 1 is
 2 + 4 λ 3 + 5λ 
,
P

 1+ λ 1+ λ 
R
⇒
Q
X
Y′
Area of ∆OAB = Area of ∆OAP + Area of ∆OPB
1
1
1
⋅ OA ⋅ OB = ⋅ OA ⋅ PL + ⋅ OB ⋅ PM
2
2
2
1
1
1
⋅ OA ⋅ OB = ⋅ OA ⋅ PL + ⋅ OB ⋅ OL
2
2
2
1
1
1
ab = ay + bx
2
2
2
⇒
ab = ay + bx
x y
or
+ =1
a b
which is the required equation of the line.
Aliter I : Equation of the line through A (a, 0 ) and B (0, b )
is
b −0
y −0 =
(x − a )
0 −a
− ay = bx − ab
or
or
bx + ay = ab
x y
+ =1
a b
which is the required equation of the line.
Aliter II : Points A (a, 0 ), P ( x , y ) and B (0, b ) are collinear,
we have
slope of AB = slope of AP
b −0 y −0
⇒
⇒ bx − ab = − ay
=
0 −a x −a
or
⇒
bx + ay = ab ⇒
x y
+ =1
a b
or
a
0 1
x y 1 =0
0 b 1
⇒
or
a (y − b ) − 0 + 1 ⋅ (bx ) = 0
bx + ay = ab
x y
or
+ =1
a b
which is the required equation of the line.
y Example 29. Find the equation of the line through
(2 , 3) so that the segment of the line intercepted
between the axes is bisected at this point.
Sol. Let the required line segment be AB.
Let O be the origin and OA = a and OB = b.
Then the coordinates of A and B are (a , 0) and (0, b )
respectively.
Chap 02 The Straight Lines
∴
and
a+0
=2 ⇒ a= 4
2
0+b
=3 ⇒ b=6
2
85
Hence, the required equations are
x y
x y
+ = 1 and
+ =1
6 8
7 7
i.e.
4 x + 3y = 24 and x + y = 7
Y
y Example 31. Find the equation of the straight line
x y
through the point P (a, b ) parallel to the line + = 1.
a b
Also find the intercepts made by it on the axes.
B(0,b)
(2,3)
X′
A(a,0)
O
x y
+ =1
a b
meets the axes in A and B respectively. So that
OA = a , OB = b
Sol. Let the line
X
Y′
Hence, the equation of the required line is
x y
+ =1
4 6
i.e.
3x + 2y = 12
Y
B'(0,b' )
y Example 30. Find the equation to the straight line
which passes through the points ( 3 , 4 ) and have
intercepts on the axes :
(a) equal in magnitude but opposite in sign
(b) such that their sum is 14
Sol. (a) Let intercepts on the axes be a and −a respectively.
∴ The equation of the line in intercept form is
x
y
+
= 1 or x − y = a
a −a
Since, Eq. (i) passes through (3, 4 ) , then
3− 4 =a
∴
a = −1
From Eq. (i), x − y + 1 = 0
which is the required equation of the line.
x y
(b) Let the equation of the line be + = 1
a b
This passes through (3, 4 ).
3 4
Therefore
+ =1
a b
It is given that a + b = 14
∴
b = 14 − a
Putting b = 14 − a in Eq. (ii), we get
3
4
+
=1
a 14 − a
⇒
...(i)
...(ii)
42 − 3a + 4a = 14a − a 2
⇒
a 2 − 13a + 42 = 0
⇒
∴
Then,
( a − 7 ) ( a − 6) = 0
a = 6, 7
b = 8, 7
(Qb = 14 − a )
P(a,b)
(0,b)B
X′
A'(a',0)
O
(a,0)A
X
Y′
Let the required parallel line meet in A ′ and B′ respectively,
so that
(say)
OA ′ = a ′
and
(say)
OB ′ = b ′
∴ Equation of required line is
x
y
...(i)
+
=1
a′ b′
Since, ∆′s OAB and OA ′ B ′ are similar, then
OA ′ OB ′
=
OA
OB
a′ b′
(say)
i.e.
=
=λ
a
b
⇒
a ′ = aλ , b ′ = bλ
Substituting these values in Eq. (i), then
x
y
...(ii)
+
=1
aλ bλ
It passes through (a, b ), then
a
b
+
=1
aλ bλ
2
⇒
= 1 or λ = 2
λ
From Eq. (ii) required equation is
x
y
+
=1
2a 2b
Evidently intercepts on the axes are 2a and 2b.
86
Textbook of Coordinate Geometry
(v) The Normal Form or Perpendicular Form
of a Line
Theorem : The equation of the straight line upon which
the length of perpendicular from the origin is p and this
normal makes an angle α with the positive direction of
X-axis is
x cos α + y sin α = p.
Proof : Let AB be a line such that the length of
perpendicular from O to the line be p
i.e.
ON = p
and
∠ NOX = α
Let P ( x , y ) be any point on the line. Draw PL
perpendicular from P on X-axis.
⇒
cosec α =
or
OR = p cosec α
Thus, AB makes intercepts p sec α and p cosec α on X-axis
and Y-axis respectively.
y
x
∴ Equation of AB is
+
=1
p sec α p cosec α
or
x cos α + y sin α = p
which is the required equation of the line AB.
Aliter II : The points Q ( p sec α, 0 ), P ( x , y ) and
R (0, p cosec α ) are collinear, then
p secα
0
1
R
or
p sec α (y − p cosec α ) − 0 + 1 ( px cosec α ) = 0
or
p (y sin α − p ) + px cos α = 0
or
x cos α + y sin α = p
which is the required equation of the line AB.
N
P(x,y)
p
X′
O
α
y
α
x
L
Q
X
A
Remarks
Y′
1. Here, p is always taken as positive and α is measured from
positive direction of X-axis in anticlockwise direction between 0
and 2 π ( i.e. 0 ≤ α < 2 π ).
2. (Coefficient of x) 2+ (Coefficient of y) 2 = cos 2 α + sin2 α = 1
Let line AB cuts X and Y-axes at Q and R respectively.
Now,
∠ NQO = 90 ° − α
∴
∠ LPQ = 90 ° − (90 ° − α ) = α
LQ LQ
In ∆PLQ,
tan α =
=
PL
y
∴
Also, in ∆ONQ,
⇒
⇒
⇒
LQ = y tan α
ON
cos α =
OQ
p
cos α =
OL + LQ
3. cosα and cos ( 90° − α ) are the direction cosines of ON.
4.
II quadrant
OQ = p sec α
OR
Also in ∆ONR, sec (90° − α ) =
ON
B
N
N
p
p
α
X′
X
O
A
α
X′
X
O
A
Y′
Y′
cos α < 0, sin α > 0, p > 0
cos α > 0, sin α > 0, p > 0
IV quadrant
III quadrant
Y
Y
B
α
α
X′
X
O
then
or
Y
B
∴
x cos α + y sin α = p
which is the required equation of the line AB.
∠ NOR = 90 ° − α
OQ OQ
Now, in ∆ONQ, sec α =
=
ON
p
I quadrant
Y
...(i)
OL cos α + LQ cos α = p
x cos α + y tan α cos α = p
(Q OL = x and LQ = y tan α )
Aliter I : Q ∠ NOQ = α
y
1 =0
p cosec α 1
x
0
Y
B
OR
p
B
X′
X
O
p
p
N
N
A
Y′
cos α < 0, sin α < 0, p > 0
A
Y′
cos α > 0, sin α < 0, p > 0
Chap 02 The Straight Lines
Corollary 1 : If α = 0 °, then equation x cos α + y sin α = p
becomes x cos 0 ° + y sin 0 ° = p
i.e. x = p (Equation of line parallel to Y-axis)
π
Corollary 2 : If α = , then equation x cos α + y sin α = p
2
π
π
becomes x cos   + y sin   = p
2
2
In ∆ONA,
cos30° =
ON p
=
OA a
Y
B
b
i.e. y = p (Equation of line parallel to X-axis).
Corollary 3 : If α = 0 °, p = 0 then equation
x cos α + y sin α = p becomes x cos 0 ° + y sin 0 ° = 0
i.e. x = 0 (Equation of Y-axis)
π
Corollary 4 : If α = , p = 0 then equation
2
π
π
x cos α + y sin α = p becomes x cos   + y sin   = 0
2
2
i.e.
(Eq. (i) of X-axis).
y =0
X′
N
60° p
30°
a
O
A
⇒
3 p
2p
or a =
=
2
a
3
and in ∆ONB, cos60° =
ON
=
OB
1 p
or
=
2 b
1
Area of ∆OAB = ab =
2
p
b
⇒
b = 2p
Q
Sol. Here, α = 60° and p = 9.
∴
2p 2 50
=
3
3
⇒
p 2 = 25
( 2p ) =
∴ Equation of the required line is
x cos 60 ° + y sin 60 ° = 9
120°
N
30°
X′
60°
O
A
X
Y′
⇒
or
 3
 1
x   +y   =9
 2
 2
x + y 3 = 18
y Example 33. Find the equation of the straight line on
which the perpendicular from origin makes an angle of
30° with X-axis and which forms a triangle of area
 50 
  sq units with the coordinates axes.
 3
2p 2
3
(given)
Reduction of General Equation to
Standard Form
Let Ax + By + C = 0 be the general equation of a straight
line where A and B are not both zero.
(i) Reduction of ‘Slope-Intercept’ Form
Given equation is Ax + By + C = 0
⇒
By = − Ax − C
 A
 C
⇒
y = −  x + − 
 B
 B
(Assuming B ≠ 0)
Comparing it with y = mx + c
coefficient of x
A
we get
slope (m ) = − = −
coefficient of y
B
Sol. Let ∠ NOA = 30°
Let ON = p > 0, OA = a , OB = b
1  2p 
 
2  3
or
p =5
(Qp > 0)
∴ Using x cos α + y sin α = p , the equation of the line AB is
x cos 30° + y sin 30° = 5
or
x 3 + y = 10
Y
60°
X
Y′
y Example 32. The length of perpendicular from the
origin to a line is 9 and the line makes an angle of
120° with the positive direction of Y -axis. Find the
equation of the line.
B
87
and
y intercept (c ) = −
C
constant term
=−
B
coefficient of y
88
Textbook of Coordinate Geometry
(ii) Reduction to ‘Intercept’ Form
Corollary 1 : Find angle between the lines
A1 x + B 1 y + C 1 = 0 and A2 x + B 2 y + C 2 = 0.
Slope of the line
A
A1 x + B 1 y + C 1 = 0 is − 1 = m 1
B1
(say)
and slope of the line
A2 x + B 2 y + C 2 = 0 is −
A2
= m2
B2
(say)
If θ is the angle between the two lines, then
A B − A2 B 1

= 1 2
A1 A2 + B 1 B 2
Corollary 2 : Find the condition of (i) parallelism (ii)
perpendicularity of the lines
A1 x + B 1 y + C 1 = 0
and
A2 x + B 2 y + C 2 = 0
(i) If the two lines are parallel, θ = 0 °
∴
tan θ = tan 0 ° = 0
⇒
or

= 0
A1 B 2 − A2 B 1
A1 A2 + B 1 B 2
A1 B 2 − A2 B 1 = 0
A1 B 1
=
A2 B 2
x y
+ =1
a b
we get, x-intercept (a) = −
C
constant term
=−
A
coefficient of x
y-intercept (b) = −
C
constant term
=−
B
coefficient of y
and
Given equation is Ax + By + C = 0 . Let its normal form
be x cos α + y sin α = p.
Clearly, equations Ax + By + C = 0 and
x cos α + y sin α = p represent the same line.
A
B
C
Therefore,
=
=
cos α sin α −p
Ap
C
Bp
sin α = −
C
⇒
cos α = −
cos 2 α + sin2 α = 1
2
(Remember)
⇒
A1 A2 + B 1 B 2 = 0
which is required condition of perpendicularity.
Remark
If two lines are coincident, then
(Remember)
2
 Ap 
 Bp 
−
 + −  = 1
 C 
 C 
(Remember)

A1 B 2 − A2 B 1

=∞
A1 A2 + B 1 B 2
A1 B 1 C 1
=
=
A2 B 2 C 2
y
x
+
= 1 (Assuming A ≠ 0, B ≠ 0)
(− C / A) (− C / B )
and
which is required condition of parallelism.
(ii) If the two lines are perpendicular, θ = 90 °
∴
tan θ = tan 90 ° = ∞
⇒
(Assuming C ≠ 0)
(iii) Reduction to ‘Normal’ Form
A B − A2 B 1

θ = tan −1  1 2
A1 A2 + B 1 B 2
⇒
⇒
Comparing with
  A1   A2 
  − B  −  − B 
m1 − m2 

1
2


tan θ =
=
 A1   A2 
1 + m 1m 2 
1 +  −

 −
  B1   B2 
∴
Given equation is Ax + By + C = 0
⇒
Ax + By = − C
B
A
y =1
x+
⇒
−C
−C
or
p2 =
⇒
p=
C2
A2 + B 2
| C|
(A + B 2 )
2
From Eq. (i), cos α = −
| C|
A
,
.
2
C
(A + B 2 )
sin α = −
| C|
B
.
.
2
C
(A + B 2 )
Putting the values of cos α, sin α and p in
x cos α + y sin α = p, we get
...(i)
89
Chap 02 The Straight Lines
 − | C|



A
B
 x + y  − | C| .

x
⋅
 C

2
2 
2
2 
C
(A + B ) 
(A + B ) 


=
⇒
| C|
( A2 + B 2 )




C
A
B
−
 x + −
y =


2
2 
2
2 
2
(A + B 2 )
(A + B ) 
(A + B ) 


This is the normal form of the line Ax + By + C = 0.
Rule : First shift the constant term on the RHS and make
it positive, if it is not so by multiplying the whole equation
by ‘− 1 ’ and then divide both sides by
(coefficient of x ) 2 + (coefficient of y ) 2
y Example 34. Reduce x + 3y + 4 = 0 to the :
(i) Slope-intercept form and find its slope and y-intercept
(ii) Intercept form and find its intercepts on the axes
(iii) Normal form and find the values of p and α
3y = − x − 4
⇒
 4
 1
y = −

 x + −


3
3
which is in the slope-intercept form y = mx + c
1
4
Where slope (m ) = −
and y-intercept (c ) = −
3
3
(ii) Given equation is
x + 3y + 4 = 0
⇒
⇒
⇒
y Example 35. Find the measure of the angle of
intersection of the lines whose equations are
3x + 4 y + 7 = 0 and 4 x − 3y + 5 = 0.
Sol. Given lines are 3x + 4y + 7 = 0, 4 x − 3y + 5 = 0. Comparing
the given lines with A1x + B1y + C 1 = 0, A 2 x + B 2y + C 2 = 0
respectively, we get
A1 = 3, B1 = 4 and A 2 = 4, B 2 = − 3
Q
A1A 2 + B1B 2 = 3 × 4 + 4 ( −3) = 0
Hence, the given lines are perpendicular.
Sol. The given equations of lines can be written as
x
3y
+
=1
−4
−4
x
y
+
=1
−4 − 4 / 3
...(ii)
A 2 = ab − b 2 , B 2 = − (ab + a 2 )
and
Let θ be the acute angle between the lines, then

A B − A 2 B1
tanθ = 1 2
A1A 2 + B1B 2
4
3
x + 3y = − 4
tan θ =
=
(RHS made positive)
Dividing both sides by ( −1)2 + ( − 3 )2 = 2 , we get

−

(ab − b ) x − (ab + a ) y + a = 0
3
A1 = ab + b 2 , B1 = − (a 2 − ab )
(iii) Given equation is x + 3y + 4 = 0
− x − 3y = 4
...(i)
2
Comparing the given lines (i) and (ii) with the lines
A1x + B1y + C 1 = 0 and A 2 x + B 2y + C 2 = 0
respectively, we get
x + 3y = − 4
where x-intercept (a ) = − 4 and y-intercept (b ) = −
⇒
(ab + b 2 ) x − (a 2 − ab ) y + b 3 = 0
2
and
x y
which is in the intercept form + = 1
a b
⇒
cos α = −
y Example 36. Find the angle between the lines
(a 2 − ab) y = (ab + b 2 ) x + b 3
and (ab + a 2 ) y = (ab − b 2 ) x + a 3
where a > b > 0.
Sol. (i) Given equation is x + 3 y + 4 = 0
⇒
1
= − cos 60° = cos (180° − 60° )
2
or
cos (180° + 60° )
∴
α = 120° or 240°
3
and
sin α = −
= − sin 60° = sin (180° + 60° )
2
or
sin (360° − 60° )
∴
α = 240° or 300°
Hence,
α = 240° , p = 2
∴ Required normal form is
x cos 240° + y sin 240° = 2
where,

1
3
 y =2
 x + −
2
 2
Which is the normal form x cos α + y sin α = p .
(ab + b 2 ) × ( − (ab + a 2 )) − (ab − b 2 ) × ( − (a 2 − ab ))
(ab + b 2 ) (ab − b 2 ) + (a 2 − ab ) (ab + a 2 )
− {a 2b 2 + a 3b + ab 3 + a 2b 2 − a 3b + a 2b 2 + a 2b 2 − b 3a }
(a 2b 2 − b 4 + a 4 − a 2b 2 )
− 4a 2b 2 4 a 2b 2
=
= 4
4
4
4
a − b  a − b
 4 a 2b 2 
∴ θ = tan −1  4

a − b4 
90
Textbook of Coordinate Geometry
y Example 37. Two equal sides of an isosceles triangle
are given by the equations 7 x − y + 3 = 0 and
x + y − 3 = 0 and its third side passes through the
point (1, − 10) . Determine the equation of the third
side.
Sol. Given equations
...(i)
7x − y + 3 = 0
and
...(ii)
x +y −3=0
represents two equal sides AB and AC of an isosceles
triangle ABC. Since its third side passes through D (1, − 10)
then its equation is
...(iii)
y + 10 = m ( x − 1)
Q
AB = AC
Let
∠ ABC = ∠ ACB = θ
then
∠ ACE = π − θ
The Distance form or Symmetric form
or Parametric form of a line
Theorem : The equation of the straight line passing
through ( x 1 , y 1 ) and making an angle θ with the positive
direction of X-axis is
x − x1 y − y1
=
=r
cos θ
sin θ
where, r is the directed distance between the points ( x , y )
and ( x 1 , y 1 ).
Proof : Let AB be a line which passes through the point
Q ( x 1 , y 1 ) and meet X-axis at R and makes an angle θ with
the positive direction of X-axis.
Y
B
Y
r
(x1,y1)Q
A
O
3
y–
0
7x–y+
=
E
q
D(1,–10)
Y′
From Eqs. (i) and (ii), slopes of AB and AC are
m1 = 7 and m 2 = − 1
respectively.
7 −m
∴
tanθ =
1 + 7m
tan ( π − θ ) =
1 + m
−1 − m
= −

1 + ( − 1) m
1 − m
⇒
1 + m
1 + m
− tanθ = − 

 ⇒ tanθ = 
1 − m
1 − m
∴
7 −m 1+m
=
1 + 7m 1 − m
⇒
⇒
( 7 − m ) ( 1 − m ) = ( 1 + 7m ) ( 1 + m )
6m 2 + 16m − 6 = 0
or
3m 2 + 8m − 3 = 0 or (3m − 1) (m + 3) = 0
1
m = ,−3
3
Hence from Eq. (iii), the third side BC has two equations
1
y + 10 = ( x − 1) and y + 10 = − 3 ( x − 1)
3
or
x − 3y − 31 = 0 and 3x + y + 7 = 0
⇒
θ
R
O
M
X
L
Y′
p–q
C
and
N
A
q
B
X′
X
x+
3=0
X′
P(x,y)
θ
Let P ( x , y ) be any point on the line at a distance r from Q.
Draws PL and QM are perpendiculars from P and Q on
X-axis respectively and draw QN perpendicular on PL.
Then,
QN = ML = OL − OM = x − x 1
and
PN = PL − NL = PL − QM = y − y 1
from ∆PQN ,
x − x1
QN x − x 1
or
...(i)
cos θ =
=
=r
cos θ
PQ
r
and
sin θ =
PN y − y 1
=
PQ
r
or
y − y1
=r
sin θ
...(ii)
From Eqs. (i) and (ii), we get
x − x1 y − y1
=
=r
cos θ
sin θ
x − x1 y − y1
Corollary 1 : Q
=
= r , then
cos θ
sin θ
x = x 1 + r cos θ

y = y 1 + r sin θ 
parametric equations of straight line AB.
Corollary 2 : If P point above Q then r is positive then
coordinates of P are ( x 1 + r cos θ, y 1 + r sin θ ) and if P
below Q then r is negative then coordinates of P are
( x 1 − r cos θ, y 1 − r sin θ ).
Chap 02 The Straight Lines
y Example 38. The slope of a straight line through
3
A ( 3 , 2) is . Find the coordinates of the points on
4
the line that are 5 units away from A.
Sol. Let straight line makes an angle θ with positive direction of
X -axis,
3
then
tanθ =
4
3
4
and cosθ =
∴
sinθ =
5
5
5
3
6
(cos θ + sin θ ) = 1
3
3
3
cos θ + sin θ =
=
6
2
⇒
⇒
1
1
3
cos θ +
sin θ =
2
2
2
π
 π

cos θ −  = cos  
 6

4
π
π
θ − = 2nπ ± ; n ∈I
4
6
π π
θ=± +
6 4
= 15° , 75°
⇒
⇒
for n = 0,
θ
4
∴ Equation of the straight line through A (3, 2) in
parametric form is
x −3 y −2
=
= ±5
cos θ sin θ
4
∴
x = 3 ± 5 cosθ = 3 ± 5 × = 3 ± 4 = 7 or −1
5
3
and y = 2 ± 5 sinθ = 2 ± 5 × = 2 ± 3 = 5 or −1
5
Hence, the coordinates of the points are (7, 5) and ( − 1, − 1) .
3π
4
with the negative direction of X-axis. Find the length of
the line segment cut off between (2 , 3) and the line
x + y − 7 = 0.
3π
with the negative direction of
Sol. Q Line makes an angle
4
X -axis.
π
∴ Line makes an angle with the positive direction of
4
X -axis.
X′
O
(4,0)
X
Y′
6
6
cosθ and y = 2 +
sinθ
3
3


6
6
Since, the point 1 +
cos θ, 2 +
sin θ lies on the line
3
3


or
∴
x =1+
x +y = 4
6
6
cos θ + 2 +
sin θ = 4
1+
3
3
A(2,3)
3π/4
X′
π/4
O
π/α
Y′
X
0
A(1,2)
r P
=
6
3
π/4
7
y–
(0,4)
Y
x+
Y
(Q0 ≤ θ < π )
y Example 40. A line through (2 , 3) makes an angle
y Example 39. Find the direction in which a straight
line must be drawn through the point (1, 2) so that
its point of intersection with the line x + y = 4 may
1
be at a distance
6 from this point.
3
Sol. Let the straight line makes an angle θ with the positive
direction of X -axis.
∴ Equation of the line through (1, 2) in parametric form is
x −1 y −2 1
6
=
=
cos θ sin θ 3
91
∴ The equation of the line through (1, 2) in parametric form is
x −2
y −3
=r
=
 π
 π
cos   sin  
 4
 4
x −2 y −3
i.e.
...(i)
=
=r
1
1
2
2
r
r
x =2+
and y = 3 +
∴
2
2
Let the line (i) meet the line x + y − 7 = 0 in P
r
r 

,3 +
∴ Coordinates of P 2 +
 lies on x + y − 7 = 0

2
2
r
r
+3+
−7 =0
2
2
then
2+
or
2r
= 2 or r = 2
2
∴
AP = 2
92
Textbook of Coordinate Geometry
y Example 41. Find the distance of the point (2 , 3)
from the line 2x − 3y + 9 = 0 measured along the line
2x − 2y + 5 = 0 .
Sol. Since, slope of the line 2x − 2y + 5 = 0 is 1, its makes an
angle
π
with positive direction of X -axis.
4
The equation of the line through (2, 3) and making an angle
π
in parametric form
4
x −2
y −3
x −2 y −3
= r or
=
=
=r
1
1
 π
 π
cos  
sin  
 4
 4
2
2
Special Corollaries
(i) Angle made by AB with positive X-axis (where
A and B are given points) :be two points and let AB
makes an angle θ with the positive direction of X-axis
and let d be the distance between A and B. Then Let
A ( x 1 , y 1 ) and B ( x 2 , y 2 )
x − x1
y − y1
cos θ = 2
, sin θ = 2
d
d
Y
B'(x2',y2' )
B(x2,y2)
r
r 

Coordinates of any point on this line are 2 +
,3 +
.

2
2
This point lies on the line 2x − 3y + 9 = 0
r 
r 


2 2 +
⇒
 − 3 3 +
 +9 =0


2
2
∴
X
where, d = ( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 = AB
y Example 42. If the line y − 3x + 3 = 0 cuts the
parabola y 2 = x + 2 at A and B, then find the value of
PA. PB { where P ≡ ( 3, 0)}.
Sol. Slope of line y − 3x + 3 = 0 is 3
If line makes an angle θ with X-axis, then tan θ = 3
∴
(y2–y1)
O
r
−
+ 4 =0
2
r =4 2
⇒
(x1,y1)A
α
θ
(x2–x1)
θ = 60°
x− 3
y −0
=r
=
cos 60° sin 60°
and x 2 = x 1 + d cos θ, y 2 = y 1 + d sin θ
If AB rotates an angle α about A, then new
coordinates of B are
x 2′ = x 1 + d cos(θ + α ),
y ′2 = y 1 + d sin(θ + α )
and here, AB = AB ′ = d .
(ii) Complex number as a rotating arrow in Argand
plane :
Let z = r (cos θ + i sin θ ) = re iθ , where i = −1
... (i)
be a complex number representing a point P in the
Argand plane.
Y
Y
A
Imaginary axis
Q(zeiφ)
60°
P(√3,0)
O
X
B
O

r r 3
2
⇒  3+ ,
 be a point on the parabola y = x + 2
2 2 

then,
3 2
r
r = 3 + + 2 ⇒ 3r 2 − 2r − 4 (2 + 3 ) = 0
4
2
∴
−4 (2 + 3 )
 = 4 (2 + 3 )
PA ⋅ PB = r1r 2 =
3
3


Then,
P(z)
φ
θ
Real axis
X
OP = | z | = r and ∠ POX = θ
Now, consider complex number z 1` = ze iφ
or
z 1 = re iθ ⋅ e i φ = r ⋅ e i ( θ + φ )
[from Eq. (i)]
Clearly the complex number z 1 represents a point Q
in the Argand plane, when
OQ = r and ∠ QOX = θ + φ
Chap 02 The Straight Lines
→
Clearly multiplication of z with e iφ rotates the vector OP
through angle φ in anti-clockwise sense. Similarly →
multiplication of z with e − iφ will rotate the vector OP in
clockwise sense.
Remark
If z1, z2 and z3 are the affixes of the three points A, B and C such
that AC = AB and ∠ CAB = θ. Therefore
→
→
AB = z2 − z1, AC = z3 − z1
→
→
Then AC will be obtained by rotating AB through an angle θ in
anticlockwise sense and therefore
Since, AC = r = 2
Put
1 4+ 2
=
2
2
3
6
and
y = 2.
=
2
2
Equation of the line AC is
x −2
1
= cot 60° =
y
3
Aliter (By special corollary (ii))
Q
A ≡ (2, 0), B ≡ (3, 1), let C ≡ ( x , y )
∴
z A = 2,z B = 3 + i , zC = x + iy , where i = −1
θ
i
zC − z A
=e
zB − zA
A(z1)
→
AC = ABei θ
iθ
( z3 − z1 ) = ( z2 − z1 ) e
or
⇒
 z − z1 
iθ
or  3
 =e
 z2 − z1 
or
AB = (2 − 3)2 + (0 − 1)2 = 2
and
slope of AB =
 4 + 2 6
, 
C ≡
2
 2
y Example 44. The centre of a square is at the origin
and one vertex is A (2 , 1). Find the coordinates of other
vertices of the square.
C
B(3,1)
Sol. [By special corollary (ii)]
15°
O
zC − 2 = (1 + i ) (cos 15° + i sin 15° )
 3 +1
 3 − 1 
zC = 2 + ( 1 + i ) 
+i 

 2 2 
 2 2
and equation of AC
y − 0 = tan 60° ( x − 2) ⇒ x 3 − y − 2 3 = 0
1−0
= 1 = tan 45°
3−2
Y
X′
5π 

Q15° = 

12 
 6
 3
1
4+ 2

= 2 +
+i  
 +i   =


2
2
 2
 2
∴
Sol. By special corollary (i)
5π
12
 3 +1

3 +1
3 − 1
3 − 1
+
= 2 +
−

 +i 
2 2
2 2 
2 2 
 2 2

y Example 43. The line joining the points A (2 , 0) and
B ( 3 , 1) is rotated about A in the anticlockwise direction
through an angle of 15°. Find the equation of the line
in the new position. If B goes to C in the new position,
what will be the coordinates of C ?
Here
x 3 −y −2 3 =0
 4 + 2 6
and coordinates of C are 
, .
2
 2
B(z2)
→
r = 2 in Eq. (i), then
x = 2+ 2⋅
or
C(z3)
93
45°
A(2,0)
X
Q
A ≡ ( 2, 1)
∴
z A = 2 + i , where i = −1
B
Y′
A(2,1)
∴
∠ BAX = 45°
Now line AB is rotated through an angle of 15°
⇒
∠ CAX = 45° + 15° = 60°
and
AB = AC = 2
Equation of line AC in parametric form is
x = 2 + r cos 60°

y = 0 + r sin 60° 
O
C
...(i)
D
94
Textbook of Coordinate Geometry
 1 
then centre of square E ≡  − , 0
 2 
1
∴
zE = −
2
Now, in ∆ AEB, ( EA = EB )
Now, in triangle AOB,
π
OA = OB, ∠ AOB = 90° =
2
∴
zB = zA e
i
π
2
= iz A = 2i − 1
∴
B ≡ ( − 1 , 2)
QO is the mid-point of AC and BD
∴
C ≡ ( −2, − 1) and D ≡ (1, − 2).
i
zB − zE
=e
zA − zE
Sol. (By special corollary (ii))
Q
A ≡ (1, 1)
∴
z A = 1 + i , where i = −1
C ≡ ( − 2, − 1)
zC = − 2 − i
and
∴
90°
E
(–1/2,0)
C
(–2,–1)
1

2
∴
then
3 3

D ≡  −1 + , − 

2 2
or
1
D ≡ ,−
2
B
A(1,1)
=i
1

= i 1 + i +

2
3 3
zB = − + i
2 2
 3 3
B ≡ − , 
 2 2
⇒
y Example 45. The extremities of the diagonal of a
square are (1, 1) , ( −2 , − 1). Obtain the other two vertices
and the equation of the other diagonal.
π
2
zB +
3

2
Hence, equation of other diagonal BD is
3
−0
1

y −0= 2
x + 
3 1 
2
− +
2 2
⇒
6x + 4y + 3 = 0
D
Exercise for Session 1
1.
The distance of the point (3, 5) from the line 2x + 3y − 14 = 0 measured parallel to the line x − 2y = 1
is
(a)
2.
7
5
(b)
7
13
(b) | α − β | = π / 2
(d) α ± β = π / 2
If each of the points ( x1, 4), ( −2, y1) lies on the line joining the points (2, − 1), (5, − 3), then the point P ( x1, y1) lies
on the line
(a) 6 (x + y ) − 25 = 0
(c) 2x + 3y − 6 = 0
4.
(d) 13
The lines x cos α + y sin α = p1 and x cos β + y sin β = p 2 will be perpendicular, if
(a) α = β
(c) α = π / 2
3.
(c) 5
(d) 2x + 6y + 1 = 0
(d) 6 (x + y ) + 25 = 0
The equation of the straight line passing through the point (4, 3) and making intercepts on the coordinate axes
whose sum is −1 is
x
y
x
y
+ = − 1 and
+ =−1
2 3
−2 1
x
y
x
y
(c) + = 1 and
+ =1
2 3
−2 1
(a)
x y
x
y
− = − 1 and
+ = −1
2 3
−2 1
x y
x
y
(d) − = 1 and
+ =1
2 3
−2 1
(b)
Chap 02 The Straight Lines
5.
If the straight lines ax + by + c = 0 and x cos α + y sin α = c enclose an angle π / 4 between them and meet
the straight line x sin α − y cos α = 0 in the same point, then
(a) a 2 + b 2 = c 2
6.
(b) 45°
(b) π / 2
(d) 90°
(c) 3 π / 4
(d) π
(b) y ( 3 + 1) + x (1 − 3 ) = 2a
(d) y ( 3 + 1) + x ( 3 − 1) = 2a
(b) x − 2y + 5 = 0
(d) 2x − y + 1 = 0
The equation of a straight line passing through the point ( −5 , 4) and which cuts off an intercept of 2 units
between the lines x + y + 1 = 0 and x + y − 1 = 0 is
(a) x − 2y + 13 = 0
(c) x − y + 9 = 0
11.
(c) 60°
A (1, 3) and C (7, 5) are two opposite vertices of a square. The equation of side through A is
(a) x + 2y − 7 = 0
(c) 2x + y − 5 = 0
10.
(d) a 2 + b 2 = 4
A square of side a lies above the X-axis and has one vertex at the origin. The side passing through the origin
makes an angle π /6 with the positive direction of X-axis. The equation of its diagonal not passing through the
origin is
(a) y ( 3 − 1) − x (1 − 3 ) = 2a
(c) y ( 3 + 1) + x (1 + 3 ) = 2a
9.
(c) a 2 + b 2 = 2c 2
The inclination of the straight line passing through the point (–3, 6) and the mid-point of the line joining the
points (4, –5) and (–2, 9) is
(a) π / 4
8.
(b) a 2 + b 2 = 2
The angle between the lines 2x − y + 3 = 0 and x + 2y + 3 = 0 is
(a) 30°
7.
95
(b) 2x − y + 14 = 0
(d) x − y + 10 = 0
Equation to the straight line cutting off an intercept 2 from negative direction of the axis of y and inclined at 30°
to the positive direction of axis of x is
(a) y + x − 3 = 0
(c) y − x 3 − 2 = 0
(b) y − x + 2 = 0
(d) y 3 − x + 2 3 = 0
12.
What is the value of y so that the line through (3, y ) and (2, 7) is parallel to the line through (–1, 4) and (0, 6) ?
13.
A straight line is drawn through the point P (2, 2) and is inclined at an angle of 30° with the X-axis. Find the
coordinates of two points on it at a distance 4 from P on either side of P.
14.
If the straight line through the point P (3, 4) makes an angle
π
with X-axis and meets the line 12x + 5y + 10 = 0
6
at Q, find the length of PQ.
15.
Find the distance of the point (2, 3) from the line 2x − 3y + 9 = 0 measured along the line x − y + 1 = 0.
16.
A line is such that its segment between the straight line 5x − y − 4 = 0 and 3x + 4y − 4 = 0 is bisected at the
point (1, 5) . Obtain the equation.
17.
The side AB and AC of a ∆ ABC are respectively 2x + 3y = 29 and x + 2y = 16. If the mid-point of BC is (5, 6),
then find the equation of BC.
18.
A straight line through A ( −15, − 10) meets the lines x − y − 1 = 0, x + 2y = 5 and x + 3y = 7 respectively at A, B
12 40 52
and C. If
+
=
, prove that the line passes through the origin.
AB AC AD
Session 2
Position of Two Points Relative to a Given Line,
Position of a Point Which Lies Inside a Triangle,
Equations of Lines Parallel and Perpendicular to
a Given Line, Distance of a Point From a Line,
Distance Between Two Parallel Lines,
Area of Parallelogram,
X¢
Position of Two Points Relative
to a Given Line
Theorem : The points P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) lie on the
same or opposite sides of the line ax + by + c = 0 according
as
ax 1 + by 1 + c
> 0 or <0 .
ax 2 + by 2 + c
∴ λ is negative
⇒
 ax + by 1 + c 
− 1
 <0
 ax 2 + by 2 + c 
⇒
 ax 1 + by 1 + c 

 >0
 a x 2 + by 2 + c 
or
f (x 1 , y 1 )
>0
f (x 2 , y 2 )
Proof : Let the line PQ be divided by the line
ax + by + c = 0 in the ratio λ : 1 (internally) at the point R.
where, f ( x , y ) ≡ ax + by + c .
 x + λx 2 y 1 + λy 2 
∴ The coordinates of R are  1
,
.
1+ λ 
 1+ λ
Case II : Let P and Q are on opposite sides of the line
ax + by + c = 0
∴ R divides PQ internally.
∴ λ is positive
The point of R lies on the line ax + by + c = 0
then
 y 1 + λy 2 
 x + λx 2 
a 1
 +c =0
 +b 
 1+ λ 
 1+ λ 
⇒
λ (ax 2 + by 2 + c ) + (ax 1 + by 1 + c ) = 0
⇒
 ax + by 1 + c 
λ =− 1

 ax 2 + by 2 + c 
(Qax 2 + by 2 + c ≠ 0 )
Case I : Let P and Q are on same side of the line
ax + by + c = 0.
∴ R divides PQ externally.
Q
R
P
P
 ax + by 1 + c 
− 1
 >0
 ax 2 + by 2 + c 
⇒
 ax 1 + by 1 + c 
 <0

 ax 2 + by 2 + c 
or
f (x 1 , y 1 )
<0
f (x 2 , y 2 )
P
R
Q
where, f ( x , y ) = ax + by + c
R
Q
⇒
Remarks
1. The side of the line where origin lies is known as origin side.
2. A point ( α, β ) will lie on origin side of the line ax + by + c = 0, if
aα + bβ + c and c have same sign.
3. A point ( α, β ) will lie on non-origin side of the line
ax + by + c = 0, if aα + bβ + c and c have opposite sign.
97
Chap 02 The Straight Lines
y Example 46. Are the points (2 , 1) and ( −3, 5) on the
same or opposite side of the line 3 x − 2y + 1 = 0 ?
Hence, the given village V does not lie on the canal.
9
Also if f ( x , y ) ≡ x + y −
2
Sol. Let f ( x , y ) ≡ 3x − 2y + 1
∴
f ( 2, 1)
3 (2) − 2(1) + 1
5
= − <0
=
f ( −3, 5) 3 ( −3) − 2 (5) + 1
18
∴
Therefore, the two points are on the opposite sides of the
given line.
y Example 47. Is the point (2 , − 7 ) lies on origin side of
the line 2x + y + 2 = 0 ?
Sol. Let f ( x , y ) ≡ 2x + y + 2
∴
f ( 2, − 7 ) = 2 ( 2) − 7 + 2 = − 1
f (2, − 7 ) < 0 and constant 2 > 0
Hence, the point (2, − 7 ) lies on non-origin side.
y Example 48. A straight canal is at a distance of
1
4 km from a city and the nearest path from the city
2
to the canal is in the north-east direction. Find
whether a village which is at 3 km north and 4 km east
from the city lies on the canal or not. If not, then on
which side of the canal is the village situated ?

4 +3−
f ( 4, 3) 
=
f (0, 0)  0 + 0 −

9 

2  = −  7 2 − 9  <0


9 
9



2
Hence, the village is on that side of the canal on which
origin or the city lies.
Position of a Point Which Lies
Inside a Triangle
Let P ( x 1 , y 1 ) be the point and equations of the sides of a
triangle are
BC : a 1 x + b 1 y + c 1 = 0
CA : a 2 x + b 2 y + c 2 = 0
and
AB : a 3 x + b 3 y + c 3 = 0
A (x′,y′)
Sol. Let O (0, 0) be the given city and AB be the straight canal.
Given, OL =
9
km
2
P (x1,y1)
North
Y
(x′′ ,y′′ )B
C (x′′′,y′′′)
L
First find the coordinates of A, B and C say,
A ≡ ( x ′ , y ′ ); B ≡ ( x ′ ′ , y ′ ′ ) and C ≡ ( x ′ ′ ′ , y ′ ′ ′ )
and if coordinates of A, B, C are given, then find equations of
BC , CA and AB.
If P ( x 1 , y 1 ) lies inside the triangle, then P and A must be
on the same side of BC , P and B must be on the same side
of AC , P and C must be on the same side of AB , then
a 1 x 1 + b 1y 1 + c 1
...(i)
>0
a 1 x ′ + b 1y ′ + c 1
V(4,3)
No
rth
-E
as
t
B
X′
45°
O
A
X
East
Y′
∴ Equation of AB
i.e. Equation of canal is
x cos 45° + y sin 45° =
or
9
2
9
x +y =
2
Let V be the given village, then V ≡ ( 4, 3)
Putting x = 4 and y = 3 in Eq. (i),
9
9
then 4 + 3 =
, i.e. 7 =
which is impossible.
2
2
...(i)
and
a 2 x 1 + b 2y 1 + c 2
>0
a 2 x ′ ′ + b 2y ′ ′ + c 2
...(ii)
a 3 x 1 + b 3y 1 + c 3
>0
a 3 x ′ ′ ′ + b 3y ′ ′ ′ + c 3
...(iii)
The required values of P ( x 1 , y 1 ) must be intersection of
these inequalities Eqs. (i), (ii) and (iii).
98
Textbook of Coordinate Geometry
Aliter (Best Method) : First draw the exact diagram of
the problem. If the point P ( x 1 , y 1 )
move on the line y = ax + b for all x 1 , then
P ≡ ( x 1 , ax 1 + b )
and the portion DE of the line y = ax + b (Excluding D and
E) lies within the triangle. Now line y = ax + b cuts any
two sides out of three sides, then find coordinates of D and
E.
D ≡ (α, β)
and
E ≡ ( γ , δ ) (say)
then
α < x1 < γ
and
β < ax 1 + b < δ
4t − 6 < 0
3
∴
t<
2
and P , C must be on the same side of AB
value of (3x − 2y + 6) at P (t , t + 1)
∴
>0
value of (3x − 2y + 6) at C (6, 1)
y Example 49. For what values of the parameter t
does the point P (t, t + 1) lies inside the triangle ABC
where A ≡ (0, 3), B ≡ ( − 2 , 0) and C ≡ (6 , 1) .
From Eqs. (i), (ii) and (iii), we get
6
3
− <t <
7
2
 6 3
i.e.
t ∈ − , 
 7 2
Sol. Equations of sides
BC : x − 8y + 2 = 0
CA : x + 3y − 9 = 0
and
AB : 3x − 2y + 6 = 0
Since, P (t , t + 1) lies inside the triangle ABC, then P and A
must be on the same side of BC
value of ( x − 8y + 2) at P (t , t + 1)
∴
>0
value of ( x − 8y + 2) at A (0, 3)
− 7t − 6
>0
− 22
or
7t + 6 > 0
∴
t>−
6
7
Y′
and P , B must be on the same side of CA
value of ( x + 3y − 9 ) at P (t , t + 1)
∴
>0
value of ( x + 3y − 9 ) at B ( − 2, 0)
4t − 6
>0
−11
D
C(6,1)
X
B(–2,0) O
Y′
...(i)
X
B(–2,0) O
or
Y
A(0,3)
X′
C(6,1)
t + 3 ( t + 1) − 9
>0
−2+0−9
Aliter : First draw the exact diagram of ∆ ABC, the point
P (t , t + 1) move on the line
y = x +1
for all t.
P
P
i.e.
...(iii)
E
Y
A(0,3)
X′
or
∴
1
or
t +4
>0
22
t + 4>0
t>−4
or
x+
t − 8 ( t + 1) + 2
>0
0 − 24 + 2
...(ii)
3t − 2 ( t + 1) + 6
>0
18 − 2 + 6
i.e.
y=
i.e.
or
Now, D and E are the intersection of
y = x + 1, x − 8y + 2 = 0
and
y = x + 1, x + 3y − 9 = 0
respectively.
 6 1
∴
D ≡ − , 
 7 7
and
 3 5
E≡ , 
 2 2
Thus, the points on the line y = x + 1 whose x-coordinates
6
3
lies between − and lie within the triangle ABC.
7
2
6
3
Hence,
− <t <
7
2
 6 3
i.e.
t ∈ − , 
 7 2
Chap 02 The Straight Lines
y Example 50. Find λ if ( λ , 2) is an interior point of
∆ ABC formed by x + y = 4, 3x − 7 y = 8 and
4 x − y = 31.
Sol. Let P ≡ ( λ , 2)
First draw the exact diagram of ∆ ABC, the point P ( λ , 2)
move on the line y = 2 for all λ .
Y
A
B
=8
D
E
P
X
O
 5 7
 1 1
A  ,  , B ( − 7, 5) and C  , 
 4 8
3 9
Q P (α, α 2 ) lies inside the ∆ ABC, then
(i) A and P must lie on the same side of BC
(ii) B and P must lie on the same side of CA
(iii) C and P must lie on the same side of AB, then
5 21
+
−1
2 8
>0
2α + 3α 2 − 1
⇒
x+
4x–y
=31
X′
7y
3x–
y=
4
C
Y′
33
2α + 3α 2 − 1
>0
or
3α 2 + 2α − 1 > 0
⇒
1

( α + 1) α −  > 0

3
⇒
1 
α ∈ ( − ∞, − 1) ∪  , ∞
3 
and
− 35 − 30 − 1
5α − 6α 2 − 1
⇒
5α − 6α 2 − 1 < 0
and 4 x − y = 31, y = 2 respectively.
⇒
1 
1

α −  α −  > 0

3 
2
 22 
 33 
D ≡  , 2 and E ≡  , 2
3 
4 
∴
Thus, the points on the line y = 2 whose x-coordinates lies
22
33
between and lie within the ∆ ABC.
3
4
22
33
Hence,
<λ<
3
4
22
33


i.e.
λ ∈ , 
 3 4
and
⇒
...(i)
>0
Now, D and E are the intersection of
3x − 7y = 8, y = 2
∴
99
α ∈ ( − ∞, 1 / 3) ∪ (1 / 2, ∞ )
1 2
+ −3
3 9
>0
α + 2α 2 − 3
..(ii)
α + 2α 2 − 3 < 0
y Example 51. Determine all values of α for which the
point (α , α 2 ) lies inside the triangle formed by the lines
2x + 3y − 1 = 0, x + 2y − 3 = 0 and 5x − 6 y − 1 = 0.
⇒
(2α + 3) (α − 1) < 0
...(iii)
∴
α ∈ ( − 3 / 2, 1)
From Eq. (i), Eq. (ii) and Eq. (iii), we get
α ∈ ( −3 / 2, − 1) ∪ (1 / 2, 1).
Aliter : Let P (α, α 2 ) first draw the exact diagram of
∆ ABC.
The point P (α, α 2 ) move on the curve y = x 2 for all α.
Sol. The coordinates of the vertices are
Now,
and
or
Y
B
∴
intersection of y = x 2
2x + 3y − 1 = 0
2x + 3x 2 − 1 = 0
x = − 1, x =
1
3
Let intersection points
P
 1 1
D ≡ ( − 1, 1) and E ≡  , 
3 9
A
intersection of y = x 2
X′
C
O
Y′
X
and
or
x + 2y − 3 = 0
x + 2x 2 − 3 = 0
∴
x = 1, x = − 3 / 2
100
Textbook of Coordinate Geometry
Y
B
Dividing it by k, then
y=x2
ax + by +
G
F
or
A
D
I
C
X′
O
X
Y′
Let intersection points
 3 9
F ≡ (1, 1) and G ≡  − , 
 2 4
and intersection of y = x 2 and 5x − 6y − 1 = 0
or
5x − 6x 2 − 1 = 0
1
1
x = ,x =
3
2
Let intersection points
 1 1
1
H ≡  ,  and I ≡  ,
3 9
2
∴
1
.
4
Thus the points on the curve y = x 2 whose x-coordinate
1
lies between −3 / 2 & − 1 and & 1 lies within the triangle
2
ABC.
1
3
Hence,
− < α < − 1 or < α < 1
2
2
1 

 3
i.e.
α ∈  − , − 1 ∪  , 1
2 

 2
Equations of Lines Parallel and
Perpendicular to a Given Line
Theorem 1: The equation of line parallel to
ax + by + c = 0 is ax + by + λ = 0, where λ is some
constant.
Proof : Let the equation of any line parallel to
ax + by + c = 0
be
a 1 x + b 1y + c 1 = 0
a1 b1
then
=
=k
a
b
∴
a 1 = ak , b 1 = bk
Then from Eq. (ii),
akx + bky + c = 0
...(i)
...(ii)
(say)
c
=0
k
ax + by + λ = 0
c

 writing λ for 

k
Hence, any line parallel to ax + by + c = 0 is
ax + by + λ = 0
where λ is some constant.
Aliter : The given line is
ax + by + c = 0
a
Its slope = −
b
Thus, any line parallel to Eq. (i) is given by
 a
y =  −  x + λ1
 b
...(i)
⇒
ax + by − bλ 1 = 0
(writing λ for − bλ 1 )
⇒
ax + by + λ = 0
where, λ is some constant.
Corollary : The equation of the line parallel to
ax + by + c = 0 and passing through ( x 1 , y 1 ) is
a ( x − x 1 ) + b (y − y 1 ) = 0
Working Rule :
(i) Keep the terms containing x and y unaltered.
(ii) Change the constant.
(iii) The constant λ is determined from an additional
condition given in the problem.
Theorem 2 : The equation of the line perpendicular to the
line ax + by + c = 0 is
bx − ay + λ = 0, where λ is some constant.
Proof : Let the equation of any line perpendicular to
... (i)
ax + by + c = 0
be
...(ii)
a 1 x + b 1y + c 1 = 0
then
aa 1 + bb 1 = 0
or
aa 1 = − bb 1
a1 b1
(say)
⇒
=
=k
b −a
∴
a 1 = bk , b 1 = − ak
Then, from Eq. (ii), bkx − aky + c 1 = 0 dividing it by k, then
c
bx − ay + 1 = 0
k
Chap 02 The Straight Lines
or
bx − ay + λ = 0
c1 

 writing λ for 

k
Hence, any line perpendicular to ax + by + c = 0 is
bx − ay + λ = 0
where, λ is some constant.
Aliter : The given line is
ax + by + c = 0
a
Its slope = −
b
bx − ay + aλ 1 = 0
bx − ay + λ = 0
x −y + λ =0
...(i)
(writing λ for aλ 1 )
Working Rule :
(i) Interchange the coefficients of x and y and changing
sign of one of these coefficients.
(ii) Changing the constant term.
(iii) The value of λ can be determined from an additional
condition given in the problem.
y Example 52. Find the general equation of the line
which is parallel to 3x − 4 y + 5 = 0. Also find such line
through the point ( −1 , 2) .
Sol. Equation of any parallel to 3x − 4y + 5 = 0 is
∴
λ = 11
Then from Eq. (i) required line is
3x − 4y + 11 = 0
which is general equation of the line.
Also Eq. (i) passes through (1, 2), then
1−2+ λ =0
x −y +1=0
Corollary 1 : The equation of the line through ( x 1 , y 1 )
and perpendicular to ax + by + c = 0 is
b ( x − x 1 ) − a (y − y 1 ) = 0
Corollary 2 : Also equation of the line perpendicular to
ax + by + c = 0 is written as
x y
− + k = 0, where k is some constant.
a b
3 ( − 1) − 4 ( 2) + λ = 0
...(i)
∴
λ =1
Then from Eq. (i), required line is
where, λ is some constant.
3x − 4y + λ = 0
which is general equation of the line.
Also Eq. (i) passes through ( −1, 2), then
y Example 53. Find the general equation of the line
which is perpendicular to x + y + 4 = 0 . Also find such
line through the point (1, 2) .
Sol. Equation of any line perpendicular to x + y + 4 = 0 is
b
Slope of perpendicular line of Eq. (i) is .
a
Thus any line perpendicular to Eq. (i) is given by
b 
y =   x + λ1
a 
⇒
or
101
y Example 54. Show that the equation of the line
passing through the point (a cos 3 θ, a sin 3 θ ) and
perpendicular to the line
x sec θ + y cosec θ = a is
x cosθ − y sinθ = a cos2θ
Sol. The given equation x sec θ + y cosec θ = a can be written as
x sin θ + y cos θ = a sin θ cos θ
∴ equation of perpendicular line of Eq. (i) is
x cos θ − y sin θ = λ
Also it is pass through (a cos 3 θ, a sin 3 θ )
∴
⇒
...(ii)
a cos 3 θ ⋅ cos θ − a sin 3 θ ⋅ sin θ = λ
λ = a (cos 4 θ − sin 4 θ )
= a (cos 2 θ + sin 2 θ ) (cos 2 θ − sin 2 θ )
= a ⋅ 1 ⋅ cos 2θ = a cos 2θ
From Eq. (ii), the required equation of the line is
x cos θ − y sin θ = a cos 2θ
Aliter : (From corollary (2) of Theorem (2)
Equation of any line perpendicular to the line
x sec θ + y cosec θ = a , is
x
y
−
=k
sec θ cosec θ
or
x cos θ − y sin θ = k
Also, it pass through (a cos 3 θ, a sin 3 θ )
∴
or
...(i)
...(i)
a cos 3 θ ⋅ cos θ − a sin 3 θ ⋅ sin θ = k
k = a (cos 4 θ − sin 4 θ )
= a (cos 2 θ + sin 2 θ ) (cos 2 θ − sin 2 θ )
= a ⋅ 1 ⋅ cos 2θ
= a cos 2θ
From Eq. (iii), the required equation of the line is
x cos θ − y sin θ = a cos 2θ
...(iii)
102
Textbook of Coordinate Geometry
Distance of a Point From a Line
Aliter I : Let PM makes an angle θ with positive
direction of X-axis.
Theorem : The length of perpendicular from a point
( x 1 , y 1 ) to the line ax + by + c = 0 is
| ax 1 + by 1 + c |
Then, equation of PM in distance form will be
x − x1 y − y1
(QPM = p )
=
=p
cos θ
sin θ
Therefore coordinates of M will be
( x 1 + p cos θ, y 1 + p sin θ )
(a 2 + b 2 )
Proof : Given line is ax + by + c = 0
y
x
⇒
+
=1
 c  c
−  − 
 a  b
Since, M lies on ax + by + c = 0, then
a ( x 1 + p cos θ ) + b (y 1 + p sin θ ) + c = 0
or
…(iii)
p (a cos θ + b sin θ ) = − (ax 1 + by 1 + c )
a
Since, slope of AB = −
b
b
Slope of PM =
∴
a
b
∴
tan θ =
(Q PM makes an angle θ with
a
positive direction of X-axis)
Y
B(0, –c/b)
P(x1,y1)
M
A(–c/a,0)
X′
A
X
Y′
(a2+b 2 )
Let the given line intersects the X-axis and Y-axis at A and
 c 
B respectively, then coordinates of A and B are  − , 0 
 a 
c

and  0, −  respectively.

b
Draw PM perpendicular to AB .
Now, Area of ∆ PAB
=
θ
B
then
and
c c  c
1


x 1  0 +  −  − − y 1  + 0 (y 1 − 0 )




b
a
b
2
1 c 
=  | ax 1 + by 1 + c |
2 ab
1 c 
=   (a 2 + b 2 ) ⋅ p
2 ab
... (ii)
From Eqs. (i) and (ii), we have
1 c 
1

c 
(a 2 + b 2 ) ⋅ p =  | ax 1 + by 1 + c |
2 ab
2 ab
p=
| ax 1 + by 1 + c |
(a 2 + b 2 )
cos θ =
C
b
(a + b 2 )
2
a
(a + b 2 )
2


a2
b2
 = − (ax 1 + by 1 + c )
p
+
 (a 2 + b 2 )
2
2 
a
b
(
+
)


or
2
2
1  c
c  
1


= ⋅ AB ⋅ PM =
 − − 0  +  0 +   ⋅ p


2  a
b  
2
sin θ =
a
Now, from Eq. (iii),
...(i)
Let PM = p
Also, area of ∆ PAB
or
b
p=−
(ax 1 + by 1 + c )
(a 2 + b 2 )
Since, p is positive
∴
p=
| ax 1 + by 1 + c |
(a 2 + b 2 )
Aliter II : Let Q ( x , y ) be any point on the line
ax + by + c = 0
Hence, the length of perpendicular from P on AB will be
least value of PQ.
Chap 02 The Straight Lines
Q Least value of PQ is PM
Y
B
∴
(x,y)
M
O
X
A
Y′
z = ( PQ ) 2
Let
= ( x − x 1 ) 2 + (y − y 1 ) 2
...(iv)
 c ax

= (x − x 1 )2 +  − −
− y1 
 b b

2
⇒
(by law of proportion)
=
dz
 c ax
  a
= 2 (x − x 1 ) + 2  − −
− y1  − 
 b b
  b
dx
 a2 
d z
 a  a
 1 + 2  = positive
2
=
2
+
2
−
=
−



 b  b
 b 
dx 2
=
∴
dz
=0
dx
 c ax
  a
− y1  −  = 0
2 (x − x 1 ) + 2  − −
 b b
  b
 a
2 ( x − x 1 ) + 2 (y − y 1 )  −  = 0
 b
c ax 

Qy = − − 

b b 
( x − x 1 ) (y − y 1 ) a ( x − x 1 ) + b (y − y 1 )
=
=
a
b
a ⋅a + b ⋅b
(ax + by + c ) − (ax 1 + by 1 + c )
=
(a 2 + b 2 )
(by law of proportion)
0 − (ax 1 + by 1 + c )
(Qax + by + c = 0 )
=
(a 2 + b 2 )
⇒
and
(x − x 1 ) = −
(y − y 1 ) = −
a2 + b2
ax 1 + by 1 + c
[from Eq. (v)]
a2 + b2
2
For maximum or minimum,
or
(ax 1 + by 1 + c ) − (ah + bk + c )
( PM ) 2 = ( x − x 1 ) 2 + (y − y 1 ) 2
Q z is minimum
∴ PQ is also minimum.
or
(a 2 + b 2 )
x 1 − h y 1 − k a ( x 1 − h ) + b (y 1 − k )
=
=
a
b
a .a +b .b
2
and
| ax 1 + by 1 + c |
Aliter III : Let M ≡ (h, k )
Since, M (h, k ) lies on AB,
..(v)
ah + bk + c = 0
Now, AB and PM are perpendicular to each other, then
(slope of PM ) × (slope of AB) = − 1
y1 − k  a 
⇒
× −  = − 1
x1 −h  b
Q ax + by + c = 0 

c ax 

 ∴y = − −

b
b 
∴
p = PM =
P(x1,y1)
Q
X′
103
 ax + by + c 
 (a 2 + b 2 )
=  1 2 12

 a +b
∴ Length of perpendicular
(ax 1 + by 1 + c )
PM = ±
(a 2 + b 2 )
Hence,
PM = p =
| ax 1 + by 1 + c )
(a 2 + b 2 )
Aliter IV : Equation of AB in normal form is
−c
b
a
y=
x+
2
2
2
2
2
(a + b )
(a + b )
(a + b 2 )
⇒
c
OL = −
(a 2 + b 2 )
Y
a (ax 1 + by 1 + c )
B
(a 2 + b 2 )
P(x1,y1)
b (ax 1 + by 1 + c )
Q
(a 2 + b 2 )
M
L
∴ From Eq. (iv),
PQ = (ax 1 + by 1 + c ) 2
(a + b )
2
2
(a 2 + b 2 ) 2
=
| ax 1 + by 1 + c |
(a 2 + b 2 )
X′
α
O
Y′
A
X
104
Textbook of Coordinate Geometry
Equation of line parallel to AB and passes through ( x 1 , y 1 )
is
a ( x − x 1 ) + b (y − y 1 ) = 0
or
ax + by = ax 1 + by 1
Normal form is
ax 1 + by 1
b
a
y=
x+
(a 2 + b 2 )
(a 2 + b 2 )
(a 2 + b 2 )
ax 1 + by 1
⇒
OQ =
∴
PM = QL = OQ − OL =
(a 2 + b 2 )
ax 1 + by 1 + c
(a 2 + b 2 )
Hence, required perpendicular distance
| ax 1 + by 1 + c |
p=
(a 2 + b 2 )
Aliter V : The equation of line through P ( x 1 , y 1 ) and
perpendicular to ax + by + c = 0 is
...(vi)
b ( x − x 1 ) − a (y − y 1 ) = 0
If this perpendicular meet the line ax + by + c = 0 in
M ( x 2 , y 2 ) then ( x 2 , y 2 ) lie on both the lines
ax + by + c = 0 and Eq. (vi), then
b ( x 2 − x 1 ) − a (y 2 − y 1 ) = 0, ax 2 + by 2 + c = 0
ax 2 + by 2 + c = a ( x 2 − x 1 ) + b(y 2 − y 1 ) + ax 1 + by 1 + c = 0
or
...(vii)
b ( x 2 − x 1 ) − a (y 2 − y 1 ) = 0
and a ( x 2 − x 1 ) + b (y 2 − y 1 ) = − (ax 1 + by 1 + c ) …(viii)
On squaring and adding Eqs. (vii) and (viii), we get
(a 2 + b 2 ) (( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 ) = (ax 1 + by 1 + c ) 2
or
PM = ( x 2 − x 1 ) 2 + (y 2 − y 1 ) 2
=
| ax 1 + by 1 + c |
(a 2 + b 2 )
Hence, length of perpendicular
| ax 1 + by 1 + c |
PM = p =
(a 2 + b 2 )
Corollary 1 : The length of perpendicular from the origin
to the line ax + by + c = 0 is
| a ⋅0 + b ⋅0 + c |
| c|
i.e.
2
2
2
(a + b )
(a + b 2 )
Corollary 2 : The length of perpendicular from ( x 1 , y 1 ) to
the line x cos α + y sin α = p is
| x 1 cos α + y 1 sin α − p |
= | x 1 cos α + y 1 sin α − p |
(cos 2 α + sin2 α )
Working Rule :
(i) Put the point ( x 1 , y 1 ) for ( x , y ) on the LHS while the
RHS is zero.
(ii) Divide LHS after Eq. (i) by (a 2 + b 2 ) , where a and b
are the coefficients of x and y respectively.
y Example 55. Find the sum of the abscissas of all the
points on the line x + y = 4 that lie at a unit distance
from the line 4 x + 3y − 10 = 0.
Sol. Any point on the line x + y = 4 can be taken as ( x 1, 4 − x 1 ).
As it is at a unit distance from the line 4 x + 3y − 10 = 0, we
get
| 4 x 1 + 3 ( 4 − x 1 ) − 10 |
=1
( 4 2 + 32 )
⇒
| x1 + 2 | = 5 ⇒ x1 + 2 = ± 5
⇒
x 1 = 3 or −7
∴ Required sum = 3 − 7 = − 4.
y Example 56. If p and p ′ are the length of the
perpendiculars from the origin to the straight lines
whose equations are x sec θ + y cos ec θ = a and
x cos θ − y sin θ = a cos 2θ , then find the value of
4p 2 + p ′ 2 .
Sol. We have, p =
∴
⇒
and
∴
|−a |
(sec θ + cosec 2θ )
2
p2 =
a2
sec θ + cosec θ
2
2
=
a 2 sin 2 θ cos 2 θ
1
4 p = a 2 sin 2 2θ
2
p′ =
…(i)
| − a cos 2θ |
(cos 2 θ + sin 2 θ )
= | − a cos 2θ |
( p ′ )2 = a 2 cos 2 2θ
…(ii)
∴ Adding Eqs. (i) and (ii), we get
4 p 2 + p ′2 = a2
y Example 57. If p is the length of the perpendicular
x y
from the origin to the line + = 1, then prove that
a b
1
1
1
+
= .
a2 b2 p2
Sol. p = length of perpendicular from origin to
x y
+ =1
a b
| 0 + 0 − 1|
1
=
=
2
2
1
1

 1
 1
 2 + 2
  + 
a
b 
b
 a
or
1
p
2
=
1
a
2
+
1
b
2
or
1
a
2
+
1
b
2
=
1
p2
Chap 02 The Straight Lines
y Example 58. Prove that no line can be drawn
through the point (4, − 5) so that its distance from
( −2 , 3) will be equal to 12.
y Example 59. Find the distance between the lines
5x − 12y + 2 = 0 and 5x − 12y − 3 = 0.
Sol. The distance between the lines
5x − 12y + 2 = 0 and 5x − 12y − 3 = 0 is
| 2 − ( − 3) |
5
=
2
2
13
(5) + ( −12)
Sol. Suppose, if possible.
Equation of line through ( 4, − 5) with slope of m is
y + 5 = m (x − 4)
⇒
mx − y − 4m − 5 = 0
| m ( − 2) − 3 − 4m − 5 |
Then,
= 12
m2 + 1
⇒
Aliter I : The constant term in both equations are 2 and −3
which are of opposite sign. Hence origin lies between them.
5
| 2| + | − 3|
=
∴ Distance between lines is
2
2
13
(5) + ( −12)
| − 6m − 8 | = 12 (m 2 + 1)
Aliter II : Putting y =0 in 5x − 12y − 3 = 0 then x =
On squaring, (6m + 8)2 = 144 (m 2 + 1)
⇒
4 (3m + 4 )2 = 144 (m 2 + 1)
⇒
(3m + 4 ) = 36 (m + 1)
⇒
2
∴
2
27m 2 − 24m + 20 = 0
105
...(i)
Since, the discriminant of Eq. (i) is ( −24 )2 − 4 ⋅27 ⋅ 20 = − 1584
which is negative, there is no real value of m . Hence no
such line is possible.
Distance between Two
Parallel Lines
Let the two parallel lines be
ax + by + c = 0 and ax + by + c 1 = 0
The distance between the parallel lines is the perpendicular
distance of any point on one line from the other line.
Let ( x 1 , y 1 ) be any point on ax + by + c = 0
...(i)
∴
ax 1 + by 1 + c = 0
Now, perpendicular distance of the point ( x 1 , y 1 ) from the
line ax + by + c 1 = 0 is
| ax 1 + by 1 + c 1 |
| c 1 − c|
[from Eq. (i)]
=
2
2
(a + b )
(a 2 + b 2 )
This is required distance between the given parallel lines.
Aliter I : The distance between the lines is
λ
d=
(a 2 + b 2 )
(i) λ = | c 1 − c |, if both the lines are on the same side of
the origin.
(ii) λ = | c 1 | + | c |, if the lines are on the opposite side of the
origin.
Aliter II : Find the coordinates of any point on one of
the given lines, preferably putting x = 0 or y = 0. Then the
perpendicular distance of this point from the other line is
the required distance between the lines.
3
5
3 
 , 0 lie on 5x − 12y − 3 = 0
5 
Hence, distance between the lines
5x − 12y + 2 = 0 and (5x − 12y − 3 = 0)
3 
= Distance from  , 0 to the line 5x − 12y + 2 = 0
5 
3
5× −0+2
5
5
=
=
2
2
13
5 + ( −12)
y Example 60. Find the equations of the line parallel to
5x − 12y + 26 = 0 and at a distance of 4 units from it.
Sol. Equation of any line parallel to 5x − 12y + 26 = 0 is
...(i)
5x − 12y + λ = 0
Since, the distance between the parallel lines is 4 units, then
| λ − 26 |
=4
(5)2 + ( −12)2
or
| λ − 26 | = 52 or λ − 26 = ± 52
or
λ = 26 ± 52
∴ λ = − 26 or 78
Substituting the values of λ in Eq. (i), we get
5x − 12y − 26 = 0
and 5x − 12y + 78 = 0
Area of Parallelogram
Theorem : Area of parallelogram ABCD whose sides
AB, BC , CD and DA are represented by a 1 x + b 1 y + c 1 = 0,
a 2 x + b 2 y + c 2 = 0, a 1 x + b 1 y + d 1 = 0
and a 2 x + b 2 y + d 2 = 0 is
p1 p2
| c1 − d 1| | c2 − d 2|
or
sin θ
a1 b1
|
|
a2 b2
where, p 1 and p 2 are the distances between parallel sides
and θ is the angle between two adjacent sides.
106
Textbook of Coordinate Geometry
Proof : Since, p 1 and p 1 are the distances between the pairs
of parallel sides of the parallelogram and θ is the angle
between two adjacent sides, then
a1x+b1y+d1=0
A
0
2 y+
c2 =
C
p1
θ
a1x+b1y+c1=0
M
θ
B
Corollaries :
1. If p 1 = p 2 , then ABCD becomes a rhombus
p2
L
Area of parallelogram ABCD
= 2 × Area of ∆ ABD
1
= 2 × × AB × p 1
2
= AB × p 1
p
p2 

= 2 × p1
Q in ∆ ABL, sin θ =


sin θ
AB 
=
p1 p2
sin θ
…(i)
Now, p 1 = Distance between parallel sides AB and DC
|c1 − d 1 |
=
(a 12 + b 12 )
and p 2 = Distance between parallel sides AD and BC
|c2 − d 2 |
=
(a 22 + b 22 )
m − m 2 
=
Also, tan θ = 1
1 + m 1m 2
−
a1  a2 
− − 
b1  b2 
 a  a 
1 + − 1  − 2 
 b1   b2 
a1 

 Q m 1 = slope of AB = − 
b1 


a2 
 and m 2 = slope of AD = − 

b2 
a b − a 2 b 1

= 1 2
a 1 a 2 + b 1 b 2
∴
sin θ =
|c1 − d 1 | |c2 − d 2 |
| a 1b 2 − a 2b 1 |
|c − d 1 | |c2 − d 2 |
= 1
a1 b1
|
|
a2 b2
∴ Area of parallelogram ABCD =
a2 x
+b
a2 x
+b
2y
+d
2 =0
D
Now, substitute the values of p 1 , p 2 and sin θ in Eq. (i)
| a 1b 2 − a 2b 1 |
(a 12 + b 12 )(a 22 + b 22 )
∴ Area of rhombus ABCD =
=
p 12
sin θ
(c 1 − d 1 ) 2
 a 2 + b 12 

| a 1 b 2 − a 2 b 1 |  12
 a 2 + b 22 
2. If d 1 and d 2 are the lengths of two perpendicular
diagonals of a rhombus, then
1
Area of rhombus = d 1 d 2
2
3. Area of the parallelogram whose sides are y = mx + a,
|a − b | |c − d |
.
y = mx + b, y = nx + c and y = nx + d is
|m − n|
y Example 61. Show that the area of the parallelogram
formed by the lines x + 3y − a = 0, 3x − 2y + 3a = 0,
20 2
a sq units.
x + 3y + 4a = 0 and 3x − 2y + 7a = 0 is
11
Sol. Required area of the parallelogram
| −a − 4a| | 3a − 7a | 20 2
=
= a sq units
1 3
11
|
|
3 –2
y Example 62. Show that the area of the parallelogram
formed by the lines
x cos α + y sin α = p, x cos α + y sin α = q ,
x cos β + y sin β = r, x cos β + y sin β = s is
|(p − q)(r − s) cosec (α − β)|.
Sol. The equation of sides of the parallelogram are
x cos α + y sin α − p = 0,
x cos α + y sin α − q = 0,
x cos β + y sin β − r = 0,
and
x cos β + y sin β − s = 0
Chap 02 The Straight Lines
∴ Required area of the parallelogram
| − p − ( −q )| | − r − ( −s )| | p − q | | r − s |
=
=
cos α sin α
| sin(β − α )|
|
|
cos β sin β
∴
|( −1 + 2)( −1 + 2)|
1 1
a
b |
|
1
1
b a
Area of the rhombus =
= |( p − q )(r − s )cosec (α − β )|
=
y Example 63. Prove that the diagonals of the
parallelogram formed by the lines
x y
x y
x y
x y
+ = 1, + = 1, + = 2 and + = 2
a b
b a
a b
b a
are at right angles. Also find its area (a ≠ b).
Sol. The distance between the parallel sides
x y
x y
+ = 1 and
+ =2
a b
a b
|2 − 1|
1
is
= p1
=
1
1
1
1
 2 + 2
 2 + 2
a
a
b 
b 
a 2b 2
|b 2 − a 2 |
(a ≠ b )
y Example 64. Show that the four lines ax ± by ± c = 0
2c 2
enclose a rhombus whose area is
.
| ab |
(say)
and the distance between the parallel sides
x y
x y
+ = 1 and
+ =2
b a
b a
|2 − 1|
1
is
= p2
=
1
1
1
1
 2 + 2
 2 + 2
a
b
a 
b 
107
(say)
Here,
p1 = p 2 .
∴ Parallelogram is a rhombus.
But we know that diagonals of rhombus are perpendicular
to each other.
Sol. The given lines are
…(i)
ax + by + c = 0
…(ii)
ax + by − c = 0
…(iii)
ax − by + c = 0
and
…(iv)
ax − by − c = 0
Distance between the parallel lines Eqs. (i) and (ii) is
2c
= p 1 (say) and distance between the parallel
2
(a + b 2 )
lines Eqs. (iii) and (iv) is
2c
(say)
= p2
2
(a + b 2 )
Here, p 1 = p 2
∴ it is a rhombus.
∴ Area of rhombus =
4c 2
|(c + c )(c + c )|
2c 2
=
=
| − 2ab | | ab |

a b
|
|
a −b
Exercise for Session 2
1.
The number of lines that are parallel to 2x + 6y − 7 = 0 and have an intercept 10 between the coordinate axes is
(a) 1
2.
7
2
(b)
7
5
(c)
7
10
(d)
9
10
(b) (–1, 1)
(c) (–1, –1)
(d) (1, –1)
If the quadrilateral formed by the lines ax + by + c = 0, a′ x + b ′ y + c′ = 0, ax + by + c′ = 0 and
a′ x + b ′ y + c′ = 0 have perpendicular diagonals, then
(a) b 2 + c 2 = b ′2 + c ′2
5.
(d) infinitely many
If the algebraic sum of the perpendicular distances from the points (2, 0), (0, 2) and (1, 1) to a variable straight
line is zero, then the line passes through the point
(a) (1, 1)
4.
(c) 4
The distance between the lines 4x + 3y = 11 and 8x + 6y = 15 is
(a)
3.
(b) 2
(b) c 2 + a 2 = c ′2 + a ′2
(c) a 2 + b 2 = a ′2 + b ′2
(d) None of these
The area of the parallelogram formed by the lines 3x − 4y + 1 = 0, 3x − 4y + 3 = 0, 4x − 3y − 1 = 0 and
4x − 3y − 2 = 0, is
(a)
1
sq units
7
(b)
2
sq units
7
(c)
3
sq units
7
(d)
4
sq units
7
108
Textbook of Coordinate Geometry
6.
Area of the parallelogram formed by the lines y = mx , y = mx + 1, y = nx and y = nx + 1equals
(a)
7.
|m + n |
(m − n )2
2
3
3 3
(b)  , 
 2 2
(b)
2
3
(b) a = 7
(b)
1
2
8 8
(c)  − , − 
 7 7
1
|m − n |
32 32
(d)  , − 
 7
7
(c) −
4
3
(d)
(c) a = 11
4
3
(d) a < 7 or a > 11
(c) 1
(d) 2
The points on the axis of x, whose perpendicular distance from the straight line
b
(a ± (a 2 + b 2 ), 0)
a
b
(c) (a + b , 0)
a
a
(b ±
b
a
(d) (a ±
b
(a)
12.
(d)
The lines y = mx , y + 2x = 0, y = 2x + k and y + mx = k form a rhombus if m equals
(a) –1
11.
1
|m + n |
If the points (1, 2) and (3, 4) were to be on the opposite side of the line 3x − 5y + a = 0, then
(a) 7 < a < 11
10.
(c)
A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3, then its y-intercept is
(a) −
9.
2
|m + n |
The coordinates of a point on the line y = x where perpendicular distance from the line 3x + 4y = 12 is 4 units,
are
3 5
(a)  , 
 7 7
8.
(b)
(b)
x y
+ = 1is a
a b
(a 2 + b 2 ), 0)
(a 2 + b 2 ), 0)
The three sides of a triangle are given by ( x 2 − y 2 ) (2x + 3y − 6) = 0. If the point ( −2, a ) lies inside and (b ,1) lies
outside the triangle, then
10
(a) a ∈  2,  ; b ∈ (−1, 1)
 3
10
(b) a ∈  −2,  ; b ∈  −1,


3
10
(c) a ∈  1,  ; b ∈ (−3, 5)
 3
(d) None of these
9

2
13.
Are the points (3, 4) and (2, –6) on the same or opposite sides of the line 3x − 4y = 8 ?
14.
If the points (4, 7) and (cos θ, sin θ), where 0 < θ < π, lie on the same side of the line x + y − 1 = 0, then prove
that θ lies in the first quadrant.
15.
Find the equations of lines parallel to 3x − 4y − 5 = 0 at a unit distance from it.
16.
Show that the area of the parallelogram formed by the lines 2x − 3y + a = 0, 3x − 2y − a = 0,2x − 3y + 3a = 0
and 3x − 2y − 2a = 0 is
2a 2
sq units.
5
17.
A line ‘L’ is drawn from P (4, 3) to meet the lines L1 : 3x + 4y + 5 = 0 and L2 : 3x + 4y + 15 = 0 at point A and B
respectively. From ‘A’ a line, perpendicular to L is drawn meeting the line L2 at A1. Similarly from point ‘B’ a line,
perpendicular to L is drawn meeting the line L1 at B1. Thus a parallelogram AABB
1
1 is formed. Find the
equation (s) of ‘L’ so that the area of the parallelogram AABB
1
1 is least.
18.
The vertices of a ∆OBC are O (0, 0), B ( −3, − 1), C ( −1, − 3). Find the equation of the line parallel to BC and
1
intersecting the sides OB and OC and whose perpendicular distance from the origin is .
2
Session 3
Point of Intersection of Two Lines, Concurrent Lines
Family of Lines, How to Find Circumcentre and
Orthocentre by Slopes
Points of Intersection of
Two Lines
Let a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 be two
non-parallel lines. If ( x 1 , y 1 ) be the coordinates of their
point of intersection,
then a 1 x 1 + b 1 y 1 + c 1 = 0 and a 2 x 1 + b 2 y 1 + c 2 = 0
Solving these two by cross multiplication, then
y1
x1
1
=
=
b 1c 2 − b 2c 1 c 1a 2 − c 2a 1 a 1b 2 − a 2b 1
we get
 b c − b 2c 1 c 1a 2 − c 2a 1 
(x 1 , y 1 ) ≡  1 2
,

 a 1b 2 − a 2b 1 a 1b 2 − a 2b 1 


≡



b1
c1
b2
c2
a1
a2
b1
b2
,
c1
a1
a1
b1
c2 

a2 
a2 

b2 
Remarks
1. Here lines are not parallel, they have unequal slopes, then,
a1b2 − a2b1 ≠ 0
2. In solving numerical questions, we should not be remember
the coordinates ( x1, y1 ) given above, but we solve the equations
directly.
Concurrent Lines
The three given lines are concurrent, if they meet in a
point. Hence to prove that three given lines are
concurrent, we proceed as follows :
I Method : Find the point of intersection of any two lines
by solving them simultaneously. If this point satisfies the
third equation also, then the given lines are concurrent.
II Method : The three lines a i x + b i y + c i = 0, i = 1 , 2 , 3 are
concurrent if
a 1 b1 c 1 
a 2 b 2 c 2 = 0


 a 3 b 3 c 3
III Method : The condition for the lines P = 0, Q = 0 and
R = 0 to be concurrent is that three constants l, m, n (not all
zeros at the same time) can be obtained such that
lP + mQ + nR = 0
Remarks
1. The reader is advised to follow method I in numerical problems.
2. For finding unknown quantity applying method II.
y Example 65. Show that the lines
2 x + 3 y − 8 = 0 , x − 5 y + 9 = 0 and 3 x + 4y − 11 = 0
are concurrent.
Sol. I Method : Solving the first two equations, we see that
their point of intersection is (1, 2) which also satisfies the
third equation
3 × 1 + 4 × 2 − 11 = 0
Hence the given lines are concurrent.
a1 b1 c 1 2 3
8
II Method : We have a 2 b 2 c 2 = 1 −5 9
a 3 b 3 c 3 3 4 −11
Applying C 3 → C 3 + C 1 + 2C 2
2 3 0
= 1 −5
0 =0
3 4 0
Hence the given lines are concurrent.
III Method : Suppose
l (2x + 3y − 8) + m ( x − 5y + 9 ) +n (3x + 4y − 11) = 0
⇒ x (2l + m + 3n ) + y (3l − 5m + 4n ) + ( −8l + 9m − 11n ) = 0
= 0 ⋅ x + 0 ⋅y + 0
On comparing,
2l + m + 3n = 0, 3l − 5m + 4n = 0, − 8l + 9m − 11n = 0
After solving, we get l = 19, m = 1, n = − 13
Hence, 19 (2x + 3y − 8) + ( x − 5y + 9 ) − 13 (3x + 4y − 11) = 0
Hence the given lines are concurrent.
110
Textbook of Coordinate Geometry
y Example 66. If the lines ax + y + 1 = 0 , x + by + 1 = 0
and x + y + c = 0 (a, b and c being distinct and
different from 1) are concurrent, then find the value of
1
1
1
.
+
+
1 − a 1 −b 1 − c
Sol. The given lines are concurrent, then
a 1 1
a 1−a 1−a
1 b 1 =0 ⇒ 1 b −1
0 =0
1 1 c
1
0
c −1
(applying C 2 → C 2 − C 1 and C 3 → C 3 − C 1 )
Expanding along first row
⇒
a ( b − 1) ( c − 1) − ( 1 − a ) ( c − 1) − ( 1 − a ) ( b − 1) = 0
⇒
a (1 − b ) (1 − c ) + (1 − a ) (1 − c ) + (1 − a ) (1 − b ) = 0
Dividing by (1 − a ) (1 − b ) (1 − c ), then
a
1
1
+
+
=0
1−a 1−b 1−c
⇒
Hence,
−1+
1
1
1
+
+
=0
1−a 1−b 1−c
1
1
1
+
+
=1
1−a 1−b 1−c
y Example 67. Show that the three straight lines
2x − 3y + 5 = 0 ,3x + 4 y − 7 = 0 and 9 x − 5y + 8 = 0
meet in a point.
Sol. If we multiply these three equations by 3, 1 and −1, we have
3 (2x − 3y + 5) + (3x + 4y − 7 ) − (9 x − 5y + 8) = 0
which is an identity.
Hence, three lines meet in a point.
Family of Lines
Theorem : Any line through the point of intersection of
the lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 can be
represented by the equation
(a 1 x + b 1 y + c 1 ) + λ (a 2 x + b 2 y + c 2 ) = 0
where λ is a parameter which depends on the other
property of line.
Proof : The equations of the lines are
...(i)
a 1 x + b 1y + c 1 = 0
and
...(ii)
a 2 x + b 2y + c 2 = 0
Multiplying µ and ν in Eqs. (i) and (ii) and adding, we get
µ (a 1 x + b 1 y + c 1 ) + ν (a 2 x + b 2 y + c 2 ) = 0
where µ, ν are any constants not both zero.
Dividing both sides by µ, then
ν
(a 1 x + b 1 y + c 1 ) + (a 2 x + b 2 y + c 2 ) = 0
µ

ν
⇒ (a 1 x + b 1 y + c 1 ) + λ (a 2 x + b 2 y + c 2 ) = 0  where, λ = 
µ

It is a first degree equation in x and y. So it represents
family of lines through the point of intersection of Eqs. (i)
and (ii).
Thus, the family of straight lines through the intersection
of lines
L1 ≡ a 1 x + b 1 y + c 1 = 0
and L2 ≡ a 2 x + b 2 y + c 2 = 0 is
(a 1 x + b 1 y + c 1 ) + λ (a 2 x + b 2 y + c 2 ) = 0
i.e.
L1 + λL2 = 0
Corollaries :
1. The equation L1 + λL2 = 0 or µL1 + νL2 = 0 represent
a line passing through the intersection of the lines
L1 = 0 and L2 = 0 which is a fixed point, where λ, µ, ν
are constants.
2. For finding fixed point, the number of constants in
family of lines are one or two. If number of constants
more than two, then convert in two or one constant
form.
y Example 68. Find the equation of the straight line
passing through the point (2 , 1) and through the point
of intersection of the lines x + 2y = 3 and 2x − 3y = 4.
Sol. Equation of any straight line passing through the
intersection of the lines x + 2y = 3 and 2x − 3y = 4 is
λ ( x + 2y − 3) + (2x − 3y − 4 ) = 0
Since, it passes through the point (2, 1)
∴
λ ( 2 + 2 − 3) + ( 4 − 3 − 4 ) = 0
⇒
λ −3=0
∴
λ =3
Now, substituting this value of λ in (i), we get
3 ( x + 2y − 3) + (2x − 3y − 4 ) = 0
i.e.
5x + 3y − 13 = 0
which is the equation of required line.
...(i)
y Example 69. The family of lines
x (a + 2b ) + y (a + 3b ) = a + b passes through the point
for all values of a and b. Find the point.
Sol. The given equation can be written as
a ( x + y − 1) + b (2x + 3y − 1) = 0
which is equation of a line passing through the point of
intersection of the lines x + y − 1 = 0 and 2x + 3y − 1 = 0.
The point of intersection of these lines is (2, − 1). Hence the
given family of lines passes through the point (2, − 1) for all
values of a and b.
Chap 02 The Straight Lines
y Example 70. If 3a + 2b + 6c = 0 the family of straight
lines ax + by + c = 0 passes through a fixed point. Find
the coordinates of fixed point.
Sol. Given, 3a + 2b + 6c = 0
a b
+ +c =0
2 3
and family of straight lines is
ax + by + c = 0
Subtracting Eqs. (i) from (ii), then
1
1


a  x −  + b y −  = 0



2
3
...(i)
or
...(ii)
(a) Line Eq. (i) is to be parallel to 7 x + 2y − 5 = 0
( 1 + 3λ )
7
then
=−
−
( 5 + 2λ )
2
2 + 6λ = 35 + 14 λ
⇒
8λ = − 33
33
⇒
λ=−
8
Substituting this value of λ in Eq. (i), we get the required
equation as 7 x + 2y − 17 = 0
(b) Line, (i) is to be perpendicular to 7 x + 2y − 5 = 0
∴
y Example 71. If 4a 2 + 9b 2 − c 2 + 12ab = 0, then the
family of straight lines ax + by + c = 0 is either
concurrent at ... or at ....
(2a + 3b )2 − c 2 = 0
...(i)
...(ii)
⇒
a ( x ± 2) + b ( y ± 3) = 0
Taking ‘+’ sign : a ( x + 2) + b (y + 3) = 0
which is equation of a line passing through the point of
intersection of the lines x + 2 = 0 and y + 3 = 0
∴ coordinates of fixed point are ( −2, − 3) .
Taking ‘–’ sign : a ( x − 2) + b (y − 3) = 0
which is equation of a line passing through the point of
intersection of the lines
x − 2 = 0 and y − 3 = 0
∴ coordinates of fixed point are (2, 3)
Hence, the family of straight lines ax + by + c = 0 is either
concurrent at ( −2, − 3) or at (2, 3) .
y Example 72. Find the equation of the line passing
through the point of intersection of the lines
x + 5y + 7 = 0, 3x + 2y − 5 = 0
and (a) parallel to the line 7 x + 2 y − 5 = 0
(b) perpendicular to the line 7 x + 2 y − 5 = 0
−
( 1 + 3λ )
7
×−   = −1
 2
( 5 + 2λ )
7 + 21 λ = − 10 − 4 λ
17
∴
λ=−
25
Substituting this value of λ is Eq. (i), we get the required
equation as
2x − 7y − 20 = 0.
Aliter :
The point of intersection of the given lines
or
Sol. Given, 4a 2 + 9b 2 − c 2 + 12ab = 0
or
c = ± (2a + 3b )
and family of straight lines is
ax + by + c = 0
Substituting the value of c from Eqs. (i) in (ii), then
ax + by ± (2a + 3b ) = 0
Sol. Any line passing through the point of intersection of the
given lines is
( x + 5y + 7 ) + λ (3x + 2y − 5) = 0
...(i)
⇒
x ( 1 + 3λ ) + y ( 5 + 2λ ) + ( 7 − 5λ ) = 0
(1 + 3λ)
Its slope = −
( 5 + 2λ )
⇒
which is equation of a line passing through the point of
intersection of the lines
1
1
x − = 0 and y − = 0
2
3
 1 1
∴ The coordinates of fixed point are  ,  .
 2 3
or
111
x + 5y − 7 = 0 and 3x + 2y − 5 = 0 is (3, − 2).
∴ Equation of line through (3, − 2) is
y + 2 = m ( x − 3)
(a) Line (ii) is parallel to 7 x + 2y − 5 = 0
7
∴
m=−
2
Hence, the equation of the required line is
7
y + 2 = − ( x − 3)
2
or
7 x + 2y − 17 = 0
(b) Line (ii) is perpendicular to 7 x + 2y − 5 = 0
 7
then m ×  −  = − 1
 2
2
7
Hence, the equation of the required line is
2
y + 2 = ( x − 3)
7
or
2x − 7y − 20 = 0
or
m=
...(ii)
112
Textbook of Coordinate Geometry
y Example 73. Find the equation of straight line which
passes through the intersection of the straight lines
3 x − 4y + 1 = 0 and 5 x + y − 1 = 0
and cuts off equal intercepts from the axes.
Solution : Equation of any line passing through the
intersection of the given lines is
( 3x − 4y + 1) + λ ( 5x + y − 1) = 0
⇒
x ( 3 + 5λ ) + y ( − 4 + λ ) + ( 1 − λ ) = 0
x
y
⇒
=1
+
 λ − 1   λ − 1

 

 3 + 5λ   λ − 4 
...(i)
Sol. Any line through the point of intersection of given lines is

x y

x y
 + − 1 + λ  + − 1 = 0

b a

a b
1 λ
1 λ
x  +  + y  +  = (1 + λ )
b a 
a b 
but given x-intercept = y-intercept
 λ − 1   λ − 1
i.e.


 =
 3 + 5λ   λ − 4 
⇒
y Example 75. A variable straight line through the
x y
point of intersection of the lines + = 1 and
a b
x y
+ = 1 meets the coordinate axes in A and B. Show
b a
that the locus of the mid-point of AB is the curve
2xy (a + b ) = ab ( x + y ).
⇒
1
1
=
3 + 5λ λ − 4
⇒
(λ ≠ 1Qif λ = 1 then line (i) pass through origin)
∴
λ − 4 = 3 + 5λ
or
4λ = − 7
7
∴
λ=−
4
Substituting the value of λ in Eq. (i), we get required
equation is 23x + 23y = 11.
y Example 74. If t 1 and t 2 are roots of the equation
t 2 + λt + 1 = 0, where λ is an arbitrary constant. Then
prove that the line joining the points (at 12 , 2at 1 ) and
(at 22 , 2at 2 ) always passes through a fixed point. Also
find that point.
Sol. Qt 1 and t 2 are the roots of the equation t 2 + λt + 1 = 0
∴
t 1 + t 2 = − λ and t 1t 2 = 1
Equation of the line joining the points (at 12 , 2at 1 ) and
(at 22 , 2at 2 ) is
y − 2at 1 =
⇒
2at 2 − 2at 1
at 22
−
at 12
( x − at 12 )
y (t 1 + t 2 ) − 2at 1t 2 − 2at 12 = 2x − 2at 12
⇒
⇒
y (t 1 + t 2 ) − 2at 1t 2 = 2x
y ( −λ ) − 2a = 2x
y 
( x + a) + λ   = 0
 2
Y
B
M(x1,y1)
X′
A
X
This meets the X-axis at
 ab (1 + λ ) 
, 0
A≡

 b + aλ
and meets the Y-axis at
 ab (1 + λ )
B ≡ 0,


a + bλ 
Let the mid-point of AB is M ( x 1, y1 ), then
ab (1 + λ )
ab (1 + λ )
and y1 =
x1 =
2 (b + aλ )
2 (a + bλ )
[from Eq. (i)]
which is equation of a line passing through the point of
y
intersection of the lines x + a = 0 and = 0.
2
∴ coordinates of fixed point are ( −a , 0) .
O
Y′
∴
2
y − 2at 1 =
( x − at 12 )
(t 2 + t 1 )
⇒
or
...(i)
 b + aλ 
 a + bλ 
x
 +y 
 = (1 + λ )
 ab 
 ab 
x
y
+
=1
 ab (1 + λ )  ab (1 + λ )

 

 b + aλ   a + bλ 
⇒
1
1 2 (b + aλ ) 2 (a + bλ )
+
+
=
x 1 y1 ab (1 + λ ) ab (1 + λ )
=
2
(b + aλ + a + bλ )
ab (1 + λ )
=
2
(a + b )(1 + λ )
ab (1 + λ )
( x 1 + y 1 ) 2 (a + b )
=
x 1y1
ab
⇒
2x 1y1 (a + b ) = ab ( x 1 + y1 )
Hence, the locus of mid point of AB is
2xy (a + b ) = ab ( x + y ) .
Chap 02 The Straight Lines
How to Find Circumcentre and
Orthocentre by Slopes
(i) Circumcentre
The circumcentre of a triangle is the point of intersection
of the perpendicular bisectors of the sides of a triangle. It
is the centre of the circle which passes through the
vertices of the triangle and so its distance from the
vertices of the triangle is the same and this distance is
known as the circumradius of the triangle.
A(x1,y1)
E
F
O
(x2,y2)B
D
C(x3,y3)
Let O ( x , y ) be the circumcentre.
If D , E and F are the mid points of BC , CA and AB
respectively and OD ⊥ BC , OE ⊥ CA and OF ⊥ AB
slope of OD × slope of BC = − 1
∴
and
slope of OE × slope of CA = − 1
and
slope of OF × slope of AB = − 1
Solving any two, we get ( x , y ).
y Example 76. Find the coordinates of the
circumcentre of the triangle whose vertices are
A ( 5 , − 1), B ( −1, 5) and C (6, 6 ) . Find its radius also.
Sol. Let circumcentre be O ′ (α, β ) and mid points of sides
 5 11
 11 5
BC , CA and AB are D  ,  , E  ,  and F (2, 2)
2 2 
 2 2
respectively. Since O ′ D ⊥ BC .
∴ Slope of O ′ D × slope of BC = − 1
11
β−
2 × 6−5 = −1
⇒
5 6+1
α−
2
Y
(–1,5)B
(α,β) O'
E
X
O
A(5,–1)
Y′
⇒
2β − 11 = − 14α + 35
⇒
14α + 2β = 46
∴
7α + β = 23
and
O ′ E ⊥ CA
∴ Slope of O ′ E × Slope of CA = − 1
5
β−
2 × −1 − 6 = − 1
⇒
11
5−6
α−
2
2β − 5 7
⇒
× = −1
2α − 11 1
…(i)
⇒
14 β − 35 = − 2α + 11
∴
2α + 14 β = 46
∴
α + 7β = 23
Solving Eqs. (i) and (ii), we get
23
α =β =
8
23
23

∴ Circumcentre =  , 
 8 8
... (ii)
∴ Circumradius = O ′ A = O ′ B = O ′ C
= O ′ C = ( α − 6) 2 + ( β − 6) 2
2

 23

 23
=  − 6 +  − 6

8

8
2
2
2
25 2
 25
 25
units.
=   +  =
 8
 8
8
(ii) Orthocentre
The orthocentre of a triangle is the point of intersection of
altitudes.
A
F
E
O
B
D
C
Remarks
F
X′
2β − 11
= −1
7 (2α − 5)
⇒
Here O is the orthocentre since AD ⊥ BC , BE ⊥ CA and
CF ⊥ AB, then OA ⊥ BC, OB ⊥ CA, and OC ⊥ AB
Solving any two we can get coordinates of O.
C(6,6)
D
113
1. If any two lines out of three lines i.e. AB, BC and CAare
perpendicular, then orthocentre is the point of intersection of
two perpendicular lines.
2. Firstly find the slope of lines BC, CA and AB.
114
Textbook of Coordinate Geometry
y Example 77. Find the orthocentre of the triangle
formed by the lines xy = 0 and x + y = 1.
F
(α,β) B
D
B(2,3)
C(4,3)
E
X′
A(1,2)
O
Y′
∴
X
Slope of AB =
3−2
=1
2−1
Let orthocentre be O ′(α, β ) then
Slope of O ′ A × slope of BC = − 1
2−β
×0= −1
1−α
0
⇒
= −1
1−α
⇒
∴
and
⇒
0
m=
13
61
…(i)
⇒
4β − 4 = 5α − 5
⇒
5α − 4β − 1 = 0
Solving Eqs. (i) and (ii), we get
α = − 13 and β = −
1−α =0
α =1
Slope of OB × slope of CA = − 1
3−β 1
× = −1
2−α 3
⇒
3 − β = 3α − 6
⇒
3α + β = 9
∴
β =6
Hence, orthocentre of the given triangle is (1, 6) .
C
Let the vertex B is (α, β ) .
(α, β ) lies on 3x − 2y + 6 = 0
∴
3α − 2β + 6 = 0
and
O ′ B ⊥ AC
∴ Slope of O ′ B × slope of AC = − 1
β − 1  4
× −  = − 1
α − 1  5
2−3 1
Slope of CA =
=
1− 4 3
and
X
O
∴ Slope of O ′ A × Slope of BC = − 1
84
−1
23
⇒
×m = −1
10
−1
23
61
⇒
m = −1
− 13
D
F
20
X′
Y′
O'
Y
)
=
5y
3−3
=0
4 −2
6=
3x–
2y+
(
y Example 78. Find the orthocentre of the triangle ABC
whose angular points are A (1, 2), B (2, 3) and C (4, 3)..
Sol. Now, Slope of BC =
10 , 84
A 23 23
E
O'
+
4x
Sol. Three sides of the triangle are x = 0, y = 0 and x + y =1. The
coordinates of the vertices are O (0, 0), A (1, 0) and B (0, 1) .
The triangle OAB is a right angled triangle having right
angle at O. Therefore O (0, 0) is the orthocentre. Since we
know that the point of intersection of two perpendicular
lines is the orthocentre of the triangle OAB .
Y
(Qα = 1)
y Example 79. The equations of two sides of a triangle
are 3x − 2y + 6 = 0 and 4 x + 5y = 20 and the
orthocentre is (1, 1). Find the equation of the third side.
Sol. Let 3x − 2y + 6 = 0 and 4 x + 5y = 20 are the equations of the
sides AB and AC. The point of intersection of AB and AC is
 10 84 
 ,  . Let slope of BC is m. Since O ′ A ⊥ BC
 23 23 
...(ii)
33
2
33
13

Since, third side passes through  − 13, −  with slope ,


2
61
therefore its equation is
33 13
y+
= ( x + 13)
2 61
⇒
122y + 33 × 61 = 26x + 2 × 169
⇒
26x − 122y − 1675 = 0
Aliter : The equation of line through A. i.e. point of
intersection of AB and AC is
...(i)
(3x − 2y + 6) + λ ( 4 x + 5y − 20) = 0
it passes through (1, 1), then
(3 − 2 + 6) + λ ( 4 + 5 − 20) = 0
⇒
7 − 11λ = 0
7
∴
λ=
11
Chap 02 The Straight Lines
Since, AD ⊥ BC
( 2 − a )  1
∴
× −  = − 1
−
( 3 − b )  2
7
( 4 x + 5y − 20) = 0
11
⇒
61x + 13y − 74 = 0
61
Slope of AD = −
∴
13
13
Slope of BC =
⇒
61
If coordinates of B (α, β ), B lies on AB
∴
3α − 2β + 6 = 0
and
O ′ B ⊥ CA
β − 1  4
then
× −  = − 1
α − 1  5
From Eq. (i), (3x − 2y + 6) +
...(ii)
⇒
5α − 4β − 1 = 0
Solving Eqs. (ii) and (iii), we get
α = − 13 and β = −
...(iii)
33
2
⇒
2 − a = − 6 + 2b
⇒
a + 2b = 8
Similarly, BE ⊥ AC , we get
a+b=0
Solving Eqs. (ii) and (iii), we get
b = 8 and a = − 8
∴
(a, b ) is ( − 8, 8).
–1
3y
+
2x
0
=
C
A
Y′
From Eq. (i),
2x + 3y − 1 − ax − by + 1 = 0
⇒
(2 − a ) x + (3 − b ) y = 0
a
1
3 y=
b
3 x+
a
D
a1x+b1y=1
C
−1− λ =0
∴
λ = −1
From Eq. (i), a 2 x + b 2y − 1 − a 3 x − b 3y + 1 = 0
∴
(a 2 − a 3 ) x + (b 2 − b 3 ) y = 0
Since, AD ⊥ BC
∴ Slope of AD × slope of BC = − 1
−
X
1
y=
B
=0
–1=0
x+by
O(0,0)
2
B
E
F
b
x+
a2
Y
O
...(i)
A
Sol. The equation of a line through A i.e. the point of intersection of AB and AC, is
...(i)
(2x + 3y − 1) + λ (ax + by − 1) = 0
It passes through O (0, 0) , then
−1− λ =0
∴
λ = −1
X′
...(iii)
y Example 81. If the equations of the sides of a
triangle are ar x + b r y = 1 ; r = 1 , 2 , 3 and the orthocentre
is the origin, then prove that
a1a 2 + b1b 2 = a 2a 3 + b 2b 3 = a 3a1 + b 3b1 .
y Example 80. If the orthocentre of the triangle
formed by the lines 2x + 3y − 1 = 0, x + 2y − 1 = 0,
ax + by − 1 = 0 is at origin, then find (a, b ).
y–1
...(ii)
Sol. The equation of the line through A, i.e. the point of
intersection of AB and AC is
(a 2 x + b 2y − 1) + λ (a 3 x + b 3y − 1) = 0
It passes through (0, 0), then
∴ Equation of third side i.e. BC is
33 13
y+
= ( x + 13)
2 61
∴
26x − 122y − 1675 = 0
x+
2
115
⇒
(a 2 − a 3 )  a1 
× −  = − 1
(b 2 − b 3 )  b1 
a1a 2 − a 3a1 = − b1b 2 + b1b 3
⇒
a1a 2 + b1b 2 = a 3a1 + b 3b1
Similarly, BE ⊥ CA , then we get
a1a 2 + b1b 2 = a 2a 3 + b 2b 3
From Eqs. (ii) and (iii), we get
a1a 2 + b1b 2 = a 2a 3 + b 2b 3 = a 3a1 + b 3b1
…(ii)
…(iii)
116
Textbook of Coordinate Geometry
Exercise for Session 3
1.
2.
The locus of the point of intersection of lines x cos α + y sin α = a and x sin α − y cos α = b (α is a parameter) is
(a) 2 (x 2 + y 2 ) = a 2 + b 2
(b) x 2 − y 2 = a 2 − b 2
(c) x 2 + y 2 = a 2 + b 2
(d) x 2 − y 2 = a 2 + b 2
If a, b , c are in AP then ax + by + c = 0 represents
(a) a straight line
(c) a family of parallel lines
3.
If the lines x + 2ay + a = 0, x + 3by + b = 0 and x + 4cy + c = 0 are concurrent, then a, b , c are in
(a) AP
(c) HP
4.
5.
3 1
(a)  , 
 4 2
1 3
(b)  , 
 2 4
3
1
(c)  − , − 
 4
2
1 3
(d)  − , − 
 2
4
If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0, (a, b and c being distinct and different from 1) are
a
b
c
concurrent, then the value of
is
+
+
a −1 b −1 c −1
1 5
(c)  − ,− 
 6 9
(b) (b , a )
(d) (−a, b )
1
(b)  ,
2
1

3
2 7
(d)  , − 
 3 5
The straight line through the point of intersection of ax + by + c = 0 and a′ x + b ′ y + c′ = 0 are parallel toY-axis
has the equation
(a) x (ab ′ − a ′b ) + (cb ′ − c ′b ) = 0
(c) y (ab ′ − a ′b ) + (c ′a − ca ′ ) = 0
10.
(b) a family of concurrent lines
(d) None of these
If the straight lines x + y − 2 = 0, 2x − y + 1 = 0 and ax + by − c = 0 are concurrent, then the family of lines
2ax + 3by + c = 0 (a, b , c are non-zero) is concurrent at
(a) (2, 3)
9.
a1 b1 c1
=
= , then u + kv = 0 represents
a 2 b 2 c2
The straight lines x + 2y − 9 = 0, 3x + 5y − 5 = 0 and ax + by − 1 = 0 are concurrent, if the straight line
35x − 22y + 1 = 0 passes through the point
(a) (a, b )
(c) (a, − b )
8.
(b) −1
(d) 2
If u ≡ a1x + b1y + c1 = 0 and v ≡ a 2x + b 2y + c2 = 0 and
(a) u = 0
(c) a family of parallel lines
7.
(b) GP
(d) AGP
The set of lines ax + by + c = 0, where 3a + 2b + 4c = 0 is concurrent at the point
(a) −2
(c) 1
6.
(b) a family of concurrent lines
(d) None of these
(b) x (ab ′ + a ′b ) + (cb ′ + c ′b ) = 0
(d) y (ab ′ + a ′b ) + (c ′a + ca ′ ) = 0
If the equations of three sides of a triangle are x + y = 1, 3x + 5y = 2 and x − y = 0, then the orthocentre of the
triangle lies on the line/lines
(a) 5x − 3y = 1
(c) 2x − 3y = 1
(b) 5y − 3x = 1
(d) 5x − 3y = 2
Chap 02 The Straight Lines
11.
Find the equation of the line through the intersection of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 and perpendicular to
6x − 7y + c = 0
(a) 199y + 120x = 125
(c) 119x + 102y = 125
12.
117
The locus of the point of intersection of the lines
(a) a circle
(c) a hyperbola
(b) 199y − 120x = 125
(d) 119x − 102y = 125
x y
x y
1
is
− = m, + =
a b
a b m
(b) an ellipse
(d) a parabola
13.
Find the condition on a and b, such that the portion of the line ax + by − 1 = 0, intercepted between the lines
ax + y + 1 = 0 and x + by = 0 subtends a right angled at the origin.
14.
If the lines (a − b − c )x + 2ay + 2a = 0, 2bx + (b − c − a )y + 2b = 0 and (2c + 1)x + 2cy + c − a − b = 0 are
concurrent, then prove that either a + b + c = 0 or (a + b + c )2 + 2a = 0.
15.
Prove that the lines ax + by + c = 0,bx + cy + a = 0 and cx + ay + b = 0 are concurrent if a + b + c = 0 or
a + bω + c ω 2 = 0 or a + bω 2 + c ω = 0, where ω is a complex cube root of unity.
16.
Find the equation of the straight line which passes through the intersection of the lines x − y − 1 = 0 and
2x − 3y + 1 = 0 and is parallel to (i) X-axis (ii) Y-axis ( iii ) 3x + 4y = 14.
17.
Let a, b , c be parameters. Then, the equation ax + by + c = 0 will represent a family of straight lines passing
through a fixed-point, if there exists a linear relation between a, b and c.
18.
Prove that the family of lines represented by x (1 + λ ) + y (2 − λ ) + 5 = 0, λ being arbitrary, pass through a fixed
point. Also find the fixed point.
19.
a
a

Prove that  −a , −  is the orthocentre of the triangle formed by the lines y = m i x +
, i = 1, 2, 3; m1, m 2, m 3


2
mi
being the roots of the equation x 3 − 3x 2 + 2 = 0.
Session 4
Equations of Straight Lines Passing Through a Given Point
and Making a Given Angle with a Given Line, A Line Equally
Inclined With Two Lines, Equation of the Bisectors, Bisector
of the Angle Containing the Origin, Equation of that Bisector
of the Angle Between Two Lines which Contains a Given Point,
How to Distinguish the Acute (Internal) and Obtuse (External)
Angle Bisectors
Equations of Straight Lines
Passing Through a Given Point
and Making a Given Angle with
a Given Line
Theorem : Prove that the equations of the straight lines
which pass through a given point ( x 1 , y 1 ) and make a
given angle α with the given straight line y = mx + c are
y − y 1 = tan (θ ± α ) ( x − x 1 )
where,
m = tan θ.
Proof : Let AB be the given line which makes an angle θ
with X-axis.
∴
m = tan θ
Let CD and EF are two required lines which make angle α
with the given line. Let these lines meet the given line AB
at Q and R respectively
Y
F
D
P(x1,y1)
α
θ
X′
A
Q
O
C
Y′
α
θ
(180°– α)
θ
R
S
B
T
E
X
∴
∠ DQS = ∠ PQR + ∠ RQS = (α + θ )
∴ Equation of line CD is
y − y 1 = tan (θ + α ) ( x − x 1 )
and
...(i)
∠ FRT = ∠ FRB + ∠ BRT
= 180 ° − α + θ
= 180 ° + (θ − α )
∴ Equation of line EF is
y − y 1 = tan (180 ° + θ − α ) ( x − x 1 )
or
y − y 1 = tan (θ − α ) ( x − x 1 )
From Eqs. (i) and (ii), we get
y − y 1 = tan (θ ± α ) ( x − x 1 )
These are the equations of the two required lines.
...(ii)
y Example 82. Find the equations of the straight lines
passing through the point (2 , 3) and inclined at π / 4
radians to the line 2x + 3y = 5 .
Sol. Let the line 2x + 3y = 5 make an angle θ with positive
X -axis.
2
Then
tanθ = −
3
π
2
Now
tan θ ⋅ tan = − × 1
4
3
2
= − ≠ ±1
3
Slopes of required lines are
π
π


tan θ +  and tan θ − 


4
4
Chap 02 The Straight Lines
∴
 π
 2
tan θ + tan  
−  + 1
 4
 3
1
π

tan θ +  =
=
=

4
5
 π
 2
1 − tan θ tan   1 −  −  (1)
 4
 3
and
 π
tan θ − tan  
 4
π

tan θ −  =


4
 π
1 + tan θ tan  
 4
(2 + 3 ) x − y = 1 + 2 3
⇒
Hence, equations of other sides are
(2 − 3 ) x − y = 1 − 2 3
(2 + 3 ) x − y = 1 + 2 3
and
y Example 84. The straight lines 3x + 4 y = 5 and
4 x − 3y = 15 intersect at a point A. On these lines, the
points B and C are chosen so that AB = AC . Find the
possible equations of the line BC passing through the
point (1, 2) .
 2
−  − 1
 3
= −5
=
 2
1 +  −  ( 1)
 3
Sol. Clearly ∠ BAC = 90°
∴ Equations of required lines are
1
y − 3 = ( x − 2) and y − 3 = − 5 ( x − 2)
5
i.e.
x − 5y + 13 = 0 and 5x + y − 13 = 0
y Example 83. A vertex of an equilateral triangle is
(2 , 3) and the opposite side is x + y = 2 . Find the
equations of the other sides.
Sol. Let A (2, 3) be one vertex and x + y = 2 be the opposite side
of an equilateral triangle. Clearly remaining two sides pass
through the point A (2, 3) and make an angle 60° with
x +y =2
Slope of x + y = 2 is −1
Q
AB = AC
Q
∴
∠ ABC = ∠ BCA = 45°
α = 45°
Q Slope of 3x + 4y = 5 is −
3
4
So, possible equations of BC are given by
y − 2 = tan (θ ± α ) ( x − 1)
Y
3x
+
(1,2)
4y
=
5
B
Q
A(2,3)
X′
O
X
45°
60°
A(3,–1)
60°
45°
C
2
y=
x+
C
X′
3
4
Let tanθ = −
Y
B
119
O
Y′
X
y=
–3
4x
P
Let
tanθ = − 1
∴
θ = 135°
∴ Equations of the other two sides are
y − 3 = tan (135° ± 60° ) ( x − 2)
i.e. sides are
y − 3 = tan (195° ) ( x − 2)
⇒
y − 3 = tan (180° + 15° ) ( x − 2)
⇒
y − 3 = tan 15° ( x − 2)
⇒
y − 3 = ( 2 − 3 ) ( x − 2)
⇒
(2 − 3 ) x − y = 1 − 2 3
and
y − 3 = tan (75° ) ( x − 2)
Y′
(taking ‘+’ sign)
(taking ‘−’ sign)
⇒
y − 3 = cot 15° ( x − 2)
⇒
y − 3 = ( 2 + 3 ) ( x − 2)
15
⇒
 tan θ ± tan α 
y −2= 
 ( x − 1)
 1 m tan θ tan α 
⇒

3
 −
4
y −2= 


 1 m  −

⇒
 −3 ± 4 
y −2= 
 ( x − 1)
 4 m ( − 3) 
±1
3

4


 ( x − 1)

( 1) 

1
( x − 1)
4 − ( − 3)
⇒
y −2=
or
x − 7y + 13 = 0
(taking upper sign)
120
Textbook of Coordinate Geometry
y −2=
and
( −3 − 4 )
( x − 1)
( 4 + ( −3))
(taking below sign)
or
7x + y − 9 = 0
Hence, possible equation of the line BC are x − 7y + 13 = 0
and 7 x + y − 9 = 0
A Line Equally Inclined with
Two Lines
Theorem : If two lines with slopes m 1 and m 2 be equally
inclined to a line with slope m, then
 m2 − m 
 m1 − m 


 =−
 1 + mm 2 
 1 + mm 1 
Proof : Let be two lines of slopes m 1 and m 2 intersecting
at a point P.
Let ∠ CPA = ∠ BPC = θ
B
C
θ
θ
A
 m − m1 
tan ( ∠ CPA ) = 

 1 + mm 1 
 m − m1 
tan θ = 

 1 + mm 1 
or
 12 − 5m 
 3 − 4m 

 =−

 5 + 12m 
 4 + 3m 
⇒
⇒
(3 − 4m ) (5 + 12m ) + ( 4 + 3m ) (12 − 5m ) = 0
63m 2 − 32m − 63 = 0
( 7m − 9 ) ( 9m + 7 ) = 0
9
7
∴
m= ,−
7
9
Putting these values of m in Eq. (i) we obtain the equations
of required lines as 9 x − 7y + 3 = 0 and x + 9y − 41 = 0.
Sol. Let m be the slope of BC. Since AB = AC .
(Qm > m 1 ) ...(i)
Therefore BC makes equal angles with AB and AC.
Y
A
 m −m 
tan ( ∠ BPC ) =  2

 1 +m 2m 
X
O
0
(Qm 2 > m ) … (ii)
 m1 − m 
 m2 − m 
 m − m1   m2 − m 
 =−


 =
 or 
 1 + mm 1   1 + m 2m 
 1 + mm 1 
 1 + mm 2 
1. The above equation gives two values of m which are the slopes
of the lines parallel to the bisectors of the angles between the
two given lines.
2. Sign of m in both brackets is same.
0
From Eqs. (i) and (ii), we get
Remarks
3=
y–
x+
 m −m 
tan θ =  2

 1 + m 2m 
or
X′
7x–y
+3=
and
⇒
y Example 86. Two equal sides of an isosceles triangle
are given by the equations 7 x − y + 3 = 0 and
x + y − 3 = 0 and its third side passes through the
point (1, − 10) . Determine the equation of the third
side.
m1
P
∴
Sol. Let m be the slope of the required line. Then its equation is
...(i)
y − 3 = m ( x − 2)
It is given that line (i) is equally inclined to the lines
3x − 4y − 7 = 0 and 12x − 5y + 6 = 0 then

 12

 3
−m 

 −m 
4
5
⇒

 = −

 1 + 12 m 
1 + 3 m


5 
4 
3 


 slope of 3x − 4y − 7 = 0 is
4 

12
 and slope of 12x − 5y + 6 = 0 is 

5
⇒
m2
m
y Example 85. Find the equations to the straight lines
passing through the point (2 , 3) and equally inclined to
the lines 3x − 4 y − 7 = 0 and 12x − 5y + 6 = 0.
θ
θ
B
C
D(1,–10)
Y′
 −1−m 
7 −m
Then 

 =−
 1 + ( − 1) m 
 1 + 7m 
⇒
⇒
( 7 − m ) ( 1 − m ) − ( 1 + 7m ) ( 1 + m ) = 0
6m 2 + 16m − 6 = 0
⇒
3m 2 + 8m − 3 = 0
Chap 02 The Straight Lines
⇒
(3m − 1) (m + 3) = 0
1
⇒
m = , −3
3
Equation of third side BC is y + 10 = m ( x − 1)
1
i.e.
y + 10 = − 3 ( x − 1)
y + 10 = ( x − 1) and
3
or
and
x − 3y − 31 = 0
3x + y + 7 = 0
y Example 87. Find the equations of the bisectors of
the angles between the straight lines 3x − 4 y + 7 = 0
and 12x + 5y − 2 = 0.
Sol. The equations of the bisectors of the angles between
3x − 4y + 7 = 0 and 12x + 5y − 2 = 0 are
( 3x − 4y + 7 )
(12x + 5y − 2)
=±
( 3) 2 + ( − 4 ) 2
(12)2 + (5)2
( 3x − 4y + 7 )
(12x + 5y − 2)
=±
5
13
or
(39 x − 52y + 91) = ± (60x + 25y − 10)
Taking the positive sign, we get
21x + 77y − 101 = 0
as one bisector.
Taking the negative sign, we get 99 x − 77y + 81 = 0
as the second bisector.
or
Equation of the Bisectors
Theorem : Prove that the equation of the bisectors of the
angles between the lines
a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0
(a 1 x + b 1 y + c 1 )
(a x + b 2 y + c 2 )
are given by
=± 2
2
2
(a 1 + b 1 )
(a 22 + b 22 )
Proof : Let the given lines be AA ′ and BB ′ whose
equations are
...(i)
a 1 x + b 1y + c 1 = 0
and
...(ii)
a 2 x + b 2y + c 2 = 0
Since bisectors of the angles between the two lines are the
locus of a point which moves in a plane such that whose
distance from two lines are equal.
=0
c2
y+
X
O
X
D¢
Y′
C′
A′
B′
Let CC ′ and DD ′ be the two bisectors of the angle between
the lines AA′ and BB′. Let P ( x , y ) be any point on CC ′, then
Length of the perpendicular from P on AA′
= length of the perpendicular from P on BB ′
| a 1 x + b 1y + c 1 | | a 2 x + b 2y + c 2 |
∴
=
(a 12 + b 12 )
(a 22 + b 22 )
or
(a 1 x + b 1 y + c 1 )
(a 12
+ b 12 )
=±
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
These are the required equations of the bisectors.
Note
The two bisectors are perpendicular to each other.
A
C
2
D
X
PP
B
+b
a 2x
a2x+b2y+c2=0 P P a1x+b1y+c1=0
X
X
X
X′
Y
0
M
c1 =
P(x,y)
D′
...(i)
...(ii)
Let P ( x , y ) be taken on the bisector of the angle which
contains the origin.
A
L
Let equations of lines be
a 1 x + b 1y + c 1 = 0
a 2 x + b 2y + c 2 = 0
where c 1 and c 2 are positive.
1 y+
C
B
Bisector of the Angle Containing
the Origin
X¢
P(x,y)
PP
X
a1 x
+b
Y
121
X
D
X
PP
X
O
Y¢
A¢
C¢'
B¢
(i) Let P ( x , y ) lies on DD ′ ,then either O (0, 0 ) and P ( x , y )
will lie on the same side of the two lines Eqs. (i) and (ii),
then
a 1 x + b 1y + c 1
>0
0 + 0 + c1
and
a 2 x + b 2y + c 2
>0
0 + 0 + c2
or
and
a 1 x + b 1y + c 1 > 0
a 2 x + b 2y + c 2 > 0
(Qc 1 , c 2 > 0 )
122
Textbook of Coordinate Geometry
If the origin O (0, 0 ) and P ( x , y ) lie on the opposite side of
the two lines Eqs. (i) and (ii), then
a 1 x + b 1y + c 1
a 2 x + b 2y + c 2
< 0 and
<0
0 + 0 + c1
0 + 0 + c2
∴ The equation of the bisector bisecting the angle
containing origin is
( − 4 x − 3y + 6) (5x + 12y + 9 )
=
( − 4 2 ) + ( − 3) 2
(5)2 + (12)2
a 1 x + b 1 y + c 1 < 0 and a 2 x + b 2 y + c 2 < 0
(Qc 1 , c 2 > 0 )
Then equation of bisectors will be
| a 1 x + b 1y + c 1 | | a 2 x + b 2y + c 2 |
=
(a 12 + b 12 )
(a 22 + b 22 )
⇒
 − 4 x − 3y + 6  5x + 12y + 9 

 =


 

5
13
⇒
− 52x − 39y + 78 = 25x + 60y + 45
⇒
77 x + 99y − 33 = 0 or 7 x + 9y − 3 = 0
and the equation of the bisector bisecting the angle not
containing origin is
( − 4 x − 3y + 6)
(5x + 12y + 9 )
=−
2
2
( − 4 ) + ( − 3) )
(52 ) + (12)2
or
Case I : If a 1 x + b 1 y + c 1 > 0 and a 2 x + b 2 y + c 2 > 0
(a 1 x + b 1 y + c 1 ) (a 2 x + b 2 y + c 2 )
then
=
(a 12 + b 12 )
(a 22 + b 22 )
Case II : If a 1 x + b 1 y + c 1 < 0 and a 2 x + b 2 y + c 2 < 0
(a x + b 1 y + c 1 )
(a x + b 2 y + c 2 )
then − 1
=− 2
(a 12 + b 12 )
(a 22 + b 22 )
i.e.
(a 1 x + b 1 y + c 1 )
(a 12 + b 12 )
=
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
Thus is both cases equation of the bisector containing the
origin, when c 1 and c 2 are positive is
(a 1 x + b 1 y + c 1 )
(a 12
+ b 12 )
=
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
and equation of the bisector of the angle between the lines
a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0
which does not contain the origin when c 1 and c 2 are
positive is
(a 1 x + b 1 y + c 1 )
(a 12
+ b 12 )
=−
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
Working Rule :
(i) First re-write the equations of the two lines so that
their constant terms are positive.
(ii) The bisector of the angle containing the origin and
does not containing the origin, then taking +ve
and – ve sign in
(a 1 x + b 1 y + c 1 )
(a 12 + b 12 )
=±
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
respectively.
y Example 88. Find the equations of angular bisector
bisecting the angle containing the origin and not
containing the origin of the lines 4 x + 3y − 6 = 0 and
5x + 12y + 9 = 0 .
Sol. Firstly make the constant terms (c 1, c 2 ) positive, then
− 4 x − 3y + 6 = 0 and 5x + 12y + 9 = 0
⇒
 5x + 12y + 9 
 − 4 x − 3y + 6

 =−





5
13
⇒
⇒
− 52x − 39y + 78 = − 25x − 60y − 45
27 x − 21y − 123 = 0 or 9 x − 7y − 41 = 0
Equation of that Bisector of the
Angle between Two Lines which
Contains a Given Point
Let the equations of the two lines be
...(i)
a 1 x + b 1y + c 1 = 0
and
...(ii)
a 2 x + b 2y + c 2 = 0
The equation of the bisector of the angle between the two
lines containing the points (h, k ) will be
(a 1 x + b 1 y + c 1 ) (a 2 x + b 2 y + c 2 )
=
(a 12 + b 12 )
(a 22 + b 22 )
(a 1 x + b 1 y + c 1 )
or
(a 12
+ b 12 )
=−
(a 2 x + b 2 y + c 2 )
(a 22 + b 22 )
according as a 1 h + b 1 k + c 1 and a 2 h + b 2 k + c 2 are of the
same sign or opposite sign.
y Example 89. Find the bisector of the angle between
the lines 2x + y − 6 = 0 and 2x − 4 y + 7 = 0 which
contains the point (1, 2).
Sol. Value of 2x + y − 6 at (1, 2) is – 2 (negative)
and value of 2x − 4y + 7 at (1, 2) is 1 (positive)
i.e. opposite sign.
∴ Equation of bisector containing the point (1, 2) is
( 2x + y − 6)
( 2x − 4y + 7 )
=−
2
2
(2 + 1 )
( 2) 2 + ( − 4 ) 2
⇒
or
2 ( 2x + y − 6) + ( 2x − 4y + 7 ) = 0
6x − 2y − 5 = 0
123
Chap 02 The Straight Lines
How to Distinguish the Acute
(Internal) and Obtuse (External)
Angle Bisectors?
Let the equations of the two lines be
a 1 x + b 1y + c 1 = 0
and
a 2 x + b 2y + c 2 = 0
where, c 1 > 0, c 2 > 0.
Q Equations of bisectors are
(a 1 x + b 1 y + c 1 )
(a x + b 2 y + c 2 )
=± 2
2
2
(a 1 + b 1 )
(a 22 + b 22 )
+ b 12 )
Conditions
(a 2 x `+ b 2 y + c 2 )
=±
(a 22 + b 22 )
Acute angle bisector
Obtuse angle bisector
a1a 2 + b1b 2 > 0
–
+
a1a 2 + b1b 2 < 0
+
–
Remarks
...(iii)
m − m2 

tan α = 1
1 + m 1 m 2
  a1   p1  
  − b  −  − q   a q − b p 
1
1
1 1
1 1
=
=
b
q
a
+


p


a

1 p 1
1 +  − 1   − 1  1 1
  b 1   q 1 
Hence, if 0 < tan α < 1, p 1 x + q 1 y + r1 = 0 is the acute
(internal) bisector and if tan α > 1, p 2 x + q 2 y + r2 = 0 is the
obtuse (external ) bisector.
Shortcut Method for finding Acute
(Internal) and Obtuse (External) Angle
Bisectors
Let the equations of the two lines be
a 1 x + b 1y + c 1 = 0
a 2 x + b 2y + c 2 = 0
Taking
c 1 > 0, c 2 > 0 and a 1 b 2 ≠ a 2 b 1
Then equations of the bisectors are
(a 12
...(i)
...(ii)
when Eq. (iii) be simplified, let the bisectors be
...(iv)
p 1 x + q 1 y + r1 = 0
and
...(v)
p 2 x + q 2 y + r2 = 0
Since the two bisectors are at right angles, the angle α
between the acute (internal ) bisector and any one of the
given lines must lie between 0 and 45° i.e. 0 < α < 45 °.
∴
0 < tan α < 1
If m 1 and m 2 are the slopes of Eqs. (i) and (iii) respectively.
p
a
Then,
m 1 = − 1 and m 2 = − 1
b1
q1
∴
(a 1 x + b 1 y + c 1 )
1. Bisectors are perpendiculars to each other.
2. `+' sign gives the bisector of the angle containing origin.
3. If a1a2 + b1b2 > 0 then the origin lies in obtuse angle and if
a1a2 + b1b2 < 0, then the origin lies in acute angle.
Explanation : Equations of given lines in normal form
will be respectively
b 1y
a1 x
c1
( Qc 1 > 0 )
−
−
=
(a 12 + b 12 )
(a 12 + b 12 )
(a 12 + b 12 )
a2 x
and
−
If
cos α = −
and
cos β = −
(a 22 + b 22 )
b 2y
−
(a 22 + b 22 )
a1
+ b 12 )
(a 12
a2
(a 22
+ b 22 )
=
c2
( Qc 2 > 0 )
(a 22 + b 22 )
b1
then sin α = −
then sin β = −
(a 12
+ b 12 )
b2
(a 22
+ b 22 )
Now, cos (α − β) = cos α cos β + sin α sin β
(a 1 a 2 + b 1 b 2 )
=
2
(a 1 + b 12 ) (a 22 + b 22 )
cos (α − β) > 0 or < 0
according as (α − β) is acute or obtuse.
i.e.
a 1 a 2 + b 1 b 2 > 0 or < 0
Hence, bisector of the angle between the lines will be
the bisector of the acute or obtuse angle according as
origin lies in the acute or obtuse angle according as
a 1 a 2 + b 1 b 2 < 0 or >0.
y Example 90. Find the equation of the bisector of
the obtuse angle between the lines 3x − 4 y + 7 = 0 and
12x + 5y − 2 = 0.
Sol. Firstly make the constant terms (c 1, c 2 ) positive
3x − 4y + 7 = 0 and − 12x − 5y + 2 = 0
Q a1a 2 + b1b 2 = (3) ( − 12) + ( − 4 ) ( − 5) = − 36 + 20 = − 16
∴
a1a 2 + b1b 2 < 0
Hence “−” sign gives the obtuse bisector.
∴ Obtuse bisector is
( 3x − 4y + 7 )
( − 12x − 5y + 2)
=−
( 3) 2 + ( − 4 ) 2
( − 12)2 + ( − 5)2
⇒
⇒
13 (3x − 4y + 7 ) = − 5 ( − 12x − 5y + 2)
21x + 77y − 101 = 0 is the obtuse angle bisector.
124
Textbook of Coordinate Geometry
y Example 91. Find the bisector of acute angle
between the lines x + y − 3 = 0 and 7 x − y + 5 = 0.
Sol. Firstly, make the constant terms (c 1, c 2 ) positive then
− x − y + 3 = 0 and 7 x − y + 5 = 0
Q
a1a 2 + b1b 2 = ( − 1) (7 ) + ( − 1) ( − 1) = − 7 + 1 = − 6
i.e.
a1a 2 + b1b 2 < 0
Hence “+” sign gives the acute bisector.
7x − y + 5
− x −y +3
∴ Acute bisector is
=+
2
2
( − 1) + ( − 1)
( 7 ) 2 + ( − 1) 2
⇒
− x − y + 3 7x − y + 5
=
2
5 2
⇒
− 5x − 5y + 15 = 7 x − y + 5
∴
12x + 4y − 10 = 0 or 6x + 2y − 5 = 0
is the acute angle bisector.
y Example 92. Find the coordinates of incentre of the
triangle. The equation of whose sides are
AB : x + y − 1 = 0,BC : 7 x − y − 15 = 0
and CA : x − y − 1 = 0.
Sol. Firstly, make the constant terms (c 1, c 2 , and c 3 ) positive
...(i)
AB : − x − y + 1 = 0
...(ii)
BC : − 7 x + y + 15 = 0
...(iii)
CA : − x + y + 1 = 0
Q The incentre of triangle is the point of intersection of
internal or acute angle bisectors.
Internal bisector of AB and BC :
− x −y +1=0
i.e.
Q
a1a 2 + b1b 2 = ( − 1) ( − 7 ) + ( − 1) (1) = 6 > 0
∴ Acute or internal bisector is
( − x − y + 1)
( − 7 x + y + 15)
=−
2
2
( − 1) + ( − 1)
( − 7 ) 2 + ( 1) 2
( − x − y + 1)
( − 7 x + y + 15)
=−
2
5 2
⇒
− 5x − 5y + 5 = 7 x − y − 15
or
12x + 4y − 20 = 0
or
3x + y − 5 = 0
Internal bisector of BC and CA :
− 7 x + y + 15 = 0
− x +y +1=0
Q a1a 2 + b1b 2 = ( −7 )( −1) + (1) (1) = 8 > 0
Q Acute or internal bisector is
( − 7 x + y + 15)
( − x + y + 1)
=−
2
2
( − 7 ) + ( 1)
( − 1) 2 + ( 1) 2
⇒
⇒
...(iv)
− 7 x + y + 15 ( x − y − 1)
=
5 2
2
⇒
− 7 x + y + 15 = 5x − 5y − 5
or
12x − 6y − 20 = 0
or
6x − 3y − 10 = 0
Finally, solve Eqs. (iv) and (v), we get
5
x = and y = 0
3
5 
Hence coordinates of incentre are  , 0 .
3 
...(v)
− 7 x + y + 15 = 0
Exercise for Session 4
1.
The straight lines 2x + 11y − 5 = 0, 24x + 7y − 20 = 0 and 4x − 3y − 2 = 0
(a) form a triangle
(b) are only concurrent
(c) are concurrent with one line bisecting the angle between the other two
(d) None of the above
2.
The line x + 3y − 2 = 0 bisects the angle between a pair of straight lines of which one has the equation
x − 7y + 5 = 0. The equation of other line is
(a) 3x + 3y − 1 = 0
3.
(b) x − 3y + 2 = 0
(c) 5x + 5y + 3 = 0
(d) 5x + 5y − 3 = 0
P is a point on either of the two lines y − 3 | x | = 2 at a distance of 5 units from their point of intersection. The
coordinates of the foot of the perpendicular from P on the bisector of the angle between them are
 4 + 5 3  4 − 5 3
(a)  0,
 or  0,
 depending on which the point P is taken
2
2

 

 4 + 5 3
(b)  0,

2


 4 − 5 3
(c)  0,

2


 5 5 3
(d)  ,

2 2 
Chap 02 The Straight Lines
4.
In a triangle ABC, the bisectors of angles B and C lie along the lines x = y andy = 0 . If A is (1, 2), then the
equation of line BC is
(a) 2x + y = 1
5.
5
5
(b)
3
5
(c)
6
5
(d)
7
5
(b) y = 1
(c) 2y = 3
(d) x = 1
(b) x − y + 1 = 0
(c) x − y = 5
(d) x − y + 5 = 0
The equation of the bisector of that angle between the lines x + y = 3 and 2x − y = 2 which contains the point
(1, 1) is
(a) ( 5 − 2 2 )x + ( 5 +
(c) 3x = 10
9.
(d) x + 3y = 1
The equation of the bisector of the acute angle between the lines 2x − y + 4 = 0 and x − 2y = 1 is
(a) x + y + 5 = 0
8.
(c) x − 2y = 3
The equation of the straight line which bisects the intercepts made by the axes on the lines x + y = 2 and
2x + 3y = 6 is
(a) 2x = 3
7.
(b) 3x − y = 5
In ∆ABC, the coordinates of the vertex A are (4, −1) and lines x − y − 1 = 0 and 2x − y = 3 are the internal
bisecters of angles B and C. Then, the radius of the incircle of triangle ABC is
(a)
6.
125
2 )y = 3 5 − 2 2
(b) ( 5 + 2 2 ) x + ( 5 −
(d) 3x − 5y + 2 = 0
2 )y = 3 5 + 2 2
Find the equations of the two straight lines through (7, 9) and making an angle of 60° with the line
x − 3y − 2 3 = 0.
10.
Equation of the base of an equilateral triangle is 3x + 4y = 9 and its vertex is at the point (1, 2). Find the
equations of the other sides and the length of each side of the triangle.
11.
Find the coordinates of those points on the line 3x + 2y = 5 which are equidistant from the lines 4x + 3y − 7 = 0
and 2y − 5 = 0.
12.
Two sides of rhombus ABCD are parallel to the lines y = x + 2 and y = 7x + 3. If the diagonal of the rhombus
intersect at the point (1, 2) and the vertex A lies on Y-axis, find the possible coordinates of A.
13.
The bisector of two lines L1 and L2 are given by 3x 2 − 8xy − 3y 2 + 10x + 20y − 25 = 0. If the line L1 passes
through origin, find the equation of line L2.
14.
Find the equation of the bisector of the angle between the lines x + 2y − 11 = 0 and 3x − 6y − 5 = 0 which
contains the point (1, –3).
15.
Find the equation of the bisector of the angle between the lines 2x − 3y − 5 = 0 and 6x − 4y + 7 = 0 which is the
supplement of the angle containing the point (2, –1).
Session 5
The Foot of Perpendicular Drawn from the
Point (x1, y1) to the Line ax + by + c = 0, Image or
Reflection of a Point (x1, y1) about a Line Mirror,
Image or Reflection of a Point In Different Cases,
Use of Image or Reflection
The Foot of Perpendicular
Drawn from the Point ( x 1 , y 1 )
to the Line ax + by + c = 0
Aliter II : Let the coordinates of M are ( x 2 , y 2 )
PM ⊥ RS
Q
and M lies on
i.e.
Let P ≡ ( x 1 , y 1 ) and let M be the foot of perpendicular
drawn from P on ax + by + c = 0.
In order to find the coordinates of M, find the equation of
the line PM which is perpendicular to RS and passes
through P ( x 1 , y 1 ), i.e. bx − ay = bx 1 − ay 1
and
or
or
or b( x − x 1 ) − a(y − y 1 ) = 0 and solving it with
ax + by + c = 0, then we get coordinates of M.
ax + by + c = 0
ax 2 + by 2 + c = 0
 y2 − y1   a 
 × −  = − 1

 x2 − x1   b
=
=
M
b
ax+
c
y+
or
Aliter I : Let the coordinates of M are ( x 2 , y 2 ) then
M ( x 2 , y 2 ) lies on ax + by + c = 0
ax 2 + by 2 + c = 0
Q PM ⊥ RS
then
(Slope of PM ) ( Slope of RS ) = − 1
⇒
...(i)
 y2 − y1   a 
 × −  = − 1

 x2 − x1   b
or
bx 2 − ay 2 = bx 1 − ay 1
Solving Eqs. (i) and (ii), we get ( x 2 , y 2 ).
(By ratio proportion method)
(ax 2 + by 2 ) − (ax 1 + by 1 )
a2 + b2
− c − (ax 1 + by 1 )
a2 + b2
[from Eq. (iii)]
x2 − x1 y2 − y1
(ax 1 + by 1 + c )
=
=−
a
b
(a 2 + b 2 )
y Example 93. Find the coordinates of the foot of the
perpendicular drawn from the point (2, 3) to the line
y = 3x + 4.
R
⇒
and
( Q PM ⊥ RS )
x2 − x1 y2 − y1
=
a
b
x 2 − x 1 y 2 − y 1 a( x 2 − x 1 ) + b(y 2 − y 1 )
=
=
a
b
(a 2 + b 2 )
)
,y 1
P(x 1
S
=0
...(iii)
...(ii)
Sol. Given line is
...(i)
3x − y + 4 = 0
Let Eq.
P ≡ (2, 3).
Let, M be the foot of perpendicular drawn from P on RS.
Then equation of PM passes through P (2, 3) and
perpendicular to RS is
x + 3y − (2 + 3 × 3) = 0
i.e.
...(ii)
x + 3y − 11 = 0
Chap 02 The Straight Lines
Y
)
,y 1
P(x 1
S
S
(x2,y2)M
P(2,3)
R
X
O
Solving Eqs. (i) and (ii), we get
1
37
x = − ,y =
10
10
or
 1 37 
M ≡ − , 
 10 10 
or
x 2 + 3y 2 − 11 = 0
Solving Eqs. (iii) and (iv), we get
1
37
x2 = − , y2 =
10
10
 1 37 
M ≡ − , 
∴
 10 10 
or
…(iii)
or
∴
 y2 − y1   a 

 × −  = − 1
 x2 − x1   b
( x 2 − x 1 ) (y 2 − y 1 )
=
a
b
x 2 − x 1 y 2 − y 1 a( x 2 − x 1 ) + b(y 2 − y 1 )
=
=
a
b
a2 + b2
=
x2 − 2 y2 − 3
7
=−
=
3
−1
10
1
37
and y 2 =
x2 = −
10
10
 1 37 
M ≡ − , 
 10 10 
Image or Reflection of a Point
( x1 , y1 ) About a Line Mirror
Let Q ≡ ( x 2 , y 2 ) be the image of P ≡ ( x 1 , y 1 ) then find
coordinates of the foot of perpendicular M drawn from the
point P ( x 1 , y 1 ) on RS and use fact that M is the mid- point
P and Q .
a(2 x − x 1 − x 1 ) + b(2y − y 1 − y 1 )
a2 + b2
Q M ( x , y ) is mid-point of P and Q 


∴ x 2 = 2 x − x 1 and y 2 = 2y − y 1 
...(iv)
=
=
Aliter II : By Ratio Proportion Method :
x 2 − 2 y 2 − 3 − (3 × 2 − 3 + 4 )
=
=
3
−1
32 + ( − 1) 2
⇒
y 2)
x 2,
Aliter I : Ratio Proportion Method :
Q
PQ ⊥ RS
( Slope of PQ ) × ( Slope of RS) = − 1
∴
Y′
Aliter I : Let the coordinates of M be ( x 2 , y 2 ) then
M ( x 2 , y 2 ) lies on 3x − y + 4 = 0
⇒
3x 2 − y 2 + 4 = 0
and Q PM ⊥ RS
∴ ( Slope of PM ) × ( Slope of RS ) = − 1
 y 2 − 3
⇒
 × ( 3) = − 1

 x 2 − 2
Q(
=0
 x + x2 y1 + y2 
,
M ≡ 1


2
2 
i.e.
R
∴
+c
by
ax+
M
y=3x
+4
X′
127
=
i.e.
− 2ax 1 − 2by 1 + 2 (ax + by )
a2 + b2
− 2ax 1 − 2by 1 + 2( − c )
a2 + b2
− 2(ax 1 + by 1 + c )
(Qax + by = − c )
(a 2 + b 2 )
x2 − x1 y2 − y1
2 (ax 1 + by 1 + c )
=
=−
a
b
(a 2 + b 2 )
Aliter II : By Distance form or Symmetric form or
parametric form :
a
Slope of RS = −
Q
b
b
Slope of PQ =
∴
a
(a2+b 2)
b
θ
a
128
Textbook of Coordinate Geometry
Let
tan θ =
∴
sin θ =
cos θ =
and
b
a
b
(a + b 2 )
a
2
 − ax − by − c 
1
1

=

2
(a + b 2 ) 

 − ax − by − c 
1
1
 =r
PQ = 2 PM = 2 p = 2 

2
2
(a + b ) 

⇒
Required image has the coordinates
( x 1 + r cos θ, y 1 + r sin θ ) .
y Example 94. Find the image of the point (4, − 13)
with respect to the line mirror 5x + y + 6 = 0.
Sol. Let, P ≡ ( 4, − 13) and Let, Q ≡ ( x 2 , y 2 ) be mirror image P
with respect to line mirror 5x + y + 6 = 0.
Let, M (α, β ) be the foot of perpendicular from P ( 4, − 13) on
the line mirror 5x + y + 6 = 0, then
α − 4 β + 13 − (5 × 4 − 13 + 6)
=
=
5
1
52 + 12
or
∴
Q
Slope of RS = − 5
(a 2 + b 2 )
Put the equation of the mirror line such that the
coefficient of y becomes negative.
Suppose if
b >0
then
ax + by + c = 0
becomes
− ax − by − c = 0
and p = PM = Directed distance from P ( x 1 , y 1 ) on
− ax − by − c = 0 ( i.e. p +ve or –ve)
∴
Aliter II :
By distance form or Symmetric form or Parametric
form : Let, P ≡ ( 4, − 13) and Q be the image of P with
respect to line mirror (RS) 5x + y + 6 = 0
α − 4 β + 13
1
=
=−
5
1
2
 3 27 
M ≡ ,− 
2
2
Q M is the mid-point of P and Q, then
3


 27 
Q ≡ ( x 2 , y 2 ) ≡ 2 × − 4, 2 ×  −  + 13
 2


2
i.e.
Q ≡ ( − 1, − 14 )
Aliter I :
By Ratio Proportion Method : Let, Q ( x 2 , y 2 ) the image
of P ( 4, − 13) with respect to line mirror 5x + y + 6 = 0, then
x 2 − 4 y 2 + 13
2(5 × 4 − 13 + 6)
=
=−
= −1
5
1
52 + 12
or
x 2 − 4 = − 5 and y 2 + 13 = − 1
∴
x 2 = − 1 and y 2 = − 14
Hence Q ≡ ( − 1, − 14 ) .
26
1
θ
5
∴
Slope of PQ =
1
= tanθ
5
1
5
and cosθ =
26
26
Now, put the equation of the mirror line such that the
coefficient of y becomes negative.
Then, 5x + y + 6 = 0 becomes − 5x − y − 6 = 0 and
p = ⊥ Directed distance from P ( 4, − 13) on
( − 5x − y − 6 = 0)
13
− 5 × 4 + 13 − 6
=−
=
2
2
26
( − 5) + ( − 1)
∴
sinθ =
26
= − 26
26
Hence, required image has the coordinates
Q ≡ ( 4 − 26 cos θ, − 13 − 26 sin θ )
∴
PQ = r = 2p = −
i.e.
5
1 

, − 13 − 26 ×

 4 − 26 ×

26
26 
i.e.
( 4 − 5, − 13 − 1)
Hence, Q ≡ ( − 1, − 14 )
Image or Reflection of a Point
in Different Cases
(i) The image or reflection of a point with
respect to X -a xis
Let P(α, β) be any point and Q ( x , y ) be its image about
X-axis, then ( M is the mid-point of P and Q )
Y
P(α,β)
X′
M
O
Q(x,y)
Y′
X
Chap 02 The Straight Lines
x = α and y = − β
∴
Q ≡ (α, − β)
i.e. sign change of ordinate.
Y
X′
The image of the line ax + by + c = 0 about X-axis is
ax − by + c = 0
(ii) The image or reflection of a point with
respect to Y-axis
Let P (α, β) be any point and Q ( x , y ) be its image about
Y-axis, then ( M is the mid-point of PQ )
Y
X′
P(α,β)
M
M
(α,β)P
Remark
(x,y)Q
O
X
X
x=a
Y′
Let P (α, β) be any point and Q ( x , y ) be its image
about the line x = a, then y = β
∴ Coordinates of M are (a, β)
Q M is the mid-point of PQ
∴ Q ≡ (2a − α, β)
The image of the line ax + by + c = 0 about the line x = λ is
a ( 2λ − x ) + by + c = 0
(v) The image or reflection of a point with
respect to the line y = b
Y′
Let P (α, β) be any point and Q ( x , y ) be its image about
the line y =b, then x = α
x = − α and y = β
∴
Q ≡ ( − α, β )
i.e. sign change of abscissae.
Y
Q(x,y)
M
Remark
The image of the line ax + by + c = 0 about Y-axis is
− ax + by + c = 0
Let P (α, β) be any point and Q ( x , y ) be its image about
the origin (O is the mid point of PQ ), then
Y
P(α,β)
O
X
Y′
x = − α and y = − β
∴
Q ≡ ( − α, − β )
i.e. sign change of abscissae and ordinate.
y=b
P(α,β)
X′
(iii) The image or reflection of a point with
respect to origin
Q(x,y)
Q(x,y)
Remark
O
X′
129
X
O
Y′
∴ Co-ordinates of M are (α, b )
Q M is the mid-point of PQ ∴ Q ≡ (α, 2b − β)
Remark
The image of the line ax + by + c = 0 about the line y = µ is
ax + b ( 2µ − y ) + c = 0.
(vi) The image or reflection of a point with
respect to the line y = x
Let P (α, β) be any point and Q ( x 1 , y 1 ) be its image about
the line y = x ( RS ), then PQ ⊥ RS
Y
P(α,β)
y=
x
S
M
Remark
The image of the line ax + by + c = 0 about origin is
− ax − by + c = 0.
(iv) The image or reflection of a point with respect
to the line x = a
Q(x1,y1)
X′
O
R
Y′
X
Textbook of Coordinate Geometry
...(i)
x 1 = α and y1 = − β
Now, let P2 ( x 2 ,y 2 ) be the image of P1( x 1, y1 ) in the Y-axis.
Then
∴ ( Slope of PQ ) × ( Slope of RS ) = −1
y1 − β
or
×1= −1
x1 −α
Y
or
x1 − α = β − y1
and mid--point of PQ lie on y = x
y1 + β  x 1 + α 
i.e.

 =

 2   2 
...(i)
X'
or
...(ii)
x1 + α = β + y1
Solving Eqs. (i) and (ii), we get x 1 = β and y 1 = α
∴ Q ≡ (β, α ) i.e. interchange of x and y.
Remark
The image of the line ax + by + c = 0 about the line y = x is
ay + bx + c = 0.
(vii) The image or reflection of a point with
respect to the line y = x tan θ
Let P (α, β) be any point and Q ( x 1 , y 1 ) be its image about
the line y = x tan θ ( RS ), then PQ ⊥ RS
∴ ( Slope of PQ ) × ( Slope of RS ) = − 1
y1 − β
or
× tan θ = − 1
x1 −α
⇒
y 1 − β = (α − x 1 ) cot θ
and mid-point of PQ lie on y = x tan θ
y1 + β  x 1 + α 
i.e.

 =
 tan θ
 2   2 
...(i)
Y
ta
=x
nθ
S
θ
R
O
Y′
Q(x1,y1)
P2 (x2,y2)
P1(x1,y1)
Y'
x 2 = − x 1, y 2 = y1
[from Eq. (i)] ...(ii)
⇒
x 2 = − α, y 2 = − β
further let P3 ( x 3 , y 3 ) be the image of P (α, β ) in the origin
O. Then
…(iii)
x 3 = − α, y 3 = − β
From Eqs. (ii) and (iii) ,we get
x 3 = x 2 and y 3 = y 2 .
Hence the image of P2 of P after successive reflection in
their X-axis and Y-axis is the same as the single reflection
of P in the origin.
y Example 96. Find the image of the point ( − 2 , − 7 )
under the transformations ( x , y ) → ( x − 2y , − 3x + y ).
Sol. Let ( x 1, y1 ) be the image of the point ( x , y ) under the given
transformation. Then
x 1 = x − 2y = ( − 2) − 2 ( − 7 ) = 12
∴
x 1 = 12 and y1 = − 3x + y = − 3 ( − 2) − 7 = − 1
∴
y1 = − 1
Hence, the image is (12, − 1).
y Example 97. The image of the point A(1, 2) by the line
mirror y = x is the point B and the image of B by the
line mirror y = 0 is the point (α , β) . Find α and β.
y
M
X′
X
O
X
or
…(ii)
y 1 + β = ( x 1 + α ) tan θ
Solving Eqs. (i) and (ii), we get
x 1 = α cos 2θ + β sin 2θ
and
y 1 = α sin 2θ − β cos 2θ
∴
Q ≡ (α cos 2θ + β sin 2θ, α sin 2θ − β cos 2θ )
y Example 95. The point P (α , β) undergoes a reflection
in the X-axis followed by a reflection in the Y -axis.
Show that their combined effect is the same as the
single reflection of P (α , β) in the origin when α , β > 0.
Sol. Let P1 ( x 1, y1 ) be the image of (α, β ) after reflection in the
X -axis. Then
Sol. Let ( x 1, y1 ) be the image of the point (1, 2) about the line
y = x.
x 1 = 2, y 1 = 1
Then
...(i)
Y
A(1,2)
x
(α,β)P
P(α,β)
y=
130
B(x1,y1)
X′
O
Y′
X
(α,β)
Also given image of B( x 1, y1 ) by the line mirror y = 0 is
(α, β ). Then α = x 1 = 2
[from Eq. (i)]
and
β = − y1 = − 1
Chap 02 The Straight Lines
α = 2 and β = − 1
Hence,
y Example 98. The point (4, 1) undergoes the following
three transformations successively :
(i) Reflection about the line y = x .
(ii) Translation through a distance 2 units along the
positive direction of X-axis.
(iii) Rotation through an angle π / 4 about the origin in
the anticlockwise direction.
Then, find the coordinates of the final position.
Aliter (Use of complex number) :
Let Q be the reflection of P ( 4, 1) about the line y = x , then
Q ≡ (1, 4 )
QQ move 2 units along the +ve direction of X-axis, if new
point is ,R then R ≡ (3, 4 ) .
If
R(3, 4 ) = R(z1 )
when
z1 = ( 3 + 4i )
then
R ′ ( x , y ) = R ′ (z 2 )
∴ Coordinates of Q is (1, 4).
Given that Q move 2 units along the positive direction of
X-axis.
Y
R'
5
5
π/4
θ
X′
P(4,1)
X
O
Y′
π
π

= (3 + 4i )  cos + i sin 

4
4
7i 
i   1
 1
= ( 3 + 4i ) 
+
+
 = −

 2
2 
2
2
 1 7 
Hence, new coordinates are  −
, .

2 2
Use of Image or Reflection
y=x
R(3,4)
(1,4)Q
∴ Coordinates of R is ( x 1 + 2, y1 )
or
R ( 3, 4 )
If OR makes an angle θ , then
4
tan θ =
3
4
3
and cos θ =
sin θ =
∴
5
5
π
After rotation of
let new position of R is R′ and
4
To make problems simpler and easier use Image or
reflection.
Types of problems : (i) If vertex of a ∆ ABC and
equations of perpendicular bisectors of AB and AC are
given, then B and C are the images or reflections of A
about the perpendicular bisectors of AB and AC
(where M and N are the mid-points of AB and AC ).
A
M
OR′ makes an angle ( π / 4 + θ ) with X-axis.


π

π
Coordinates of R ′ OR ′ cos  + θ , OR ′ sin  + θ 

4

4


 1
R ′ OR ′ 
cos θ −
 2

 1
OR′ sin 
cos θ +
 2
1

sin θ ,

2
1

sin θ 

2
⇒
  3
4 
4   3
R′ 5 
−
+
,5 




 5 2 5 2
5 2 5 2
⇒
 1 7 
, 
R′  −

2 2
N
B
OR = OR ′ = 32 + 4 2 = 5
∴
π

Q ∠ROR ′ = 

4
z 2 = z1e iπ / 4
∴
Sol. Let Q ( x 1, y1 ) be the reflection of P about the line y = x .
Then
x 1 = 1

y 1 = 4
i.e.
131
C
y Example 99. The base of a triangle passes through a
fixed point ( f , g ) and its sides are respectively bisected
at right angles by the lines y + x = 0 and y − 9 x = 0.
Determine the locus of its vertex.
Sol. Let A ≡ (α, β ) the image of A (α, β ) about y + x = 0 is B,
then B ≡ ( − β, − α ) and if image of A (α, β ) about y − 9 x = 0
is C ( x 2 , y 2 ), then
x 2 − α y 2 − β −2(β − 9α )
=
=
−9
1
1 + 81
∴
x2 =
9 β − 40 α
40 β + 9 α
and y 2 =
41
41
132
Textbook of Coordinate Geometry
 9β − 40α 40β + 9α 
,
C ≡


41
41 
∴
Y
C
h − 3 k + 1 − 2 (3 + 4 + 10)
= −2
=
=
1
−4
12 + ( − 4 ) 2
then
y–9x=
0
A(α,β)
i.e.
h = 1, k = 7
∴
L ≡ (1, 7 )
Q F be the mid-point of AB.
X′
Y
C
X
O
y+
L(1,7)
x=
0
B/2
(f,g)D
B (10,5)
B/2
E
F(13/2,2)
B
X′
Y′
X
O
A(3,–1)
Hence, B, D , C are collinear, then

 −β
1
 f
2
9β − 40α
 41
⇒
Y′

−α
1
g
1 = 0
40β + 9α 1


41
4 ( α 2 + β 2 ) + ( 4 g + 5 f ) α + ( 4 f − 5g ) β = 0
Hence, locus of vertex is
4 ( x 2 + y 2 ) + ( 4 g + 5 f ) x + ( 4 f − 5g ) y = 0
(ii) The images or reflections of vertex A of a ∆ ABC
about the angular bisectors of angles B and C lie on
the side BC. (By congruence) A1 and A2 are the
images of A about the angle bisectors BE and CF
respectively, where M and N are the mid-points of
AA1 and AA2 .
A
E
M
F
N
B/2
B
C/2
B/2
A2
C/2
A1
C
y Example 100. Find the equations of the sides of the
triangle having ( 3 , − 1) as a vertex, x − 4 y + 10 = 0 and
6 x + 10y − 59 = 0 being the equations of an angle
bisector and a median respectively drawn from
different vertices.
Sol. Let BE be the angle bisector and CF be the median. Given
equations of BE and CF are x − 4y + 10 = 0 and
6x + 10y − 59 = 0 respectively.
Since, image of A with respect to BE lie on BC . If image of
A is L (h , k ).
Let
F ≡ ( α, β )
then
B ≡ (2α − 3, 2β + 1)
Q B lie on BE, then
(2α − 3) − 4 (2β + 1) + 10 = 0
i.e.
2α − 8β + 3 = 0
and F lie on CF, then
6α + 10β − 59 = 0
Solving Eqs. (i) and (ii), we get
13
α = ,β = 2
2
 13 
then
F ≡  , 2
2 
...(i)
...(ii)
and
B = (10, 5)
Equation of AB is
2+1
y +1=
( x − 3)
13
−3
2
6
⇒
y + 1 = ( x − 3)
7
or
6x − 7y − 25 = 0
Equation of BC is
7 −5
( x − 10)
y −5=
1 − 10
2
( x − 10)
9
or
2x + 9y − 65 = 0
Q CA is the family of lines of CB and CF
then (2x + 9y − 65) + λ (6x + 10y − 59 ) = 0
it pass through A (3, − 1)
then (6 − 9 − 65) + λ (18 − 10 − 59 ) = 0
4
∴
λ=−
3
⇒
y −5 = −
...(iii)
Chap 02 The Straight Lines
From Eq. (iii), we get equation of AC is
18x + 13y − 41 = 0
(iii) Optimization (Minimization or Maximization)
(a) Minimization : Let A and B are two given points on
the same side of ax + by + c = 0. Suppose we want to
determine a point P on ax + by + c = 0 such that
PA + PB is minimum. Then find the image of A or B
about the line ax + by + c = 0 (say A′ or B ′ ) then join
B ′ with A or A′ with B wherever it intersects
ax + by + c = 0 is the required point.
133
(b) Maximization : Let A and B are two given points on
the same side of ax + by + c = 0. Suppose we want to
determine a point P on ax + by + c = 0 such that
| PA − PB | is maximum, then find the equation of line
AB wherever it intersects ax + by + c = 0 is the
required point.
A
B
P
y+c=
ax+b
0
A
c =0
B ax+by+
P
By triangle inequality
Difference of two sides of a triangle < Third side
i.e.
|PA − PB| = | AB| (maximum value)
B'
y Example 102. Find a point P on the line
3x + 2y + 10 = 0 such that |PA − PB| is maximum where
A is (4, 2) and B is (2, 4).
A'
∴
or
PA + PB = PA + PB ′
PA + PB = PA ′ + PB
Sol. Let,
Remark
By triangle in equality
Sum of two sides of a triangle > Third side
i.e.|PA + PB| = |PA + PB′| = | AB′| (minimum value).
y Example 101. Find a point R on the X-axis such that
PR + RQ is the minimum, when P ≡ (1, 1) and Q ≡ ( 3, 2) .
Sol. Since P and Q lie on the same side of X-axis.
The image of Q (3, 2) about X -axis is Q ′ (3, − 2) then the
equation of line PQ ′ is
−2−1
( x − 1)
y −1=
3−1
⇒
3x + 2y − 5 = 0
Y
Remark
L ( x , y ) = 3x + 2y + 10
∴
L ( 4, 2) = 12 + 4 + 10 = 26
and
L (2, 4 ) = 6 + 8 + 10 = 24
∴ A and B lie on the same side of the line
3x + 2y + 10 = 0
Equation of line AB is
4 −2
(x − 4)
y −2=
2− 4
...(i)
or
...(ii)
x +y−6=0
This line Eq. (ii) meets Eq. (i) at P ≡ ( − 22, 28) which is the
required point.
Aliter
P be ( x 1, y1 ) and ∠ APB = θ
( PA )2 + ( PB )2 − ( AB )2
cos θ =
2PA ⋅ PB
Let
then
Y
Q (3,2)
B
P (1,1)
X′
O
θ
X
R
X′
Y′
Q' (3,–2)
5 
This line meets X-axis at R  , 0 which is the required
3 
point.
P (x1,y1)
O
Y′
A
X
134
Textbook of Coordinate Geometry
since
⇒
cos θ ≤ 1
( PA )2 + ( PB )2 − ( AB )2
≤1
2PA ⋅ PB
⇒
( PA − PB )2 ≤ ( AB )2
⇒
⇒
| PA − PB | ≤ | AB|
| PA − PB | ≤ 2 2
i.e. P lies on the line AB as well as on the given line.
P
∴ Equation of AB is
4 −2
(x − 4)
y −2=
2− 4
B
⇒
y −2= − x + 4
...(i)
⇒
x +y =6
and given line
...(ii)
3x + 2y + 10 = 0
Solving Eqs. (i) and (ii), we get P ( − 22, 28).
Maximum value of | PA − PB | is 2 2
when, θ = 0.
Exercise for Session 5
1.
The coordinates of the foot of the perpendicular from (2, 3) to the line 3x + 4y − 6 = 0 are
14 17
(b)  ,− 
 25 25 
14 27
(d)  , 
 25 25 
14 27
(a)  − , − 
 25 25 
14 17 
(c)  −
, 
 25 25 
2.
If the foot of the perpendicular from the origin to a straight line is at the point (3, − 4). Then the equation of the
line is
(a) 3x − 4y = 25
(c) 4x + 3y − 25 = 0
3.
The coordinates of the foot of the perpendicular from (a, 0) on the line y = mx +
1
(a)  0,− 

a
4.
(b) 3x − 4y + 25 = 0
(d) 4x − 3y + 25 = 0
a
(b)  0, 
 m
(c)
8.
(d)
1 2
(a2 + b 22 − a12 − b12 )
2
The image of the point (3, 8) in the line x + 3y = 7 is
(b) (4, 1)
(c) (−1, − 4)
(d) (−4, − 1)
The image of the point (4, –3) with respect to the line y = x is
(a) (−4, − 3)
7.
(b) (a12 + b12 − a22 − b 22
1 2
(a1 + a22 + b12 + b 22 )
2
(a) (1, 4)
6.
1
(d)  0, 
 a
If the equation of the locus of a point equidistant from the points (a1, b1) and (a 2, b 2 ) is
(a1 − a 2 ) x + (b1 − b 2 ) y + c = 0, then the value of c is
(a) a12 − a22 + b12 − b 22
5.
a
(c)  0, − 

m
a
are
m
(b) (3, 4)
(c) (−4, 3)
(d) (−3, 4)
The coordinates of the image of the origin O with respect to the straight line x + y + 1 = 0 are
1
1
(a)  − , − 
 2
2
(b) (−2, − 2)
(c) (1, 1)
(d) (−1, − 1)
If ( −2, 6) is the image of the point (4, 2) with respect to the line L = 0, then L ≡
(a) 6x − 4y − 7 = 0
(c) 3x − 2y + 5 = 0
(b) 2x − 3y − 5 = 0
(d) 3x − 2y + 10 = 0
A
Chap 02 The Straight Lines
9.
10.
The image of P (a, b ) on the line y = − x is Q and the image of Q on the line y = x is R. Then the mid-point of
PR is
(b) 

(c) (a − b , b − a )
(d) (0, 0)
The nearest point on the line 3x − 4y = 25 from the origin is
(b) (3, − 4)
(c) (3, 5)
(d) (−3, 5)
Consider the points A (0, 1) and B (2, 0), P be a point on the line 4x + 3y + 9 = 0. The coordinates of P such that
| PA − PB | is maximum are
12 17
(a)  − , 
 5 5
12.
a + b b + 2
,

2
2 
(a) (a + b , a + b )
(a) (3, 4)
11.
135
84 13
(b)  − , 
 5 5
6 17
(c)  − , 
 5 5
(d) (0, − 3)
Consider the points A (3, 4) and B (7, 13). If P is a point on the line y = x such that PA + PB is minimum, then the
coordinates of P are
12 12
(a)  , 
 7 7
13 13
(b)  , 
 7 7
31 31
(c)  , 
 7 7
(d) (0, 0)
13.
The image of a point P(2, 3) in the line mirror y = x is the point Q and the image of Q in the line mirror y = 0 is
the point R( x , y ). Find the coordinates of R.
14.
The equations of perpendicular bisector of the sides AB and AC of a ∆ ABC are x − y + 5 = 0 and x + 2y = 0
respectively. If the point A is (1, –2), find the equation of line BC.
15.
In a ∆ ABC, the equation of the perpendicular bisector of AC is 3x − 2y + 8 = 0. If the coordinates of the point A
and B are (1, − 1) and (3, 1) respectively, find the equation of the line BC and the centre of the circumcircle of
∆ ABC.
16.
Is there a real value of λ for which the image of the point ( λ , λ − 1) by the line mirror 3x + y = 6λ is the point
( λ2 + 1, λ ) ? If so find λ.
Session 6
Reflection of Light, Refraction of Light, Conditions of
Collinearity if Three Given Points be in Cyclic Order
Reflection of Light
Y
R
A'
N
R
A(3,0)
O
X′
P(1,–1)
Point of incidence
P
=0
ray
x
r
i
0
I
Reflecting surface
=
y–3
x–2
X
y–
5
nt
ide
x
3x
–2
Inc
of
Angle
incid
I
Normal
N
Angle
of ref
Re
lectio
fle
n
c te
dr
ay
ence
When a ray of light falls on a smooth polished surface
(Mirror) separating two media, a part of it is reflected back
into the first medium.
3x − 2y − 5 = 0
∴
x = 1, y = − 1
∴ Coordinates of P are (1, − 1).
Y′
IP is the incident ray and PR is the reflected ray. A
perpendicular drawn to the surface, at the point of
incidence P is called the normal. Hence PN is the normal.
The angle between the incident ray and the normal
( ∠ IPN ) is called the angle of incidence which is
represented by ∠ i
i.e. ∠ IPN = ∠i = Angle of incidence and the angle
between the reflected ray and the normal ( ∠ IPR ) is called
the angle of reflection which is represented by ∠r.
i.e. ∠ IPR = ∠ r = Angle of reflection.
Since, slope of line mirror is 3/2.
Slope of PN = − 2 / 3 and
∴
slope of IP = 1 / 2, line PN is equally inclined to IP and PR,
then

 1  2 
 2 
 m − −  
 − −  
 3
 3

 =− 2

1  2 
 2 


1
1
+
−
m
+
−
 
 


 3 


2  3 
Laws of Reflection :
⇒
(i) The incident ray, normal and the reflected ray to a
surface at the point of incidence all lie in the same
plane.
(ii) The angle of incidence = angle of reflection
i.e.
∠i = ∠r
y Example 103. A ray of light is sent along the line
x − 2y − 3 = 0. Upon reaching the line 3x − 2y − 5 = 0,
the ray is reflected from it. Find the equation of the
line containing the reflected ray.
Sol. To get coordinates of point P, we solve the given equation
of lines together as
x − 2y − 3 = 0
Let slope of reflected ray be m.
⇒
∴
⇒
3m + 2
7
=−
3 − 2m
4
12m + 8 = − 21 + 14m
2m = 29
29
m=
2
∴ Equation of reflected ray y + 1 =
29
( x − 1)
2
⇒
2y + 2 = 29 x − 29
⇒
29 x − 2y − 31 = 0.
Aliter (Image method) : Take A (3, 0) be any point on IP
and if A ′ (α, β ) be the image of A about the mirror line
3x − 2y − 5 = 0, then
α − 3 β − 0 − 2 ( 9 − 0 − 5)
=
=
3
9+4
−2
137
Chap 02 The Straight Lines
15
16
and β =
13
13
 15 16
A′ ≡  , 
 13 13
α=
∴
Refraction of Light
29
( x − 1) or 29 x − 2y − 31 = 0
2
ay
y +1=
r
nt
de
⇒
ci
In
∴ Equation of A ′ P is the equation of the reflected ray then
its equation is,
 16

 + 1
 13

( x − 1)
y +1=

 15
1
−



 13
When a ray of light falls on the boundary separating the
two transparent media, there is a change in direction of
ray. This phenomenon is called refraction.
(–5,6)A'
P
B(7,2)
O
X
2x+
X′
(–1,–2)B'
Angle of refraction
y
A(3,10)
Laws of Refraction
(i) The incident ray, normal and the refracted ray to the
surface separating the two transparent media all lie in
the same plane.
(ii) The ratio of sine of angle of incidence to the sine of
the angle of refraction is constant for the two given
media. The constant is called the refractive index of
medium 2 with respect to medium 1.
sin i
i.e.
1 µ2 =
sin r
y Example 105. A ray of light is sent along the line
x − 6 y = 8. After refracting across the line x + y = 1 it
enters the opposite side after turning by 15° away from
the line x + y = 1. Find the equation of the line along
which the refracted ray travels.
0
=
y–6
Sol. The point of intersection of x − 6y = 8 and x + y = 1 is
A ≡ (2, − 1). Let the required ray have the slope = m, then
Y′
and
Deviation
n ra
Y
r
actio
∴
i.e.
Boundary
Medium 2
Refr
= −4
α = − 5, β = 6
A ′ ≡ ( − 5, 6)
Angle of incidence
i
Medium 1
y Example 104. A light beam, emanating from the point
(3, 10) reflects from the straight line 2x + y − 6 = 0 and,
then passes through the point (7, 2). Find the
equations of the incident and reflected beams.
Sol. Let images of A and B about the line 2x + y − 6 = 0 are
A ′ (α, β ) and B′( γ , δ ) respectively.
α − 3 β − 10 − 2 (6 + 10 − 6)
Then,
=
=
2
1
22 + 12
Normal
∴
Y
γ − 7 δ − 2 − 2 (14 + 2 − 6)
=−4
=
=
2
1
22 + 12
∴
γ = − 1, δ = − 2
i.e.
B′ ≡ ( − 1, − 2).
∴ Equation of incident ray AB′ is
10 + 2
( x + 1) or 3x − y + 1 = 0
y +2=
3+1
X
O
A
(2,–1)
8
x–6y=
15°
m
y=
x+
1
and equation of reflected ray A ′ B is
2−6
( x + 5)
y −6=
7 +5
1
⇒
y − 6 = − ( x + 5)
3
or
x + 3y − 13 = 0
X′
Y′
m − 1
6
tan 15° =
 m
1+

6
138
Textbook of Coordinate Geometry
⇒
 6m − 1
2− 3 = ± 

 m +6
then,
6m − 1
= 2 − 3 or 3 − 2
m +6
70 − 37 3
13
37 3 − 70
or
m=
61
Let the angle between x + y = 1 and the line through
70 − 37 3
be α, then
A(2, − 1) with the slope
13
At 3 + Bt 2 + Ct + D = 0
m=
⇒
70 − 37 3

− ( − 1)

83 − 37 3
13

tan α =
 =
 1 −  70 − 37 3   37 3 − 57
13



=
83 − 37 3
37 3 − 57
and if angle between x + y = 1 and the line through
37 3 − 70
be β, then
A (2, − 1) with the slope
61
37 3 − 70

− ( − 1)

 37 3 − 9 
61

tan β =
 =


37
3
70
−
131 − 37 3
1 − 


61



37 3 − 9
=
131 − 37 3
Here tan α > tan β, ∴α > β
70 − 37 3
therefore the slope of the refracted ray =
13
∴ The equation of the refracted ray is
(70 − 37 3 )
( x − 2)
y +1=
13
13y + 13 = (70 − 37 3 ) x − 140 + 74 3
⇒
or
Then
lf (t ) + m g(t ) + n = 0
where,
t = a, b , c
i.e. a, b, c are the roots of the Eq. (i).
In this case Eq. (i) must be cubic in t.
(70 − 37 3 ) x − 13y − 153 + 74 3 = 0
Conditions of Collinearity if
Three given points are in
Cyclic Order
Let the three given points
A ≡ ( f (a ), g(a )), B ≡ ( f (b ), g(b ))
and C ≡ ( f (c ), g(c )) lie on the line
lx + my + n = 0, where l, m and n are constants.
...(i)
(say)
B
C
, ab + bc + ca =
A
A
D
and
abc = −
A
which are the required conditions.
a +b +c = −
then
y Example 106. If the points
 a 3 a 2 − 3  b 3 b 2 − 3
 c 3 c 2 − 3
 and 
, 
 are

,
,
,
 a − 1 a − 1  b − 1 b − 1 
c − 1 c − 1 
collinear for three distinct values a, b , c and different
from 1, then show that
abc − (bc + ca + ab ) + 3 (a + b + c ) = 0.
Sol. Let the three given points lie on the line
where l , m and n are constants.
lx + my + n = 0,
 t 2 − 3
 t3 
Then, l 
 +n =0
 +m 
 t −1
 t − 1
⇒
lt 3 + mt 2 + nt − (3m + n ) = 0
for t = a, b, c
i.e. a, b, c are the roots of
lt 3 + mt 2 + nt − 3m − n = 0
then
and
a+b+c = −
n
m
, ab + bc + ca =
l
l
 3m + n 
abc = 

 l 
Now, abc − (bc + ca + ab ) + 3 (a + b + c )
 3m + n  n 3m
=
=0
 − −
 l  l
l
Hence, abc − (bc + ca + ab ) + 3 (a + b + c ) = 0
y Example 107. If t 1 , t 2 and t 3 are distinct, the points
(t 1 , 2at 1 + at 13 ), (t 2 , 2at 2 + at 23 ) and (t 3 , 2at 3 + at 33 ) are
collinear, then prove that t 1 + t 2 + t 3 = 0.
Sol. Let the three given points lie on the line lx + my + n = 0,
where l, m and n are constants. Then,
l (t ) + m (2at + at 3 ) + n = 0
⇒
(am ) t 3 + (2am + l ) t + n = 0
for t = t 1, t 2 , t 3
i .e ., t 1, t 2 , t 3 are the roots of Eq. (i), then
t1 + t 2 + t 3 = 0
...(i)
Chap 02 The Straight Lines
139
Exercise for Session 6
1.
A ray of light passing through the point (1, 2) is reflected on the X-axis at a point P and passes through the
point (5, 3). The abscissae of the point P is
(a) 3
2.
13
3
(b) 8x + y = 9
13
4
(c) 7x − y = 6
(d) None of these
(b) x (2 + 3 ) + y = 2 + 3
(d) x + (2 + 3)y = (2 + 3 )
(b) 2x + y − 10 ≥ 0
(d) −2x + y − 3 ≥ 0
Let O be the origin and let A(1, 0), B(0, 1) be two points. If P ( x , y ) is a point such that xy > 0 and x + y < 1then
(a) P lies either inside in ∆OAB or in third quadrant
(c) P lies inside the ∆OAB
6.
(d)
All the points lying inside the triangle formed by the points (0, 4), (2, 5) and (6, 2) satisfy
(a) 3x + 2y + 8 ≥ 0
(c) 2x − 3y − 11≥ 0
5.
13
5
A ray of light travelling along the line x + y = 1is incident on the X-axis and after refraction it enters the other
side of the X-axis by turning π / 6 away from the X-axis. The equation of the line along which the refracted ray
travels is
(a) x + (2 − 3) y = 1
(c) (2 − 3 )x + y = 1
4.
(c)
The equation of the line segment AB is y = x . If A and B lie on the same side of the line mirror 2x − y = 1, then
the image of AB has the equation
(a) x + y = 2
3.
(b)
(b) P cannot be inside in ∆OAB
(d) None of these
A ray of light coming along the line 3x + 4y − 5 = 0 gets reflected from the line ax + by − 1 = 0 and goes along
the line 5x − 12y − 10 = 0 then
64
112
,b =
115
15
64
8
,b =
(c) a =
115
115
(a) a =
64
8
,b =
115
115
64
−8
(d) a = −
,b =
115
115
(b) a = −
7.
Two sides of a triangle have the joint equation x 2 − 2xy − 3y 2 + 8y − 4 = 0. The third side, which is variable,
always passes through the point ( −5,−1). Find the range of values of the slope of the third side, so that the origin
is an interior point of the triangle.
8.
Determine the range of values of θ ∈[0, 2π ] for which (cos θ, sin θ ) lies inside the triangle formed by the lines
x + y − 2 = 0, x − y − 1 = 0 and 6x + 2y − 10 = 0.
9.

 3 
3
Let P(sin θ, cos θ ), where 0 ≤ θ ≤ 2π be a point and let OAB be a triangle with vertices (0, 0), 
, 0 and 0,
.
 2
 2 
Find θ if P lies inside the ∆ OAB.
10.
Find all values of θ for which the point (sin2 θ, sin θ ) lies inside the square formed by the line xy = 0 and
4xy − 2x − 2y + 1 = 0.
11.
Determine whether the point ( −3, 2) lies inside or outside the triangle whose sides are given by the equations
x + y − 4 = 0, 3x − 7y + 8 = 0, 4x − y − 31 = 0.
12.
A ray of light is sent along the line x − 2y + 5 = 0, upon reaching the line 3x − 2y + 7 = 0, the ray is reflected
from it. Find the equation of the line containing the reflected ray.
140
Textbook of Coordinate Geometry
Shortcuts and Important Results to Remember
1. Area of parallelogram formed by the lines
a1 x + b1 y + c1 = 0, a2 x + b2 y + d1 = 0, a1 x + b1 y + c 2 = 0 and
a2 x + b2 y + d 2 = 0 is
6. If A ≡ ( x1, y1 ), B ≡ ( x2 , y2 ) and C ≡ ( x3 , y3 ) are the vertices of
a ∆ ABC then the equations of medians AD, BE and CF are
y ( x2 + x3 − 2 x1 ) − x ( y2 + y3 − 2 y1 )
| c1 − c 2 ||d1 − d 2 |
a1 b1
|
|
a2 b2
= y1 ( x2 + x3 ) − x1 ( y2 + y3 );
y ( x3 + x1 − 2 x2 ) − x ( y3 + y1 − 2 y2 )
= y2 ( x3 + x1 ) − x2 ( y3 + y1 )
2. Area of parallelogram formed by the lines
y = m1 x + c1, y = m2 x + d1, y = m1 x + c 2 and y = m2 x + d 2
is
| c1 − c 2 ||d1 − d 2 |
| m1 − m2 |
and y ( x1 + x2 − 2 x3 ) − x ( y1 + y2 − 2 y3 ) = y3 ( x1 + x2 )
− x3 ( y1 + y2 ) respectively.
A
E
3. If A ≡ ( x1, y1 ), B ≡ ( x2 , y2 ) and C ≡ ( x3 , y3 ) are the vertices of
a ∆ ABC, then angle A is acute or obtuse according as
( x1 − x2 ) ( x1 − x3 ) + ( y1 − y2 ) ( y1 − y3 ) > 0
or
<0
( x2 − x3 ) ( x2 − x1 ) + ( y2 − y3 ) ( y2 − y1 ) > 0 or
<0
F
Similarly, for ∠ B
G
B
C
D
Where, G is the centroid of ∆ ABC.
and for ∠ C
( x3 − x1 ) ( x3 − x2 ) + ( y3 − y1 ) ( y3 − y2 ) > 0
<0
or
4. If the origin lies in the acute angle or obtuse angle
between the lines
a1 x + b1 y + c1 = 0
and
a2 x + b2 y + c 2 = 0
according as
(a1a2 + b1b2 ) c1c 2 < 0 or > 0.
7. If A ≡ ( x1, y1 ), B ≡ ( x2 , y2 ) and C ≡ ( x3 , y3 ) are the vertices of
a ∆ ABC then the equations of the altitudes AL, BM and
CN are
y ( y2 − y3 ) + x ( x2 − x3 ) = y1 ( y2 − y3 ) + x1 ( x2 − x3 );
y ( y3 − y1 ) + x ( x3 − x1 ) = y2 ( y3 − y1 ) + x2 ( x3 − x1 )
A
5. If A ≡ ( x1, y1 ), B ≡ ( x2 , y2 ) and C ≡ ( x3 , y3 ) are the vertices of
a ∆ ABC then the equations of the right bisectors
(perpendicular bisectors) of the sides BC, CA and AB are
M
N
O
 x 2 − x32   y22 − y32 
y ( y2 − y3 ) + x ( x 2 − x 3 ) =  2
 +
;
 2   2 
A
B
C
L
and y ( y1 − y2 ) + x ( x1 − x2 ) = y3 ( x1 − x2 ) + x3 ( x1 − x2 )
respectively. Where O is the orthocentre of ∆ ABC.
8. If sides of a triangle ABC are represented by
BC : a1 x + b1 y + c1 = 0,
O
B

y ( y3 − y1 ) + x ( x3 − x1 ) = 

and
CA : a2 x + b2 y + c 2 = 0
C
x32
−
2
x12 

 +
 
y32
and
−
2
y12 


 x 2 − x22   y12 − y22 
y ( y1 − y2 ) + x ( x1 − x2 ) =  1
 +

 2   2 
respectively. Where O is the circumcentre of ∆ ABC.
AB : a3 x + b3 y + c 3 = 0
then | BC|:|CA|:| AB|
a2
= (a12 + b12 ) |
a3
: (a22 + b22 )|
a3
b2
b3
b3
a1 b1
|
|: (a32 + b32 )|
a1 b1
a2
b2
|
JEE Type Solved Examples :
Single Option Correct Type Questions
n
This section contains 10 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
ONLY ONE is correct.
Ex. 1 A rectangle ABCD has its side AB parallel to y = x
and vertices A, B and D lie on y =1, x = 2 and x = − 2
respectively, then locus of vertex C is
(b) x − y = 5
(d) x + y = 5
( λ2 + 2 λ + 1 ) x + λy − 2 λ2 − 2 = 0
or
( λ2 + 1 )( x − 2 ) + λ (2 x + y ) = 0
∴ For fixed point
x − 2 = 0 and 2 x + y = 0
∴ Fixed point is (2, − 4 )
∴ Equation of required line is y + 4 = 2( x − 2 )
or
y = 2x − 8
l
(a) x = 5
(c) y = 5
or
Ex. 3 A man starts from the point P ( −3, 4 ) and reaches
point Q(0, 1) touching X-axis at R such that PR + RQ is
minimum, then the point R is
l
Sol. (c) Since AB is parallel to y = x.
∴ Equation of AB is y = x + a
Q A lies on y = 1
∴
A ≡ (1 − a, 1 )
(say)
3 
(a)  , 0
5 
 3 
(b)  − , 0
 5 
 2 
(c)  − , 0
 5 
(d) ( −2, 0)
Sol. (b) Let R = (α , 0 )
C
Y
For PR + PQ to be minimum it should be the path of light and
thus we have
y=x
B
D
P(–3, 4)
Q(0, 1)
y=1
A
q
X¢
X
O
A
(–3, 0)
x=2
x=–2
Y¢
∴
⇒
Q
∴
Q
∴
⇒
∴
Again, B lies on x = 2
B = (2, 2 + a )
Equation of AD is
y − 1 = − [ x − (1 − a )] or y = 2 − x − a
D lies on x = − 2
D ≡ ( −2, 4 − a )
Let
C ≡ (h, k )
Diagonals of rectangle bisects to each other
h + 1 −a =2 −2 ⇒ a =1 + h
and
k + 1 =2 + a + 4 −a
k =5
Locus of C is y = 5
Ex. 2 The line ( λ + 1) 2 x + λy − 2 λ 2 − 2 = 0 passes
through a point regardless of the value λ. Which of the
following is the line with slope 2 passing through the point?
l
(a) y = 2x − 8
(c) y = 2x − 4
Sol. (a)
Q
(b) y = 2x − 5
(d) y = 2x + 8
( λ + 1 ) 2 x + λy − 2 λ2 − 2 = 0
⇒
q
R
(a, 0)
B
(0, 0)
X
∆APR ~ ∆BQR
3
AR PA
α+3 4
=
⇒
= ⇒α=−
5
RB QB
0 −α 1
 3 
R ≡  − , 0
 5 
Hence,
Ex. 4 If the point P (a , a 2 ) lies inside the triangle formed
by the lines x = 0, y = 0 and x + y = 2, then exhaustive range
of ‘a’ is
l
(a) (0, 1)
(b) (1, 2 )
(c) ( 2 − 1, 1)
(d) ( 2 − 1, 2)
Sol. (a) Since the point P (a, a ) lies on y = x 2
2
Solving,
y = x2
and
x + y = 2, we get
x2 + x − 2 = 0
⇒
( x + 2 )( x − 1 ) = 0
⇒
x = − 2, 1
It is clear from figure,
A ≡ (1, 1 )
also a > 0 for I quadrant.
∴
a ∈( 0, 1 )
142
Textbook of Coordinate Geometry
Ex. 5 If 5a + 4b + 20c = t , then the value of t for which the
line ax + by + c − 1 = 0 always passes through a fixed point is
l
(a) 0
(c) 30
(b) 20
(d) None of these
ax
by
Sol. (b) Equation of line
+
+ 1 = 0 has two independent
c −1 c −1
parameters. It can pass through a fixed point if it contains
only one independent parameter. Now there must be one
a
b
and
independent of a, b and c so
relation between
c −1
c −1
a
b
can be expressed in terms of
and straight line
that
c −1
c −1
contains only one independent parameter. Now that given
5a
4b
t − 20c
relation can be expressed as
.
+
=
c −1 c −1
c −1
Ex. 8 Through the point P(α, β), when αβ > 0, the straight
x y
line + =1 is drawn so as to form with
a b
coordinate axes a triangle of area ∆ . If ab > 0, then the least
value of ∆ is
l
(a) αβ
(c) 4αβ
(b) 2αβ
(d) 8αβ
Sol. (b) Given line is
x y
+ =1
a b
∴
A ≡ (a, 0 ), B ≡ ( 0, b )
…(i)
Y
B(0, b)
RHS is independent of c if t = 20.
P(a, b)
Ex. 6 If the straight lines, ax + amy + 1 = 0,
bx + (m + 1)by + 1 = 0 and cx + (m + 2 )cy + 1 = 0, m ≠ 0 are
concurrent, then a , b, c are in
l
(a) AP only for m = 1
(c) GP for all m
X¢
(b) AP for all m
(d) HP for all m
Q Area of ( ∆AOB ) = ∆
1
(Qab > 0)
∴
| ab| = ∆ ⇒ ab = 2 ∆
2
Since, the line (i) passes through the point P(α , β ).
2∆ 
α β
α aβ

+
=1
+ =1 ⇒
∴
Q b = 

a
a b
a 2∆
Sol. (d) The three lines are concurrent if
1
am
a
b m + 1 1 = 0


c (m + 2 )c 1
Applying C 2 → C 2 − mC1, then
a 0 1
b b 1 = 0


c 2c 1
or a(b − 2c ) − 0 + 1(2bc − bc ) = 0
2ac
or b =
, which is independent of m .
a+c
Ex. 7 If a ray travelling the line x =1 gets reflected the
line x + y =1, then the equation of the line along which the
reflected ray travels is
(b) x − y = 1
(d) None of these
Sol. (a) Reflected ray is X-axis.
x=1
(0, 1)
45°
45°
O
X
(1, 0) x+y=1
∴ Equation y = 0
α 2β − 2a∆ + 2 ∆α = 0
Qa is real
∴
⇒
D≥0
4 ∆2 − 4β (2 ∆α ) ≥ 0 or ∆ ≥ 2αβ
Ex. 9 The coordinates of the point P on the line
2 x + 3y + 1 = 0, such that | PA − PB | is maximum,
where A is (2, 0) and B is (0, 2) is
l
l
Y
⇒
∴ Least value of ∆ is 2αβ.
∴ a, b, c are in HP for all m.
(a) y = 0
(c) x = 0
X
A(a, 0)
Y¢
(a) (5, − 3 )
(c) ( 9, − 7 )
(b) (7, − 5)
(d) (11, − 9)
Sol. (b) | PA − PB| ≤ | AB |
Maximum value of | PA − PB | is | AB |, which is possible only
when P , A, B are collinear
if P ( x, y ), then equation AB is
x y
+ =1
2 2
…(i)
⇒
x + y =2
Now solving Eq. (i)
and
…(ii)
2 x + 3y + 1 = 0
Then, we get,
x = 7, y = − 5
∴
P ≡ (7, − 5 )
Chap 02 The Straight Lines
The line of this family which is farthest from ( 4, − 3 ) is the line
through (1, 1) and perpendicular to the line joining (1, 1) and
( 4, − 3 )
∴ The required line is
3
y − 1 = (x − 1)
4
or
3 x − 4y + 1 = 0
Ex. 10 Equation of the straight line which belongs to the
system of straight lines a ( 2 x + y − 3 ) + b (3 x + 2y − 5 ) = 0
and is farthest from the point (4, −3) is
l
(a) 4 x + 11y − 15 = 0
(c) 7 x + y − 8 = 0
143
(b) 3 x − 4y + 1 = 0
(d) None of these
Sol. (b) The system of straight lines
a (2 x + y − 3 ) + b (3 x + 2y − 5 ) = 0 ) passes through the point
of intersection of the lines 2 x + y − 3 = 0 and 3 x + 2y − 5 = 0
i.e. (1, 1)
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
Ex. 11 The vertices of a square inscribed in the triangle
with vertices A(0, 0 ), B( 2, 1) and C(3, 0 ), given that two of its
vertices are on the side AC, are
l
3 
(a)  , 0
2 
3 3 
(b)  , 
 2 4
 9 3
(c)  , 
 4 4
9 
(d)  , 0
4 
Sol. (a, b, c, d)
Let PQRS be a square inscribed in ∆ABC and
(say)
PQ = QR = RS = SP = λ
Let
P ≡ (a, 0 ),
∴
Q ≡ (a + λ , λ ), R ≡ (a + λ , λ ) and S ≡ (a, λ )
x y
+ =1 cuts the coordinate axes at A(a , 0 )
a b
x y
and B(0, b ) and the line + = −1 at A ′ ( −a ′ , 0 ) and
a′ b′
B ′(0, − b ′ ). If the points A, B, A ′ , B ′ are concyclic, then the
orthocentre of the triangle ABA′ is
l
Ex. 12 Line
(b) (0, b ′ )
 bb ′ 
(d) 0,

 a 
(a) (0, 0)
 −aa ′ 
(c) 0,


b 
Sol. (b, c)
Q A, B, A′ , B′ are concyclic then,
Y
Y
B(0, b)
B(2, 1)
S
X¢
A (0, 0)
P
R
Q
X¢
C(3, 0)
A¢(–a¢, 0)
O
A(a, 0)
X
B¢(0,–b¢)
Y¢
Now equation of AB is
x − 2y = 0
and equation of BC is
x + y −3 = 0
Q S lies on AB, then
a − 2λ = 0
and R lies on BC, then
a + λ + λ − 3 = 0 or a + 2 λ − 3 = 0
3
3
From Eqs. (iii) and (iv), we get a = , λ =
2
4
3 
9 
Hence,
P ≡  , 0 , Q ≡  , 0 ,
2 
4 
 9 3
 3 3
R≡ , , S = , 
 4 2
 2 4
X
Y¢
…(i)
…(ii)
…(iii)
…(iv)
OA ⋅ OA′ = OB ⋅ OB′
or
(a ) ⋅ ( −a′ ) = (b ) ⋅ ( −b′ )
or
aa′ = bb′
The equation of altitude through A′ is
a
y − 0 = ( x + a′ )
b
It intersects the altitude
aa′
x = 0 at y =
b
 aa′ 
∴ Orthocentre is  0,
 or ( 0, b′ )
 b 
…(i)
[from Eq. (i)]
144
Textbook of Coordinate Geometry
Ex. 13 Two straight lines u = 0 and v = 0 passes through
7
the origin and angle between them is tan −1   . If the ratio
 9
9
of the slope of v = 0 and u = 0 is , then their equations are
2
l
y=Ö3 x
Y
B
C
(a) y = 3 x and 3y = 2x
(b) 2y = 3 x and 3y = x
(c) y + 3 x = 0 and 3y + 2x = 0
(d) 2y + 3 x = 0 and 3y + x = 0
2
2
Sol. (a, b, c, d)
X¢
Let the slope of u = 0 be m, then the slope of v = 0 is
Therefore,
or
9m
.
2
Hence, coordinates of B are ( 3 + 1, 3 + 1 ) and
( − 3 − 1, − 3 − 1 )
Ex. 15 A and B are two fixed points whose coordinates
are (3, 2) and (5, 4) respectively. The coordinates of a point P,
if ABP is an equilateral triangle are
l
⇒
9m 2 + 9m + 2 = 0 or 9m 2 − 9m + 2 = 0
2 1
2 1
⇒
m = − , − or m = ,
3 3
3 3
Therefore, the equation of lines are
(i) 2 x + 3y = 0 and 3 x + y = 0
(ii) x + 3y = 0 and 3 x + 2y = 0
(iii) 2 x = 3y and 3x = y
(iv) x = 3y and 3 x = 2y
(a) ( 4 − 3 , 3 + 3 )
(b) ( 4 + 3 , 3 − 3 )
(c) (3 − 3 , 4 + 3 )
(d) (3 + 3 , 4 − 3 )
Sol. (a, b)
Q
P
Ex. 14 Two sides of a rhombus OABC (lying entirely in
the first or third quadrant) of are equal to 2 sq units are
x
, y = 3 x . Then the possible coordinates of B is/are
y=
3
(O being the origin)
(c) (3 + 3 , 3 + 3 )
(d) ( 3 − 1, 3 − 1)
AB = AP = BP = 2 2
∴ Coordinates of P are (3 + 2 2 cos105 °, 2 + 2 2 sin 105 ° )
l
Sol. (a, b)
Here, ∠COA = 30 °
Let OA = AB = BC = CO = x
Q Area of rhombus OABC
1
= 2 × × x × x sin 30 °
2
x2
=
=2
2
∴
x =2
X
O
Coordinates of A and C are ( 3, 1 ) and (1, 3 ) in I quadrant and
in III quadrant are ( − 3, − 1 ) and (−1, − 3 )
 −7m  7

=
2
2 + 9m  9
(b) ( −1 − 3 , − 1 − 3 )
A
Y¢
 m − 9m 

2 = 7
9m 9

1+m×

2
(a) (1 + 3 , 1 + 3 )
y= x
Ö3
D
B(5, 4)
60°
45°
A(3, 2)
[given]
or
(3 − ( 3 − 1 ), 2 + 3 + 1 )
or
(4 − 3, 3 + 3 )
If P below AB, then coordinates of P are
(3 + 2 2 cos15 °, 2 − 2 2 sin 15 °)
or
[(3 + 3 + 1, 2 − ( 3 − 1 )]
or
(4 + 3, 3 − 3 )
X
Chap 02 The Straight Lines
145
JEE Type Solved Examples :
Paragraph Based Questions
n
This section contains 2 solved paragraphs based upon
each of the paragraph 3 multiple choice questions have
to be answered. Each of these question has four choices
(a), (b), (c) and (d) out of which ONLY ONE is correct.
16. (d) Q M lie on y = 1
∴ Length of ⊥ from M to OA is 1.
17. (c) λ = Perimeter of region R
= OA + AD + DE + EO
=3 + 2 + 1 + 2
Paragraph I
(Q. Nos. 16 to18)
Let d (P , OA ) ≤ min⋅ {d (P , AB ), d (P , BC ), d (P , OC )}, where d
denotes the distance from the point to the corresponding the
line and R be the region consisting of all those points P inside
the rectangle OABC such that O ≡ (0, 0 ), A ≡ (3, 0 ), B ≡ (3, 2 )
and C ≡ (0, 2 ). Let M be the peak of region R.
=4+2 2
18. (a) ∆ = Area of region R
1
= (OA + ED ) × 1
2
1
= (3 + 1 ) = 2
2
16. Length of the perpendicular from M to OA is
(a) 4
(b) 3
(c) 2
(Q. Nos. 19 to 21)
17. If λ be the perimeter of region R, then λ is
(a) 4 − 2
(b) 4 + 2
A variable straight line ‘L’ is drawn through O(0, 0 ) to meet
this lines L1 : y − x − 10 = 0 and L 2 : y − x − 20 = 0 at the
points A and B respectively.
2
1
1
,
19. A point P is taken on ‘L’ such that
=
+
OP OA OB
then the locus of P is
(c) 4 + 2 2 (d) 10
18. If ∆ be the area of region R, then ∆ is
(a) 2
(b) 4
(c) 6
(d) 8
Sol. Let P ≡ ( x, y )
Q d ( P , OA ) ≤ min ⋅ {d ( P , AB ), d ( P , BC ), d ( P , OC )}
(a) 3 x + 3y − 40 = 0
(c) 3 x − 3y − 40 = 0
Y
(0, 2) C
B(3, 2)
O
A(3, 0)
X
(a) (y − x )2 = 25
(b) (y − x )2 = 50
(c) (y − x )2 = 100
(d) (y − x )2 = 200
21. A point P is taken on ‘L’ such that
1
1
1
, then locus of P is
=
+
2
2
(OP )
(OA )
(OB ) 2
Y¢
⇒
| y | ≤ min {|3 − x |, |2 − y |, | x| }
As the rectangle OABC lies in I quadrant,
∴
y ≤ min ⋅ [3 − x, 2 − y . x ]
We draw the graph of
y = 3 − x, y = 2 − y , y = x
or
x + y = 3, y = 1, y = x
(a) (y − x )2 = 32
(b) (y − x )2 = 64
(c) (y − x )2 = 80
(d) (y − x )2 = 100
Sol. Let the equation of line ‘L’ through origin is
x−0 y −0
=
=r
cosθ
sin θ
y=
x+
y=
x
Y
3
∴
Let
(1, 1) E
X¢
D (2, 1) y=1
X
O
A
Y¢
∴
E ≡ (1, 1 ) and D ≡ (2, 1 )
(b) 3 x + 3y + 40 = 0
(d) 3 x − 3y + 40 = 0
20. A point P is taken on ‘L’ such that (OP ) 2 = OA ⋅ OB,
then the locus of P is
P (x, y)
X¢
Paragraph II
(d) 1
P ≡ (r cosθ, r sin θ )
OA = r1 and OB = r2
∴
A ≡ (r1 cosθ, r1 sin θ )
and
B ≡ (r2 cosθ, r2 sin θ )
A lies on
∴
⇒
L1 :y − x − 10 = 0
r1 sin θ − r1 cosθ − 10 = 0
10
r1 =
sin θ − cosθ
…(i)
146
Textbook of Coordinate Geometry
r sin θ − r cosθ r sin θ − r cosθ
+
10
20
y −x y −x
or
2=
+
10
20
∴ Locus of P is 3 x − 3y + 40 = 0
⇒ B lies on L1 : y − x − 20 = 0
or 2 =
L
B
Y
[Q P ≡ (r cos, r sin θ )]
20. (d) Q (OP ) 2 = OA ⋅ OB
⇒
r 2 = r1 ⋅ r2
10
20
⋅
⇒ r2 =
(sin θ − cosθ ) (sin θ − cosθ )
P
10
y–
=
0
x–
20
=
0
A
y–
x–
or
q
X¢
O
⇒
19. (d) Q
⇒
(r sin θ − r cosθ ) 2 = 200
∴ Locus of P is (y − x ) 2 = 200
X
21. (c) Q
Y¢
∴
[from Eqs. (i) and (ii)]
⇒
r2 sin θ − r2 cosθ − 20 = 0
20
r2 =
sin θ − cosθ
…(ii)
⇒
2
1
1
2 1 1
= +
=
+
⇒
r r1 r2
OP OA OB
2 sin θ − cosθ sin θ − cosθ
[from Eqs. (i) and (ii)]
=
+
r
10
20
or
1
1
1
=
+
(OP ) 2 (OA ) 2 (OB ) 2
1
1
1
=
+
r 2 r12 r22
1 (sin θ − cosθ ) 2 (sin θ − cosθ ) 2
=
+
100
400
r2
400 = 4(r sin θ − r cosθ ) 2 + (r sin θ − r cosθ ) 2
∴ Locus of P is (y − x ) 2 = 80
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
This section contains 2 examples. The answer to each example is a single digit integer, ranging from 0 to 9
(both inclusive).
Ex. 22 P ( x , y ) is called a natural point if x , y ∈N . The
total number of points lying inside the quadrilateral formed
by the lines 2 x + y = 2, x = 0, y = 0 and x + y = 5 is
l
Sol. (6) First, we construct the graph of the given quadrilateral.
Y
5
4
y=
2
5
x=0
x+
3
1
X¢
1
O
Y¢
2
3
2x+y
=2
4 5
y=0
X
It is clear from the graph that there are six points lying inside
the quadrilateral.
Ex. 23 The distance of the point ( x , y ) from the origin is
defined as d = max⋅ {| x |, | y | }. Then the distance of the
common point for the family of lines
x (1 + λ ) + λy + 2 + λ = 0 (λ being parameter) from the
origin is
l
Sol. (2) Given family of lines is
x(1 + λ ) + λy + 2 + λ = 0
⇒
(x + 2) + λ(x + y + 1) = 0
for common point or fixed point
x+2=0
and
x+y +1 =0
or
x = − 2, y = 1
∴ Common point is ( −2, 1 )
or
d = max{| −2|, |1| }
= max{2, 1 } = 2
Chap 02 The Straight Lines
147
JEE Type Solved Examples :
Matching Type Questions
n
l
This section contains 2 examples. Examples 24 and 25
has four statements (A, B, C and D) given in Column I
and four statements (p, q, r and s) in Column II. Any
given statement in Column I can have correct matching
with one or more statement(s) given in Column II.
Ex. 24 Consider the following linear equations x and y
ax + by + c = 0
bx + cy + a = 0
cx + ay + b = 0
Column I
Column II
(A) a + b + c ≠ 0 and
a 2 + b 2 + c 2 = ab + bc + ca
(p) Lines are sides of a triangle
(B) a + b + c = 0 and
a 2 + b 2 + c 2 ≠ ab + bc + ca
(q)
Lines are different and
concurrent
(C) a + b + c ≠ 0 and
a 2 + b 2 + c 2 ≠ ab + bc + ca
(r)
Number of pair ( x , y ) satisfying
the equations are infinite
(D) a + b + c = 0 and
a 2 + b 2 + c 2 = ab + bc + ca
(s)
Lines are identical
Sol. (A) → (r , s ); (B) → (q ); (C) → ( p ); (D) → (r )
(A) if a + b + c ≠ 0 and a 2 + b 2 + c 2 = ab + bc + ca
1
or {(a − b ) 2 + (b − c ) 2 + (c − a ) 2 } = 0
2
or
a − b = 0, b − c = 0, c − a = 0
or
a = b = c ⇒ All the lines an identical
and number of pair ( x, y ) are infinite.
(B) If a + b + c = 0 and a 2 + b 2 + c 2 ≠ ab + bc + ca
⇒a + b + c = 0, but a, b, c are not simultaneously equal.
Hence, lines are different and concurrent.
(C) If a + b + c ≠ 0 and a 2 + b 2 + c 2 ≠ ab + bc + ca
a b c

⇒ ∆ =
b c a ≠ 0 and a, b, c are not all simultaneously equal.
c a b
∴ Lines are sides of a triangle.
(D) If a + b + c = 0 and a 2 + b 2 + c 2 = ab + bc + ca
⇒ ∆ = 0 and a = b = c
∴ Equations are satisfied for any ( x, y ) .
Ex. 25 The equation of the sides of a triangle are
x + 2y + 1 = 0, 2 x + y + 2 = 0 and px + qy + 1 = 0 and area of
triangle is ∆ .
l
Column I
Column II
(A) p = 2, q = 3, then 8∆ is divisible by
(p)
3
(B)
p = 3, q = 2, then 8∆ is divisible by
(q)
4
(C)
p = 3, q = 4, then 10∆ is divisible by
(r)
6
(D) p = 4, q = 3, then 20∆ is divisible by
(s)
9
Sol. (A) → ( p ); (B) → ( p, q, r ); (C) → ( p, r ); (D) → ( p, s )
Q
∆=
D2
, where
2| C1 C 2 C 3 |
1 2 1

D =
2 1 2 = 3( p − 1 )
p q 1
and C1, C 2, C 3 are co-factors of third column, then
C1 = 2q − p, C 2 = 2 p − q, C 3 = − 3
3( p − 1 ) 2
∴
∆=
2|2q − p||2 p − q|
(A) for p = 2, q = 3
3
⇒
∆=
8
∴
8∆ = 3
(B) for p = 3, q = 2
3
∆=
⇒
2
∴
8 ∆ = 12
(C) for p = 3, q = 4
6
⇒
∆=
10
∴
10 ∆ = 6
(D) for p = 4, q = 3
27
∆=
⇒
20
∴
20 ∆ = 27
148
Textbook of Coordinate Geometry
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions (Ex. Nos. 26 and 27) are Assertion-Reason
type examples. Each of these examples contains two
statements.
Statement I (Assertion) and Statement II (Reason)
Each of these examples also has four alternative choices.
Only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement I is true, statement II is true; statement II
is a correct explanation for statement I.
(b) Statement I is true, statement II is true; statement II
is not a correct explanation for statement I.
(c) Statement I is true, statement II is false.
(d) Statement I is false, statement II is true.
l
Ex. 26 Consider the lines, L1 :
L3 :
L1 ≡ 3 x + 4y = 0,
L2 ≡ 5 x − 12y = 0 and L3 ≡ y − 15 = 0
3 + 32
Length of ⊥ from P to L1 =
=7
5
|5 − 96|
Length of ⊥ from P to L2 =
=7
13
|8 − 15|
and Length of ⊥ from P to L3 =
=7
1
∴ Statement II is true
Y
B (–20, 15)
x y
x y
+ = 1; L 2 : + = 1;
3 4
4 3
P
L1=0
L2, L4 are parallel.
∴ Distance between L2 and L4 =
1
12
=
5
1
1


 + 
 16 9
∴ Distance between L1 and L3 = Distance between L2 and L4 .
∴ Quadrilateral formed by L1, L2, L3, L4 is a rhombus.
Hence, statement I is true and statement II is false.
Ex. 27
Statement I : Incentre of the triangle formed by the lines
whose sides are 3 x + 4y = 0; 5 x − 12y = 0 and y − 15 = 0 is the
point P whose coordinates are (1, 8).
Statement II : Point P equidistant from the three lines
forming the triangle.
X¢
A (36, 15)
L 3 =0
L2=0
x y
x y
+ = 2 and L 4 : + = 2
3 4
4 3
Statement I : The quadrilateral formed by these four lines is a
rhombus.
Statement II : If diagonals of a quadrilateral formed by any
four lines are unequal and intersect at right angle, then it is a
rhombus.
Sol. (c) Q L1, L3 are parallel.
1
12
∴ Distance between L1 and L3 =
= and
5
1
1
 + 
 9 16
l
Sol. (b) Let
X
O (0, 0)
Y¢
1
Also,
Area of ∆OPA = × OA × 7
2
1
= × 39 × 7 = ∆1
2
1
Area of ∆OPB = × OB × 7
2
1
= × 25 × 7 = ∆ 2
2
1
and
Area of ∆APB = × AB × 7
2
1
= × 56 × 7 = ∆ 3
2
7
∆1 + ∆ 2 + ∆ 3 = (39 + 25 + 56 )
∴
2
7 × 120 1
=
= × 56 × 15 = Area of ∆AOB
2
2
⇒ P inside the triangle.
Hence, both statements are true and statement II is not correct
explanation of statement I.
Chap 02 The Straight Lines
149
Subjective Type Examples
In this section, there are 15 subjective solved examples.
l Ex. 28. If x-coordinates of two points B and C are the
roots of equation x 2 + 4 x + 3 = 0 and their y-coordinates are
the roots of equation x 2 − x − 6 = 0. If x-coordinate of B is
less than x-coordinate of C and y-coordinate of B is greater
than the y-coordinate of C and coordinates of a third point
A be (3, − 5 ). Find the length of the bisector of the interior
angle at A .
x 2 + 4x + 3 = 0 ⇒
Sol. Q
x2 − x − 6 = 0 ⇒
and
x = − 1, − 3
x = − 2, 3
Also given that x and y-coordinates of B are respectively less
than and greater than the corresponding coordinates of C.
∴
B ≡ ( − 3, 3 ) and C ≡ ( − 1, − 2 )
AB = (3 + 3 ) 2 + ( − 5 − 3 ) 2 = 10
and
AC = (3 + 1 ) 2 + ( − 5 + 2 ) 2 = 5
∴
AB 2
=
AC 1
Sol. Equation of the line through point of intersection D of lines
u − p = 0 and v − r = 0 is
...(i)
(u − p ) + λ ( ν − r ) = 0
it is also passing through u = q and v = s , then Eq. (i) becomes
(q − p ) + λ (s − r ) = 0
(q − p )
…(ii)
λ=−
∴
(s − r )
u=p
D
C
v=r
Now
Ex. 29. Let the sides of a parallelogram be
u = p , u = q , v = r and v = s where, u = lx + my + n and
v = l ′ x + m ′ y + n ′ . Show that the equation of the diagonal
through the point of intersection of u = p and v = r and u = q
u v 1
and v = s , is given by p r 1 = 0


q s 1
l
v=s
n
Y
B(–3,3)
A
B
u=q
From Eqs. (i) and (ii), we get
X′
– 5,– 1 D
3
3
(u − p ) −
X
O
⇒ u (s − r ) − p (s − r ) − v (q − p ) + r (q − p ) = 0
(–1,–2)C
x
⇒
x
A(3,–5)
⇒
Y′
Let AD be the bisector of ∠ BAC, then
BD AB 2
=
=
DC AC 1
Thus D divides BC internally in the ratio 2 : 1
 2 ( − 1 ) + 1 ( − 3 ) 2 ( − 2 ) + 1 (3 )
∴
D≡
,



2+1
2+1
Thus,
Now,
 5 1
D ≡ − , − 
 3 3

AD = 3 +

=
(q − p )
(v − r ) = 0
(s − r )
2
5

 + − 5 +


3
1

3
2
196 196 14 2
units.
+
=
9
9
3
u (r − s ) − v ( p − q ) + ps − qr = 0
u v 1
p r 1 = 0


q s 1
l Ex. 30. The vertices B and C of a triangle ABC lie on the
lines 3y = 4 x and y = 0 respectively and the side BC passes
 2 2
through the point  ,  . If ABOC is a rhombus, O being
3 3
the origin, find the equation of the line BC and the
coordinates of A.
Sol. Let the side of the rhombus be a
∴
OB = BA = AC = CO = a
4x 

Co-ordinates of A is a + x1, 1 

3 
∴
(OB ) 2 = ( BA ) 2 = ( AC ) 2 = (CO ) 2 = a 2 ⇒(OB ) 2 = a 2
150
Textbook of Coordinate Geometry
⇒
x12 +
∴
Y
5x
25 x12
16 x12
= a2 ⇒ a = 1
= a2 ⇒
9
9
3
 5 x1 
 8 x1 4 x1 
, 0 , A ≡ 
,
C ≡
,
 3 
 3
3 
(0,b)B
P(a,b)
 4x 
B ≡  x1, 1  , O ≡ ( 0, 0 )

3 
Q
X′
Y
a
O
Q PQ ⊥ AB
∴ Slope of PQ × Slope of AB = − 1
 b − y1 
 b − 0
⇒

 × 
 = −1
 a − x1 
 0 − a
A
( 32 , 32 )
3 y=
X′
C(a,0)
X
⇒
Y′
∴ Equation of BC
4 x1
−0
5 x1 

y −0= 3
x −

5 x1 
3 
x1 −
3
10 x1
y = − 2x +
⇒
3
 2 2
it is passing through  ,  , then
 3 3
...(1)
ax1 − by1 = a 2 − b 2
x y
Q Equation of AB is + = 1
a b
x
y
But Q lies on AB then 1 + 1 = 1
a
b
⇒
bx1 + ay1 = ab
From Eqs. (ii) and (iii), we get
b3
a3
, y1 = 2
x1 = 2
2
a + b2
a +b
Now,
2
4 10 x1
=− +
3
3
3
10 x1
2=
⇒
3
3
x1 =
∴
5
Hence, coordinates are
 8 4
C ≡ (1, 0 ), A ≡  , 
 5 5
3
B≡ ,
5
x12/3 + y12/3 =
2/ 3
+y
2/ 3
=c
[from Eq. (i)]
2/ 3
Y
(0,b)B
P(a,b)
Q
X′
A(a,0)
O
Ex. 31 The ends AB of a straight line segment of constant
length c slide upon the fixed rectangular axes OX and OY
respectively. If the rectangle OAPB be completed, then show
that the locus of the foot of perpendicular drawn from P to
AB is x 2 /3 + y 2 /3 = c 2 /3 .
In ∆ PQA,
sin θ =
∴
Sol. Let A ≡ (a , 0 ) , B ≡ ( 0 , b ) then P ≡ (a , b )
Now, in ∆QAM,
QA = c sin 2 θ
QM
QM
sin θ =
=
QA c sin 2 θ
AB = c
let Q ≡ ( x1, y1 )
Y′
∴
a2 + b2 = c
and
.
Aliter :
Since,
AB = c
Let
∠ BAO = θ
∴
OA = c cosθ and OB = c sin θ
Q
OB = PA = c sinθ
4
 , O ≡ ( 0, 0 )
5
a2 + b2 = c2
...(iii)
(a 2 + b 2 )
(a 2 + b 2 ) 2/3
Hence, required locus is x
l
or
...(ii)
= (a 2 + b 2 )1/3 = c 2/3
From Eq. (i), equation of BC is
y = − 2x + 2
⇒
2x + y = 2
Since
X
Y′
(x1, 4x31)
4x
B
A(a,0)
O
...(i)
and
⇒
QA
QA
=
PA c sin θ
QM = c sin 3 θ
MA
MA
cosθ =
=
QA c sin 2 θ
MA = c sin 2 θ cosθ
X
Chap 02 The Straight Lines
If coordinates of Q be ( x1, y1 )
then
x1 = OM = OA − MA = c cos θ − c sin 2 θ cosθ
x1 = c cos3 θ
and
y1 = MQ = c sin θ
...(ii)
From Eqs. (i) and (ii),
x12/3 + y12/3 = c 2/3 cos2 θ + c 2/3 sin 2 θ = c 2/3
⇒
Ex. 33 In a ∆ ABC, A ≡ (α , β), B ≡ (1 , 2 ), C ≡ ( 2 , 3 ) and
point A lies on the line y = 2 x + 3 , where α, β ∈I . If the area
of ∆ ABC be such that [ ∆ ] = 2 , where [.] denotes the greatest
integer function, find all possible coordinates of A.
l
...(i)
3
Sol. Q (α , β ) lies on y = 2 x + 3
then
β = 2α + 3
Thus ,the coordinates of A are (α , 2α + 3 )
3 
1 α 2α + 3
+ 
1 2
+
2
|
∆ = |
2  2 3 α 2α + 3
2 1
x12/3 + y12/3 = c 2/3
Hence, locus of Q is x 2/3 + y 2/3 = c 2/3
= | 2α − 2α − 3 + 3 − 4 + 4α + 6 − 3α |
1
∆ = |α + 2|
2
1

But
(given)
[ ∆ ] = |α + 2| = 2
2

|α + 2|
∴
2≤
<3
2
⇒
4 ≤ |α + 2 | < 6
⇒
4 ≤ α + 2 < 6 or − 6 < α + 2 ≤ − 4
⇒
2 ≤ α < 4 or − 8 < α ≤ − 6
∴
α = 2, 3, − 7, − 6
then
β = 7, 9, − 11, − 9
Hence, coordinates of A are (2 , 7 ), (3 , 9 ), ( − 7 , − 11 ) and
( − 6, − 9 ).
Ex. 32 A square lies above the X-axis and has one vertex
at the origin. The side passing through the origin makes an
angle α (0 < α < π / 4 ) with the positive direction of the
X-axis. Prove that the equation of its diagonals are
y (cos α − sin α) = x (sin α + cos α)
and
y (sin α + cos α) + x (cos α − sin α) = a
where, is the length of each side of the square.
l
Sol. Here, OA = AB = BC = CO = a
Y
B
C
π/4
X′
A
α
O
X
π

Equation of OB is y − 0 = tan  + α  ( x − 0 )
4


π

π
y cos  + α  = x sin  + α 

4

4
⇒ y (cos α − sin α ) = x (cos α + sin α )
Hence, equation of diagonal OB is
x (cos α + sin α ) + y (sin α − cos α ) = 0
Coordinates of A are (a cos α , a sin α )
Diagonal AC is perpendicular to the diagonal OB,
1
Slope of AC = −
slope of OB
=−
1
π

= − cot  + α 
4

π

tan  + α 
4

Hence, equation of diagonal AC is

π
y − a sin α = − cot  + α  ( x − a cos α )

4
⇒
 cos α − sin α 
y − a sin α = − 
 ( x − a cos α )
 cos α + sin α 
⇒
y (cos α + sin α ) − a sin α cos α − a sin 2 α
= − x (cos α − sin α ) + a cos2 α − a sin α cos α
⇒
x (cos α − sin α ) + y (cos α + sin α ) = a
Ex. 34 Find the values of non-negative real numbers
λ 1 , λ 2 , λ 3 , µ 1 , µ 2 , µ 3 such that the algebraic sum of the
perpendiculars drawn from points
( λ 1 , 4 ), ( λ 2 , 5 ), ( λ 3 , − 3 ), ( 2 , µ 1 ), (3 , µ 2 ) and ( 7, µ 3 ) on a
variable line passing through ( 2 ,1) is zero.
l
Y′
⇒
151
Sol. Let the equation of the variable line be ax + by + c = 0. It is
given that
(aλ 1 + 4b + c ) (aλ 2 + 5b + c )
⇒
+
(a 2 + b 2 )
(a 2 + b 2 )
+
+
(aλ 3 − 3b + c )
(a + b )
(3a + bµ 2 + c )
2
2
(a 2 + b 2 )
+
+
(2a + bµ1 + c )
(a 2 + b 2 )
(7a + bµ 3 + c )
(a 2 + b 2 )
=0
⇒ a ( λ 1 + λ 2 + λ 3 + 12 ) + b (µ1 + µ 2 + µ 3 + 6 ) + 6c = 0
 µ1 + µ 2 + µ 3 + 6 
 λ + λ 2 + λ 3 + 12
⇒ a 1
 + c = 0 …(i)
 +b 




6
6
But the line passes through (2,1), therefore
...(ii)
2a + b + c = 0
From Eqs. (i) and (ii), we get
λ 1 + λ 2 + λ 3 + 12
µ + µ2 + µ3 + 6
= 2 and 1
=1
6
6
⇒ λ 1 + λ 2 + λ 3 = 0 and µ1 + µ 2 + µ 3 = 0
[Q λ i , µi ≥ 0 for all i ]
⇒ λ 1 = λ 2 = λ 3 and µ1 = µ 2 = µ 3
(say)
λ1 = λ 2 = λ 3 = α
and
(say)
µ1 = µ 2 = µ 3 = β
where α ≥ 0 and β ≥ 0.
152
Textbook of Coordinate Geometry
Ex. 35 The three sides of a triangle are
Lr ≡ x cos θ r + y sin θ r − pr = 0, where r= 1,2,3. Show that
the orthocentre is given by
L1 cos (θ 2 − θ 3 ) = L 2 cos (θ 3 − θ 1 ) = L3 cos (θ 1 − θ 2 ) .
y
k
y
k
− = x − h or x − = h −
m
m
m m
x
y
=1
−
k
k


h −  m h − 


m
m
x
y
+
=1
k  (k − mh )

h − 

m
⇒
l
⇒
Sol. The given lines are
L1 ≡ x cosθ1 + y sin θ1 − p1 = 0
L2 ≡ x cosθ 2 + y sin θ 2 − p 2 = 0
L3 ≡ x cosθ 3 + y sin θ 3 − p 3 = 0
Now, equation of AD is L2 + λL3 = 0
…(i)
⇒ ( x cosθ 2 + y sin θ 2 − p 2 ) + λ ( x cosθ 3 + y sin θ 3 − p 3 ) = 0
⇒ x (cosθ 2 + λ cosθ 3 ) + y (sin θ 2 + λ sin θ 3 ) − ( p 2 + λp 3 ) = 0
(cosθ 2 + λ cosθ 3 )
Slope of AD = −
∴
(sin θ 2 + λ sin θ 3 )
cosθ1
and
Slope of BC = −
sin θ1
⇒
Y
P
(h,k)
X′
L3 =0
E
O
k
and OP = k − mh
m
1 
k
1
Area of triangle OPQ = . OQ. OP = . h −  (k − mh )
2 
m
2
L2 =0
B
D
L1 =0
⇒
∴
OQ = h −
C
To minimise,
⇒
∴
Ex. 36 Let (h, k) be a fixed point, where h > 0, k > 0.
A straight line passing through this point cuts the positive
direction of the co-ordinate axes at the points P and Q. Find
the minimum area of the triangle OPQ ,O being the origin.
Sol. Equation of any line passing through the fixed point (h, k )
and having slope m can be taken as
...(i)
y − k = m (x − h )
mh 2 k 2
−
2
2m
…(ii)
dA
=0
dm
k
h2
k2
+
=0 ⇒m=±
0−
h
2
2m 2
2k 2
d 2A
=−
2
2m 3
dm
d 2A
dm 2
Now from Eq. (i),
cos (θ1 − θ 2 )
L2 −
L3 = 0
cos (θ 3 − θ1 )
l
A (m ) = hk −
⇒
cosθ1 cosθ 2 + λ cosθ 3 cosθ1
= − sin θ1 sin θ 2 − λ sin θ 3 sin θ1
cos (θ1 − θ 2 ) + λ cos (θ 3 − θ1 ) = 0
cos (θ1 − θ 2 )
λ=−
cos (θ 3 − θ1 )
...(ii)
∴
L2 cos (θ 3 − θ1 ) = L3 cos (θ1 − θ 2 )
Similarly, we can obtain equation of altitude BE as
...(iii)
L3 cos (θ1 − θ 2 ) = L1 cos (θ 2 − θ 3 )
From Eqs. (ii) and (iii), we get
L1 cos (θ 2 − θ 3 ) = L2 cos (θ 3 − θ1 ) = L3 cos(θ1 − θ 2 )
1
k2
= 2hk − mh 2 − 
2
m
⇒
Since, AD ⊥ BC
Slope of BC × slope of AD = − 1
∴
cosθ1  (cosθ 2 + λ cosθ 3 ) 
−
× −
= −1
⇒
sin θ1  (sin θ 2 + λ sin θ 3 ) 
⇒
X
Y′
∴
F
Q
O
A
=
m = − k /h
h3
>0
k
(Qh > 0, k > 0)
k
k
Hence for m = − , A (m ) is minimum . Put m = − in Eq. (ii),
h
h
we get minimum area.
hk hk
+
= 2hk
⇒ Minimum area of ∆OPQ = hk +
2
2
Ex. 37 The distance between two parallel lines is unity. A
point P lies between the lines at a distance a from one of
them. Find the length of a side of an eqilateral triangle PQR,
vertex Q of which lies on one of the parallel lines and vertex
R lies on the other line.
l
Sol. Let
and
then
Given
PQ = QR = RP = r
∠ PQL = θ
∠ XQR = θ + 60 °
PL = a and RN = 1 unit
Chap 02 The Straight Lines
Now, let slope of AB is m
Y
C
R(r,θ+60°)
Y
60°
θ
P(r,θ)
N
(0,0) Q
L
α
X
X′
X
Y′
PL a
sin θ =
=
QP r
a = r sinθ
r sin (θ + 60 ° ) =
⇒
⇒
⇒
or
⇒
⇒
∴
...(i)
RN 1
=
QR r
1 a
3
a2 
× 1 − 2  = 1
r  × +
2
r 
2 r
[from Eq. (i)]
3r 2 = 4 (a 2 − a + 1 )
2
r=
(a 2 − a + 1 )
3
Ex. 38 A ray of light travelling along the line OA
(O being origin) is reflected by the line mirror x − y + 1 = 0,
is the point of incidence being A (1, 2 ) the reflected ray,
travelling along AB is again reflected by the line mirror
x − y = 2 , the point of incidence being B. If this reflected ray
moves along BC, find the equation of the line BC.
l
and slope of normal to x − y + 1 = 0 is
− 1 = m2
 2 − (− 1) 
 m − (− 1) 

 =−

 1 + 2( − 1 )
 1 + m (− 1)
⇒
3
a
+
(r 2 − a 2 ) = 1
2
2
a
3
(r 2 − a 2 ) = 1 −
2
2
2
a
3 2
(r − a 2 ) = 1 +
−a
4
4
3r 2 − 3a 2 = 4 + a 2 − 4a
2−0
= 2 = m1
1−0
Q OA and AB are equally inclined to normal of x − y + 1 = 0
then,
 m1 − m2 
 m − m2 

 =−

 1 + m1m2 
 1 + mm2 
⇒
sin (θ + 60 ° ) = 1
(sin θ cos 60 ° + cosθ sin 60 ° ) = 1

1
3
r  sin θ +
cosθ = 1

2
2
Sol. Since, slope of OA =
B
A'(x1,y1)
In ∆PQL
⇒
α
O
Y′
∴
and in ∆QRN,
α
α
x–
y=
2
A(1,2)
x–
y+
1=
0
X′
153
(say)
(say)
3 (1 − m ) = m + 1
1
or
m=
2
∴ Equation of AB is
1
y − 2 = (x − 1)
2
or
x − 2y + 3 = 0
Now, solving x − 2y + 3 = 0
and
x −y =2
then, we get x = 7, y = 5
i.e.
B ≡ ( 7, 5 )
Q BC is parallel to OA
∴ Equation of BC is
y − 5 = m1 ( x − 7 )
⇒
y − 5 = 2 (x − 7)
i.e.
2x − y − 9 = 0
Aliter :
If image of A (1, 2 ) with respect to line mirror x − y = 2 be
A′ ( x1, y1 ), then
x1 − 1 y1 − 2 − 2 (1 − 2 − 2 )
=
=
1
−1
1+1
or
x1 = 4, y1 = − 1
i.e.
A′ ≡ ( 4, − 1 )
Q BC is parallel to OA
then equation of BC = equation of A′ C is
y − y1 = 2 ( x − x1 )
⇒
y + 1 = 2 (x − 4)
or
2x − y − 9 = 0
154
Textbook of Coordinate Geometry
Ex. 39 Consider two lines L1 ≡ x − y = 0 and L 2 ≡ x + y = 0
and a moving point P ( x , y ). Let d (P , Li ), i =1, 2 represents the
distance of the point ‘P’ from Li . If point ‘P’ moves in certain
l
2
region ‘R’ in such a way Σ d (P , Li ) ∈[ 2 , 4 ].
Ex. 40 A rectangle PQRS has its side PQ parallel to the
line y = mx and vertices P , Q and S on the lines y = a , x = b
and x = − b respectively. Find the locus of the vertex R.
l
Sol. Q PQ is parallel to y = mx
∴ Equation of PQ is y = mx + λ
Q Diagonals bisect to each other
i =1
Find the area of region ‘R’.
2
Sol. Q Given
Σ d (P , Li ) ∈ [2, 4]
Y
i =1
R(h,k)
2
Σ d (P , Li ) ≤ 4
⇒
2≤
⇒
2 ≤ d ( P , L1 ) + d ( P , L2 ) ≤ 4
| x − y| | x + y|
2≤
+
≤4
2
2
2 2 ≤ | x − y| + | x + y| ≤ 4 2
⇒
⇒
i =1
X′
2 2 ≤ 2 x ≤ 4 2 or 2 ≤ x ≤ 2 2
2 2 ≤ − 2y ≤ 4 2
or
− 2 ≥ y ≥ − 2 2 or − 2 2 ≤ y ≤ − 2
2 2 ≤ 2y ≤ 4 2 or 2 ≤ y ≤ 2 2
Case IV : If x − y < 0, x + y < 0
then
2 2 ≤ − (x − y ) − (x + y ) ≤ 4 2
2 2 ≤ − 2x ≤ 4 2
B'
X′
–2√2
0
D
0
y=
–2√2
Y′
or
− 2 ≥ x ≥ −2 2
or
−2 2 ≤ x ≤ − 2
Combining all cases, we get
x ∈ [ − 2 2, − 2 ] ∪ [ 2, 2 2 ]
and
y ∈ [ − 2 2, − 2 ] ∪ [ 2, 2 2 ]
Hence, area of the required region
= ( 4 2 ) 2 − (2 2 ) 2
= 32 − 8 = 24 sq units.
⇒
mk − m 2b − am − m 2h = − h + b
⇒
(m 2 − 1 ) h − mk + b (m 2 + 1 ) + am = 0
Ex. 41. For points P ≡ ( x 1 , y 1 ) and Q ≡ ( x 2 , y 2 ) of the
coordinate plane, a new distance d (P , Q ) is defined by
d (P , Q ) = | x 1 − x 2 | + | y 1 − y 2 |. Let O ≡ (0, 0 ) and A ≡ (3 , 2 ).
Prove that the set of the points in the first quadrant which
are equidistant (with respect to the new distance) from O and
A consists of the union of a line segment of finite length and
an infinite ray. Sketch this set in a labelled diagram.
l
D'
x+
C
–√2
X
2√2
√2
–√2
X
∴ Locus of R is
(m 2 − 1 ) x − my + b (m 2 + 1 ) + am = 0
A'
√2
C'
A
x–
y=
2√2
B
O
∴ x-coordinate of P is −h.
Suppose y-coordinate of Q and S are λ 1 and λ 2 respectively.
Q Eq. (i) pass through P ( − h, a )
then
a = − mh + λ
∴
λ = a + mh
⇒
y = mx + a + mh
Q also lie on it, then
λ 1 = mb + a + mh
∴
Q ≡ (b, mb + a + mh )
Also, slope of PQ × slope of QR = − 1
(k − (mb + a + mh ))
m×
= −1
(h − b )
Case III : If x − y < 0, x + y > 0
then
2 2 ≤ − (x − y ) + (x + y ) ≤ 4 2
Y
mx
Y′
⇒
⇒
y=
P(–h,a)
Case II : If x − y > 0, x + y < 0
then
2 2 ≤ (x − y ) − (x + y ) ≤ 4 2
⇒
Q(b,λ1)
(–b,λ2)S
Case I : If x − y > 0, x + y > 0
then
2 2 ≤ (x − y ) + (x + y ) ≤ 4 2
⇒
...(i)
Sol. Let P ( x, y ) be any point in the first quadrant, we have
x > 0, y > 0
d (P , Q ) = | x − 0| + | y − 0| = | x | + | y | = x + y
and
d (P , A ) = | x − 3| + | y − 2|
Given,
d (P , Q ) = d (P , A )
...(i)
x + y = | x − 3| + |y − 2|
Chap 02 The Straight Lines
Case I : 0 ≤ x < 3, 0 ≤ y < 2
then Eq. (i) becomes
x + y =3 − x + 2 −y
or
x + y =5/2
Case II : 0 ≤ x < 3, y ≥ 2
then Eq. (i) becomes
x + y =3 − x + y −2
1
or
x=
2
Case III : x ≥ 3, 0 ≤ y < 2
then Eq. (i) becomes
x + y = x −3 + 2 −y
1
⇒
y =−
2
Case IV : x ≥ 3, y ≥ 2
x + y = x −3 + y −2
⇒
0 = −5
Combining all cases, then
x + y = 5 / 2; 0 ≤ x < 3, 0 ≤ y < 2
1
and
x = ; 0 ≤ x < 3, y ≥ 2
2
Y
3
x=1
2
2
x+y =5
2
1
X′
O
1
2
3
X
Y′
The labelled diagram is given in adjoining figure.
155
Ex. 42. A line through the variable point A (k +1, 2k )
meets the lines 7 x + y − 16 = 0, 5 x − y − 8 = 0, x − 5y + 8 = 0
at B, C , D respectively, prove that AC , AB, AD are in HP.
l
Sol. Given lines are
.
..(i)
7 x + y − 16 = 0
...(ii)
5x − y − 8 = 0
...(iii)
x − 5y + 8 = 0
Let the equation of line passing through A (k + 1, 2k ) making
an angle θ with the + ve direction of X -axis, be
x − (k + 1 ) y − 2k
(if AB = r1, AC = r2, AD = r3 )
=
= r1, r2, r3
cos θ
sin θ
∴
(Impossible)
(Impossible)
B ≡ [(k + 1 ) + r1 cos θ, 2k + r1 sin θ ]
C ≡ [(k + 1 ) + r2 cos θ, 2k + r2 sin θ ]
D ≡ [(k + 1 ) + r3 cos θ, 2k + r3 sin θ ]
Points B, C , D satisfying Eqs. (i), (ii) and (iii) respectively
9 (1 − k )
then
r1 =
7 cos θ + sin θ
3 (1 − k )
r2 =
5 cos θ − sin θ
9 (1 − k )
and
r3 =
5 sin θ − cos θ
1
1 (5 cos θ − sin θ ) (5 sin θ − cos θ )
∴
+ =
+
r2 r3
3 (1 − k )
9 (1 − k )
15 cos θ − 3 sin θ + 5 sin θ − cos θ
=
9 (1 − k )
14 cos θ + 2 sin θ 2
=
=
r1
9 (1 − k )
Hence r2, r1, r3 are in HP
i.e. AC , AB, AD are in HP.
#L
The Straight Lines Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c), (d) out of which
ONLY ONE is correct.
1. The straight line y = x − 2 rotates about a point where it
cuts X-axis and becomes perpendicular on the straight
line ax + by + c = 0, then its equation is
(a) ax + by + 2a = 0
(c) ax + by + 2b = 0
(b) ay − bx + 2b = 0
(d) None of these
2
2
1
2m
2. If
+
+
= , then orthocentre of the
1! 9 ! 3! 7 ! 5! 5! n !
triangle having sides x − y + 1 = 0, x + y + 3 = 0 and
2x + 5y − 2 = 0 is
(a) (2m − 2n, m − n )
(c) (2m − n, m + n )
(b) (2m − 2n, n − m )
(d) (2m − n, m − n )
3. If f ( x + y ) = f ( x ) f (y ) ∀ x , y ∈ R and f (1) = 2 , then area
enclosed by 3| x | + 2| y | ≤ 8 is
(a) f ( 4 ) sq units
(c)
1
f (6 ) sq units
3
1
f (6 ) sq units
2
1
(d) f (5 ) sq units
3
(b)
4. The graph of the function
y = cos x cos( x + 2) − cos ( x + 1) is
2
(a) a straight line passing through ( 0, − sin 2 1 ) with slope 2
(b) a straight line passing through (0, 0)
(c) a parabola with vertex (1, − sin 2 1 )

π
(d) a straight line passing through the point  , − sin 2 1 are

2
parallel to the X-axis.
5. A line passing through the point (2, 2) and the axes
enclose an area λ. The intercepts on the axes made by
the line are given by the two roots of
(a) x 2 − 2| λ | x + | λ | = 0
(b) x 2 + | λ | x + 2| λ | = 0
(c) x − | λ | x + 2| λ | = 0
(d) None of these
2
6. The set of value of ‘b’ for which the origin and the point
(1, 1) lie on the same side of the straight line
a 2 x + aby + 1 = 0 ∀a ∈ R, b > 0 are
(a) b ∈(2, 4 )
(c) b ∈[ 0, 2 ]
(b) b ∈( 0, 2 )
(d) None of these
7. Line L has intercepts a and b on the co-ordinates axes,
when the axes are rotated through a given angle;
keeping the origin fixed, the same line has intercepts p
and q, then
(a) a 2 + b 2 = p 2 + q 2
(c) a 2 + p 2 = b 2 + q 2
1
1
1
1
+ 2 = 2 + 2
2
a
b
p
q
1
1
1
1
(d) 2 + 2 = 2 + 2
a
p
b
q
(b)
8. If the distance of any point ( x , y ) from the origin is
defined as d ( x , y ) = max {| x |, | y | }, d ( x , y ) = a non-zero
constant, then the locus is
(a) a circle
(c) a square
(b) a straight line
(d) a triangle
9. If p 1 , p 2 , p 3 be the perpendiculars from the points
(m 2 , 2m ), (mm ′ , m + m ′ ) and (m ′ 2 , 2m ′ ) respectively on
the line x cos α + y sin α +
sin 2 α
= 0, then p 1 , p 2 , p 3
cos α
are in
(a) AP
(c) HP
(b) GP
(d) None of these
10. ABCD is a square whose vertices A, B, C and D are (0, 0),
(2, 0), (2, 2) and (0, 2) respectively. This square is rotated
in the xy plane with an angle of 30° in anti-clockwise
direction about an axis passing through the vertex A the
equation of the diagonal BD of this rotated square is ……
. If E is the centre of the square, the equation of the
circumcircle of the triangle ABE is
(a) 3 x + (1 − 3 )y = 3 , x 2 + y 2 = 4
(b) (1 + 3 ) x − (1 − 2 )y = 2, x 2 + y 2 = 9
(c) (2 − 3 ) x + y = 2( 3 − 1 ), x 2 + y 2 − x 3 − y = 0
(d) None of the above
11. The point (4, 1) undergoes the following three successive
transformations
(i) reflection about the line y = x − 1.
(ii) translation through a distance 1 unit along the positive
direction of X-axis.
π
(iii) rotation through an angle about the origin in the
anti-clockwise direction 4
Then, the coordinates of the final point are
 7 7
(a) ( 4, 3 )
(b)  , 
 2 2
(c) ( 0, 3 2 )
(d) (3, 4)
12. If the square ABCD, where A(0, 0), B(2, 0), C(2, 2) and
D(0, 2) undergoes the following three transformations
successively
(i) f1( x, y ) → (y , x )
(ii) f 2( x, y ) → ( x + 3y , y )
x −y x + y
(iii) f 3( x, y ) → 
,

 2
2 
then the final figure is a
(a) square
(c) rhombus
(b) parallelogram
(d) None of these
Chap 02 The Straight Lines
13. The line x + y = a meets the axes of x and y at A and B
respectively. A triangle AMN is inscribed in the triangle
OAB, O being the origin, with right angle at N, M and N
lie respectively on OB and AB. If the area of the triangle
3
AN
is
AMN is of the area of the triangle OAB, then
8
BN
equal to
(a) 1
(b) 2
(c) 3
157
20. Suppose that a ray of light leaves the point (3, 4), reflects
off the Y-axis towards the X-axis, reflects off the X-axis,
and finally arrives at the point (8, 2). The value of x is
Y
(3, 4)
(d) 4
(8, 2)
(0, y)
14. If P (1, 0), Q ( −1, 0) and R (2, 0) are three given points, then
the locus of point S satisfying the relation
(SQ ) 2 + (SR ) 2 = 2(SP ) 2 is
(a) a straight line parallel to X -axis
(b) a circle through the origin
(c) a circle with centre at the origin
(d) a straight line parallel to Y -axis
sin α
cos α 
15. If A 
− 1,
− 1 and B(1, 1), α ∈ [− π, π ] are two
 3

2
points on the same side of the line 3x − 2y + 1 = 0, then α
belongs to the interval
3π   π 

(a)  − π , −
∪  , π

4   4 
(c) φ
(b) [ −π , π ]
(d) None of these
16. The line x + y = 1 meets X-axis at A and Y-axis at B, P is
the mid-point of AB. P1 is the foot of the perpendicular
from P to OA; M 1 is that of P1 from OP ; P2 is that of M 1
from OA ; M 2 is that of P2 from OP ; P3 is that of M 2
from OA and so on. If Pn denotes the nth foot of the
perpendicular on OA form M n − 1 , then OPn is equal to
1
(a)
2n
(c) 2n − 1
1
(b) n
2
(d) 2n + 3
17. The line x = c cuts the triangle with corners (0, 0); (1, 1)
and (9, 1) into two regions. For the area of the two
regions to be the same, then c must be equal to
5
2
7
(c)
2
(a)
(b) 4
1
3
(c) 4
2
3
(d) 5
1
3
21. m, n are two integers with 0 < n < m. A is the point (m, n )
on the cartessian plane. B is the reflection of A in the
line y = x . C is the reflection of B in the Y -axis, D is the
reflection of C in the X -axis and E is the reflection of D
in the Y -axis. The area of the pentagon ABCDE is
(a) 2m(m + n )
(c) m(2m + 3n )
(b) m(m + 3n )
(d) 2m(m + 3n )
22. A straight line L with negative slope passes through the
point (8, 2) and cuts the positive coordinates axes at
points P and Q. As L varies, the absolute minimum value
of OP + OQ is (O is origin)
(a) 10
(b) 18
(c) 16
(d) 12
23. Drawn from origin are two mutually perpendicular lines
forming an isosceles triangle together with the straight
line 2x + y = a, then the area of this triangle is
a2
sq units
2
2
a
(c)
sq units
5
(a)
(b)
a2
sq units
3
(d) None of these
24. The number of integral values of m for which the
x-coordinate of the point of intersection of the lines
3x + 4y = 9 and y = mx + 1 is also an integer is
(d) 3 or 15
(a) 2
are concurrent, then the straight line 5x + 2y = 1, passes
through the point
(b) ( −a, b )
(d) ( −a, − b )
19. The ends of the base of the isosceles triangle are at (2, 0)
and (0, 1) and the equation of one side is x = 2, then the
orthocentre of the triangle is
 3 3
(a)  , 
 4 2
3 
(c)  , 1
4 
1
2
(b) 3
18. If the straight lines x + 2y = 9, 3x − 5y = 5 and ax + by = 1
(a) (a, − b )
(c) (a, b )
(a) 4
X
(x, 0)
O
5 
(b)  , 1
4 
4 7 
(d)  , 
 3 12
(b) 0
(c) 4
(d) 1
25. A ray of light coming from the point (1, 2) is reflected at
a point A on the X -axis and then passes through the
point (5, 3). The coordinates of the point A are
 13 
(a)  , 0
5 
5 
(b)  , 0
 13 
(c) ( −7, 0 )
(d) None of these
26. Consider the family of lines
5x + 3y − 2 + λ(3x − y − 4 ) = 0 and
x − y + 1 + µ(2x − y − 2) = 0. Equation of straight line
that belong to both families is ax + by − 7 = 0, then
a + b is
(a) 1
(b) 3
(c) 5
(d) 7
158
Textbook of Coordinate Geometry
27. In ∆ABC equation of the right bisectors of the sides AB
and AC are x + y = 0 and x − y = 0 respectively. If
A ≡ (5, 7 ), then equation of side BC is
(a) 7y = 5 x
(c) 5y = 7 x
30. In the adjacent figure combined equation of the incident
and refracted ray is
Y
(b) 5x = y
(d) 5y = x
28. Two particles start from the point (2, −1), one moving 2
units along the line x + y = 1 and the other 5 units along
the line x − 2y = 4. If the particles move towards
increasing y, then their new positions are
60°
(a) (2 − 2, 2 − 1 ); (2 2 + 2, 5 − 1 )
X¢
O
(b) (2 2 + 2, 5 − 1 ); (2 2, 2 + 1 )
30°
(d) (2 − 2, 5 − 1 ); ( 2 − 1, 2 2 + 2 )
4
(x
3
4
(b) ( x − 2 ) 2 + y 2 −
(x
3
y
(c) ( x − 2 ) 2 + y 2 +
(x
3
y
(d) ( x − 2 ) 2 + y 2 −
(x
3
(a) ( x − 2 ) 2 + y 2 +
29. Let P be (5, 3) and a point R on y = x and Q on the X -axis
be such that PQ + QR + RP is minimum, then the
coordinates of Q are
#L
X
Y¢
(c) (2 + 2, 2 + 1 ); (2 2 + 2, 5 + 1 )
 17 
(a)  , 0
8 
 17 
(c)  , 0
2 
(2, 0)
P
 17 
(b)  , 0
4 
(d) (17, 0 )
− 2 )y = 0
− 2 )y = 0
− 2) = 0
− 2) = 0
The Straight Lines Exercise 2 :
More than One Correct Option Type Questions
n
The section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c), and (d) out of
which MORE THAN ONE may be correct.
x y
31. The point of intersection of the lines + = 1 and
a b
x y
+ = 1 lies on
b a
(a) x − y = 0
(b) ( x + y )(a + b ) = 2ab
(c) (lx + my )(a + b ) = (l + m )ab
(d) (lx − my )(a + b ) = (l − m )ab
32. The equations (b − c )x + (c − a )y + a − b = 0 and
(b 3 − c 3 )x + (c 3 − a 3 )y + a 3 − b 3 = 0 will represent the
same line, if
(a) b = c
(c) a = b
(b) c = a
(d) a + b + c = 0
33. The area of a triangle is 5. Two of its vertices are (2, 1)
and (3, −2). The third vertex lies on y = x + 3. The
coordinates of the third vertex cannot be
 −3 3 
(a)  , 
 2 2
 7 13
(c)  , 
2 2 
 3 −3 
(b)  , 
4 2 
 −1 11
(d)  , 
 4 4
34. If the lines x − 2y − 6 = 0, 3x + y − 4 = 0 and
λx + 4y + λ 2 = 0 are concurrent, then
(a) λ = 2
(b) λ = − 3
(c) λ = 4
(d) λ = − 4
35. Equation of a straight line passing through the point of
intersection of x − y + 1 = 0 and 3x + y − 5 = 0 are
perpendicular to one of them is
(a) x + y + 3 = 0
(c) x − 3y − 5 = 0
(b) x + y − 3 = 0
(d) x − 3y + 5 = 0
36. If one vertex of an equilateral triangle of side a lies at the
origin and the other lies on the line x − 3y = 0, the
coordinates of the third vertex are
(a) ( 0, a )
 3a −a 
(b) 
,  (c) ( 0, − a )
 2
2
 − 3a a 
(d) 
, 
 2
2
37. If the line ax + by + c = 0, bx + cy + a = 0 and
cx + ay + b = 0 are concurrent (a + b + c ≠ 0) then
(a) a 3 + b 3 + c 3 − 3abc = 0 (b) a = b
(c) a = b = c
(d) a 2 + b 2 + c 2 − bc − ca − ab = 0
38. A(1, 3) and C(7, 5) are two opposite vertices of a square.
The equation of a side through A is
(a) x + 2y − 7 = 0
(c) 2 x + y − 5 = 0
(b) x − 2y + 5 = 0
(d) 2 x − y + 1 = 0
Chap 02 The Straight Lines
39. If 6a 2 − 3b 2 − c 2 + 7ab − ac + 4bc = 0, then the family of
43. Consider the equation y − y 1 = m( x − x 1 ). If m and x 1
lines ax + by + c = 0 is concurrent at
(a) ( −2, − 3 )
(c) (2, 3 )
are fixed and different lines are drawn for different
values of y 1 , then
(b) (3, − 1 )
(d) ( −3, 1 )
(a) the lines will pass through a fixed point
(b) there will be a set of parallel lines
(c) all the lines intersect the line x = x1
(d) all the lines will be parallel to the line y = x1
40. Consider the straight lines x + 2y + 4 = 0 and
4 x + 2y − 1 = 0. The line 6x + 6y + 7 = 0 is
(a) bisector of the angle including origin
(b) bisector of acute angle
(c) bisector of obtuse angle
(d) None of the above
44. Let L 1 ≡ ax + by + a 3 b = 0 and L 2 ≡ bx − ay + b 3 a = 0
be two straight lines. The equations of the bisectors of
the angle formed by the foci whose equations are
λ 1 L 1 − λ 2 L 2 = 0 and λ 1 L 1 + λ 2 L 2 = 0, λ 1 and λ 2 being
non-zero real numbers, are given by
41. Two roads are represented by the equations y − x = 6
and x + y = 8. An inspection bungalow has to be so
constructed that it is at a distance of 100 from each of
the roads. Possible location of the bungalow is given by
(a) (100 2 + 1, 7 )
(b) (1 − 100 2, 7 )
(c) (1, 7 + 100 2 )
(d) (1, 7 − 100 2 )
(a) L1 = 0
(c) λ 1L1 + λ 2L2 = 0
x −3 y +5
x −3 y +5
are
and
=
=
cos φ sin φ
cos θ sin θ
x −3 y +5
x −3 y +5
and
, then
=
=
cos α sin α
β
γ
two intersecting lines
diagonal has the equation x − y = a, then another vertex
of the square can be
#L
(b) L2 = 0
(d) λ 2L1 − λ 1L2 = 0
45. The equation of the bisectors of the angles between the
42. If (a, b ) be an end of a diagonal of a square and the other
(a) (a − b, a )
(c) ( 0, − a )
159
θ+φ
2
(c) γ = cosα
(a) α =
(b) (a, 0 )
(d) (a + b, b )
(b) β = − sin α
(d) β = sin α
The Straight Lines Exercise 3 :
Paragraph Based Questions
n
The section contains 5 Paragraphs based upon each of
the paragraphs 3 multiple choice questions have to be
answered. Each of these questions has four choices (a), (b),
(c), and (d) out of which ONLY ONE is correct.
(a) a line segment of finite length
(b) a line of infinite length
(c) a ray of finite length
(d) a ray of infinite length
Paragraph I
48. Let T ( x , y ), such that T is equidistant from point O and C
(Q. Nos. 46 to 48)
with respect to new distance and if T lie in first quadrant,
then T consists of the union of a line segment of finite
length and an infinite ray whose labelled diagram is
For points P ≡ ( x1 , y1 ) and Q = ( x 2 , y 2 ) of the coordinate
plane, a new distance d ( P , Q ) is defined by
d ( P , Q ) = | x1 − x 2 | + | y1 − y 2 |
Let O ≡ ( 0, 0), A ≡ (1, 2), B ≡ ( 2, 3) and C ≡ ( 4, 3) are four
fixed points on x- y plane.
46. Let R( x , y ), such that R is equidistant from the point O
(b)
(a)
1/2
O 1/2
(b) x + 2y = 3
(d) 2 x + 2y = 3
47. Let S( x , y ), such that S is equidistant from points O and B
with respect to new distance and if x ≥ 2 and 0 ≤ y < 3,
then locus of S is
3.5
3
and A with respect to new distance and if 0 ≤ x < 1 and
0 ≤ y < 2, then R lie on a line segment whose equation is
(a) x + y = 3
(c) 2 x + y = 3
Y
Y
3.5
X
3
O
Y
X
Y
3.5
2.5
(c)
(d)
1/2
O 1/2
3.5
X
O
2.5
X
160
Textbook of Coordinate Geometry
Paragraph II
How to find the image or reflection of a curve?
P
(Q. Nos. 49 to 51)
In a triangle ABC, if the equation of sides AB, BC and CA are
2x − y + 4 = 0, x − 2 y − 1 = 0 and x + 3 y − 3 = 0 respectively.
S=0
49. Tangent of internal angle A is equal to
(a) −7
1
(c)
2
L=0
(b) −3
M
(d) 7
S′=0
50. The equation of external bisector of angle B is
(a) x − y − 1 = 0
(c) x + y − 5 = 0
(b) x − y + 1 = 0
(d) x + y + 5 = 0
51. The image of point B w.r.t the side CA is
 3 26
(a)  − , 
 5 5
 3 26
(b)  − , − 
 5
5
 3 26
(c)  , − 
5
5
 3 26
(d)  , 
5 5 
Paragraph III
(Q. Nos. 52 to 54)
 2 −2
A (1, 3) and C  − ,  are the vertices of a triangle ABC and
 5 5
the equation of the angle bisector of ∠ABC is x + y = 2.
(b) 7 x + 3y + 4 = 0
(d) 7 x − 3y − 4 = 0
53. Coordinates of vertex B are
 3 17
(a)  , 
 10 10
 5 9
(c)  − , 
 2 2
Let the given curve be S : f ( x, y ) = 0 and line mirror
L : ax + by + c = 0. We take a point P on the given curve in
parametric form. Suppose Q be the image or reflection of point
P about line mirror L = 0, which again contains the same
parameter. Let Q ≡ ( φ( t ), ψ ( t )) , where t is parameter. Now let
x = φ( t ) and y = ψ( t )
Eliminating t, we get the equation of the reflected curve S ′.
55. The image of the line 3x − y = 2 in the line y = x − 1 is
(a) x + 3y = 2
(c) x − 3y = 2
(b) 3 x + y = 2
(d) x + y = 2
56. The image of the circle x 2 + y 2 = 4 in the line x + y = 2
is
52. Equation of BC is
(a) 7 x + 3y − 4 = 0
(c) 7 x − 3y + 4 = 0
Q
 17 3 
(b)  , 
 10 10
 9 5
(d)  , − 
 2 2
(a) x 2 + y 2 − 2 x − 2y = 0
2
57. The image of the parabola x 2 = 4y in the line x + y = a is
(a) ( x − a ) 2 = 4(a − y )
(b) (y − a ) 2 = 4 (a − x )
(c) ( x − a ) = 4(a + y )
(d) (y − a ) 2 = 4(a + x )
2
Paragraph V
54. Equation of AB is
(a) 3 x + 7y = 24
(b) 3 x + 7y + 24 = 0
(c) 13 x + 7y + 8 = 0
(d) 13 x − 7y + 8 = 0
Paragraph IV
(Q. Nos. 55 to 57)
Let S ′ = 0 be the image or reflection of the curve S = 0 about
line mirror L = 0. Suppose P be any point on the curve S = 0
and Q be the image or reflection about the line mirror L = 0,
then Q will lie on S ′ = 0.
(b) x 2 + y 2 − 4 x − 4y + 6 = 0
(c) x + y − 2 x − 2y + 2 = 0 (d) x 2 + y 2 − 4 x − 4y + 4 = 0
2
(Q. Nos. 58 to 60)
In a ∆ABC, the equation of the side BC is 2 x − y = 3 and its
circumcentre and orhtocentre are (2, 4 ) and (1, 2 ) respectively.
58. Circumradius of ∆ABC is
(a)
61
5
(b)
51
5
(c)
41
5
(d)
43
5
9
4 61
(c)
9
61
(d)
9
5 61
59. sin B ⋅ sin C =
(a)
9
2 61
(b)
60. The distance of orthocentre from vertex A is
(a)
1
5
(b)
6
5
(c)
3
5
(d)
2
5
161
Chap 02 The Straight Lines
#L
The Straight Lines Exercise 4 :
Single Integer Answer Type Questions
n
The section contains 10 questions. The answer to eaeh
question is a single digit integer, ranging from 0 to 9
(both inclusive).
61. The number of possible straight lines passing through
66. In a plane there are two families of lines : y = x + n,
y = − x + n, where n ∈ {0, 1, 2, 3, 4 }. The number of squares
of the diagonal of length 2 formed by these lines is
67. Given A(0, 0) and B ( x , y ) with x ∈(0, 1) and y > 0. Let the
slope of line AB be m1 . Point C lies on line x = 1 such
that the slope of BC is equal to m 2 , where 0 < m 2 < m1 .
If the area of triangle ABC can be expressed as
(m1 − m 2 ) f ( x ) and the largest possible value of f ( x ) is
1
λ, then the value of is
λ
(2, 3) and forming a triangle with the coordinate axes,
whose area is 12 sq units, is
62. The portion of the line ax + 3y − 1 = 0, intercepted
between the lines ax + y + 1 = 0 and x + 3y = 0 subtend a
right angle at origin, then the value of | a | is
63. Let ABC be a triangle and A ≡ (1, 2), y = x be the
perpendicular bisector of AB and x − 2y + 1 = 0 be the
angle bisector of ∠C. If the equation of BC is given by
ax + by − 5 = 0, then the value of a − 2b is
64. A lattice point in a plane is a point for which both
coordinates are integers. If n be the number of lattice
points inside the triangle whose sides are x = 0, y = 0 and
9 x + 223y = 2007, then tens place digit in n is
68. If ( λ , λ + 1) is an interior point of ∆ABC, where A ≡ (0, 3),
B ≡ ( −2, 0) and C ≡ (6, 1), then the number of integral
values of λ is
69. For all real values of a and b, lines
(2a + b )x + (a + 3b )y + (b − 3a ) = 0 and λx + 2y + 6 = 0 and
λx + 2y + 6 = 0 are concurrent, then the value of | λ | is
70. If from point (4, 4) perpendiculars to the straight lines
3x + 4y + 5 = 0 and y = mx + 7 meet at Q and R and area
of triangle PQR is maximum, then the value of 3m is
65. The number of triangles that the four lines y = x + 3,
y = 2x + 3, y = 3x + 2 and y + x = 3 form is
#L
The Straight Lines Exercise 5 :
Matching Type Questions
n
The section contains 5 questions. Questions 1, 2 and 3
have four statement (A, B, C and D) given in Column I
and four statements (p, q, r and s) in Column II and
questions 74 and 75 have three statements (A, B and C)
given in Column I and five statements (p, q, r, s and t) in
Column II. Any given statement in Column I can have
correct matching with one or more statement (s) given in
Column II.
71. Let L 1 , L 2 , L 3 be three straight lines a plane and n be the
number of circles touching all the lines.
Column I
72. Match the Columns
Column I
(A) Lines x − 2y − 6 = 0, 3 x + y − 4 = 0 and
λx + 4y + λ2 = 0 are concurrent, then the
value of | λ| is
(p)
2
(B) The variable straight lines
(q)
3 x (a + 1) + 4y (a − 1) − 3 (a − 1) = 0 for
different value of ‘a’ passes through a fixed
point ( p , q ) if λ = p − q, then the value of 4 | λ |
3
λ
= 0 passing through
2
the intersection of x − y + 1 = 0 and
3 x + y − 5 = 0, is perpendicular to one of
them, then the value of | λ + 1| is
(r)
4
(D) If the line y − x − 1 + λ = 0 is equidistant from (s)
the points (1, − 2 ) and ( 3 , 4 ), then the value of
| λ| is
5
(C)
Column II
(A) The lines are concurrent, then n + 1 is a (p) natural number
(B) The lines are parallel, then 2n + 3 is a
(q) prime number
(C) Two lines are parallel, then n + 2 is a
(r) composite number
(D) The lines are neither concurrent nor
parallel, then n + 2 is a
(s) perfect number
Column II
If the line x + y − 1 −
162
Textbook of Coordinate Geometry
73. Consider the triangle formed by the lines y + 3x + 2 = 0,
3y − 2x − 5 = 0 and 4y + x − 14 = 0
Column I
3
(t)
4
Column II
(A) If ( 0 , λ ) lies inside the triangle, then
integral values are less than | 3λ|
(p)
4
(B) If (1, λ ) lies inside the triangle, then
integral values are less than | 3λ|
(q)
5
(C) If ( λ , 2 ) lies inside the triangle, then
integral values of | 6λ| are
(r)
6
(D) If ( λ , 7 / 2 ) lies inside the triangle, then
integral value of | 6λ| are
(s)
7
75. Match the following
Column I
Column I
(B) If the vertices of a triangle are (6, 0), (q) ( λ , µ ) lies on
(0, 6) and (6, 6). If distance between
x 2 − y 2 = 64
circumcentre and orthocentre and
distance between circumcentre and
centroid are λ unit and µ unit
respectively, then
Column II
(A) The area bounded by the curve
max. {| x | , | y | } = 1 is
(p)
0
(B) If the point (a , a ) lies between the lines
| x + y | = 6, then [| a | ] is (where [. ]
denotes the greatest integer function)
(q)
1
(C) Number of integral values of b for which
the origin and the point (1, 1) lie on the
same side of the st. line a 2 x + aby + 1 = 0
for all a ∈ R ~ { 0 } is
(r)
2
Column II
(p) ( λ , µ ) lies on x = 3y
(A) If the distance of any point ( x , y )
from origin is defined as
d ( x , y ) = 2 | x | + 3| y |. If perimeter
and area of figure bounded by
d ( x , y ) = 6 are λ unit and µ sq units
respectively, then
74. Match the following
#L
(s)
(C) The ends of the hypotenuse of a
right angled triangle are (6, 0) and
(0, 6). If the third vertex is ( λ , µ ),
then
(r) ( λ , µ ) lies on
x 2 + y 2 − 6 x − 6y = 0
(s) ( λ , µ ) lies on
x 2 − 16y = 16
(t) ( λ , µ ) lies on
x 2 − y 2 = 16
The Straight Lines Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos 76 to 83) are Assertion-Reason type
questions. Each of these question contains two statements.
Statement I (Assertion) and
Statement II (Reason)
Each of these questions has four alternative choices, only
one of which is the correct answer.
You have to select the correct choice.
(a) Statement I is true, statement II is true; statement II
is a correct explanation for statement I
(b) Statement I is true, statement II is true; statement II
is not a correct explanation for statement I
(c) Statement I is true, statement II is false
(d) Statement I is false, statement II is true
76. Statement I The lines x (a + 2b ) + y(a + 3b ) = a + b are
concurrent at the point (2, − 1)
Statement II The lines x + y − 1 = 0 and 2x + 3y − 1 = 0
intersect at the point (2, − 1)
77. Statement I The points (3, 2) and (1, 4) lie on opposite
side of the line 3x − 2y − 1 = 0
Statement II The algebraic perpendicular distance
from the given point to the line have opposite sign.
78. Statement I If sum of algebraic distances from points
A(1, 2), B (2, 3), C(6, 1) is zero on the line ax + by + c = 0,
then 2a + 3b + c = 0
Statement II The centroid of the triangle is (3, 2)
79. Statement I Let A ≡ (0, 1) and B ≡ (2 , 0) and P be a point
on the line 4 x + 3y + 9 = 0, then the co-ordinates of P
 12 17 
such that | PA − PB | is maximum is  − ,  .
 5 5
Statement II | PA − PB | ≤ | AB |
80. Statement I The incentre of a triangle formed by the
 π
 π
line x cos   + y sin   = π,
9
9
 8π 
 8π 
x cos   + y sin  
9
9
 13π 
 13π 
= π and x cos 
 + y sin 
 = π is (0, 0).
 9 
 9 
Statement II Any point equidistant from the given
three non-concurrent straight lines in the plane is the
incentre of the triangle.
Chap 02 The Straight Lines
81. Statement I Reflection of the point (5, 1) in the line
x + y = 0 is ( −1 , − 5).
Statement II Reflection of a point P(α, β ) in the line
 α + α′ β + β′
ax + by + c = 0 is Q (α ′ , β ′ ), if 
,
 lies on
 2
2 
the line.
82. Statement I The internal angle bisector of angle C of a
triangle ABC with sides AB, AC and BC as y = 0,
3x + 2y = 0, and 2x + 3y + 6 = 0, respectively, is
5x + 5y + 6 = 0.
163
Statement II The image of point A with respect to
5x + 5y + 6 = 0 lies on the side BC of the triangle.
83. Statement I If the point (2a − 5, a 2 ) is on the same side of
the line x + y − 3 = 0 as that of the origin, then a ∈(2, 4 ).
Statement II The point ( x 1 , y 1 ) and ( x 2 , y 2 ) lie on the
same or opposite sides of the line ax + by + c = 0, as
ax 1 + by 1 + c and ax 2 + by 2 + c have the same or
opposite signs.
The Straight Lines Exercise 7 :
Subjective Type Questions
n
In this section, there are 15 subjective questions.
84. If A ( x 1 , y 1 ) , B ( x 2 , y 2 ) and C ( x 3 , y 3 ) are the vertices
of a triangle, then show that the equation of the line
joining A and the circumcentre is given by
 x y 1
 x y 1


(sin 2 B ) x 1 y 1 1 + (sin 2 C )x 1 y 1 1 = 0




x 2 y 2 1
x 3 y 3 1
pass through a fixed point for all θ. What are the
coordinates of this fixed point and its reflection in the
line x + y = 2 ? Prove that all lines through reflection
point can be represented by equation
(2 cos θ + 3 sin θ) x + (3 cos θ − 5 sin θ) y
= ( 2 − 1) (5 cos θ − 2 sin θ)
90. P is any point on the line x − a = 0. If A is the point (a , 0)
the lines
and PQ , the bisector of the angle OPA, meets the X-axis
in Q. Prove that the locus of the foot of the
perpendicular from Q on OP is
3x − 4y + 1 = 0, 8x + 6y + 1 = 0 .
( x − a) 2 ( x 2 + y 2 ) = a 2y 2 .
85. Find the coordinates of the point at unit distance from
86. A variable line makes intercepts on the coordinate axes,
the sum of whose squares is constant and equal to k 2 .
Show that the locus of the foot of the perpendicular
from the origin to this line is
( x 2 + y 2 ) 2 ( x −2 + y −2 ) = k 2 .
87. A variable line intersects n lines
y = mx , (m = 1 , 2 , 3 , ..., n ) in the points
A 1 , A 2 , A 3 , ...., A n respectively.
n
1
If Σ
= c (constant). Show that line passes through
p = 1 OA p
a fixed point. Find the coordinates of this fixed point
(O being origin).
88. Given n straight lines and a fixed point O. A straight line
is drawn through O meeting these lines in the points
R1 , R 2 , R 3 , ...., Rn and a point R is taken on it such that
n
n
1
= Σ
r
=
1
OR
ORr
Prove that the locus of R is a straight line.
89. Prove that all lines represented by the equation
(2 cos θ + 3 sin θ) x + (3 cos θ − 5 sin θ) y
= 5 cos θ − 2 sin θ
91. Having given the bases and the sum of the areas of a
number of triangles is constant, which have a common
vertex. Show that the locus of this vertex is a straight
line.
92. A (3, 0) and B (6, 0) are two fixed points and U (α, β ) is a
variable point on the plane. AU and BU meet the y-axis
at C and D respectively and AD meets OU at V. Prove
that CV passes through (2, 0) for any position of U in the
plane.
93. A variable line is drawn through O to cut two fixed
straight lines L 1 and L 2 in R and S. A point P is chosen
m +n m
n
on the variable line such that
=
+
. Show
OP
OR OS
that the locus of P is a straight line passing through the
point of intersection of L 1 and L 2 .
94. A line through A ( − 5, − 4 ) meets the lines
x + 3y + 2 = 0, 2x + y + 4 = 0 and x − y − 5 = 0 at the
points B, C and D respectively, if
2
2
 15 
 10 
 6 
  +
 =

 AB 
 AC 
 AD 
find the equation of the line.
2
164
Textbook of Coordinate Geometry
95. Two fixed straight lines X -axis and y = mx are cut by a
variable line in the points A (a , 0) and B (b, mb )
respectively. P and Q are the feet of the perpendiculars
drawn from A and B upon the lines y = mx and X -axis.
Show that, if AB passes through a fixed point (h, k ), then
PQ will also pass through a fixed point. Find the fixed
point.
96. Find the equation of straight lines passing through point
 1
 be the vertices of a

3
triangle. Let R be the region consisting of all those
97. Let O (0, 0), A (2 , 0) and B 1,
points P inside ∆OAB which satisfy
d ( P, OA ) ≤ min {d ( P, OB ), d ( P, AB )}
where d denotes the distance from the point to the
corresponding line. Sketch the region R and find its area.
98. Two triangles ABC and PQR are such that the
perpendiculars from A to QR, B to RP and C to PQ are
concurrent. Show that the perpendicular from P to BC , Q
to CA and R to AB are also concurrent.
(2 , 3) and having an intercept of length 2 units between
the straight lines 2x + y = 3 , 2x + y = 5 .
#L
The Straight Lines Exercise 8 :
Questions Asked in Previous 13 Year’s Exams
n
This section contains questions asked in IIT-JEE, AIEEE,
JEE Main & JEE Advanced from year 2005 to 2017.
99. The line parallel to the X-axis and passing through the
intersection of the lines ax + 2by + 3b = 0 and
bx − 2ay − 3a = 0, where (a, b ) ≠ (0, 0) is [AIEEE 2005, 3M]
3
from it
2
2
(b) below the X -axis at a distance of from it
3
3
(c) above the X -axis at a distance of from it
2
2
(d) above the X -axis at a distance of from it
3
(a) Statement I is true, statement II is true; statement II is not
a correct explanation for statement I
(b) Statement I is true, statement II is true; statement II is not
a correct explanation for statement I
(c) Statement I is true, statement II is false
(d) Statement I is false, statement II is true
103. Let P = ( −1, 0), Q = (0, 0) and R = (3, 3 3 ) be three point.
(a) below the X -axis at a distance of
The equation of the bisector of the angle PQR is
[AIEEE 2007, 3M]
(a)
3
x+y =0
2
(c) 3 x + y = 0
100. A straight line through the point A(3, 4 ) is such that its
(b) 3 x − 4y + 7 = 0
(d) 3 x + 4y = 25
x
2
101. If (a, a 2 ) falls inside the angle made by the lines y = ,
x > 0 and y = 3x , x > 0, then a belong to
 1
(a)  0, 
 2
1 
(c)  , 3
2 
(d) x +
3
y =0
2
104. Consider the lines given by
L 1 : x + 3y − 5 = 0
L 2 : 3x − ky − 1 = 0
L 3 : 5x + 2y − 12 = 0
intercept between the axes is bisected at A. Its equation
is
[AIEEE 2006, 4.5M]
(a) x + y = 7
(c) 4 x + 3y = 24
(b) x + 3 y = 0
Match the statements/Expressions in Column I with the
statements/Expressions in Column II
Column I
[AIEEE 2006, 6M]
Column II
(b) (3, ∞ )
(A) L1 , L 2 , L 3 are concurrent, if
(p)
k = −9
1

(d)  −3, − 

2
(B) one of L1 , L 2 , L 3 is parallel to at
least one of the other two, if
(q)
k=−
(C) L1 , L 2 , L 3 form a triangle, if
(r)
k=
(D) L1 , L 2 , L 3 do not form a triangle, if
(s)
k =5
102. Lines L 1 : y − x = 0 and L 2 : 2x + y = 0 intersect the line
L 3 : y + 2 = 0 at P and Q respectively. The bisector of the
acute angle between L 1 and L 2 intersects L 3 at R.
Statement II In any triangle, bisector of an angle
divides the triangle into two similar triangles.
5
6
[IIT-JEE 2008, 6M]
[IIT-JEE 2007, 3M]
Statement I The ratio PR : RQ equals 2 2 : 5 because
6
5
105. The perpendicular bisector of the line segment joining
P (1, 4) and Q (k , 3) has y-intercept −4. Then a possible
value of k is
[AIEEE 2008, 3M]
(a) 1
(b) 2
(c) −2
(d) −4
Chap 02 The Straight Lines
111. A ray of light along x + 3y = 3 gets reflected upon
106. The lines p ( p 2 + 1)x − y + q = 0 and
( p + 1) x + ( p + 1)y + 2q = 0 are perpendicular to a
common line for
[AIEEE 2009, 4M]
reaching X -axis, the equation of the reflected ray is
(a) exactly one values of p
(b) exactly two values of p
(c) more than two values of p (d) no value of p
(a) y = x + 3
(b) 3y = x − 3
(c) y = 3 x − 3
(d) 3y = x − 1
2
2
2
x y
+ = 1 passes through the point
5 b
(13, 32). The line K is parallel to L and has the equation
x y
+ = 1. Then the distance between L and K is
c 3
[AIEEE 2010, 4M]
107. The line L given by
(a) 17
23
(c)
17
17
15
23
(d)
15
(b)
[IIT-JEE 2011, 3M]
(a) y + 3 x + 2 − 3 3 = 0 (b) y − 3 x + 2 + 3 3 = 0
(c) 3y − x + 3 + 2 3 = 0 (d) 3y + x − 3 + 2 3 = 0
109. The lines L 1 : y − x = 0 and L 2 : 2x + y = 0 intersect the
line L 3 : y + 2 = 0 at P and Q respectively. The bisector of
the acute angle between L 1 and L 2 intersects L 3 at R .
[AIEEE 2011, 4M]
Statement I : The ratio PR : RQ equals 2 2 : 5
Statement II : In any triangle, bisector of an angle
divides the triangle into two similar triangles.
(a) Statement I is true, statement II is true; statement II is not
a correct explanation for statement I.
(b) Statement I is true, statement II is false.
(c) Statement I is false, statement II is true.
(d) Statement I is true, statement II is true; statement II is a
correct explanation for statement I
110. If the line 2x + y = k passes through the point which
divides the line segment joining the points (1, 1) and
(2, 4) in the ratio 3 : 2, then k equals
[AIEEE 2012, 4M]
29
5
(c) 6
(b) 5
(d)
11
5
112. For a > b > c > 0, the distance between (1, 1) and the
point of intersection of the lines ax + by + c = 0 and
bx + ay + c = 0 is less than 2 2. Then
[JEE Advanced 2013, 3M]
(a) a + b − c > 0
(c) a − b + c > 0
(b) a − b + c < 0
(d) a + b − c < 0
Q(6, − 1) and R(7, 3). The equation of the line passing
through (1, − 1) and parallel to PS is [JEE Main 2014, 4M]
an angle 60° to the line 3x + y = 1. If L also intersects
(a)
[JEE Main 2013, 4M]
113. Let PS be the median of the triangle with vertices P(2, 2),
108. A straight line L through the point (3, − 2) is inclined at
the X -axis, then the equation of L is
165
(a) 4 x + 7y + 3 = 0
(c) 4 x − 7y − 11 = 0
(b) 2 x − 9y − 11 = 0
(d) 2 x + 9y + 7 = 0
114. Let a, b, c and d be non-zero numbers. If the point of
intersection of the lines 4ax + 2ay + c = 0 and
5bx + 2by + d = 0 lies in the fourth quadrant and is
equidistant from the two axes, then [JEE Main 2014, 4M]
(a) 3bc − 2ad = 0
(c) 2bc − 3ad = 0
(b) 3bc + 2ad = 0
(d) 2bc + 3ad = 0
115. For a point P in the plane, let d 1 ( P ) and d 2 ( P ) be the
distance of the point P from the lines x − y = 0 and
x + y = 0 respectively. The area of the region R
consisting of all points P lying in the first quadrant of
the plane and satisfying 2 ≤ d 1 ( P ) + d 2 ( P ) ≤ 4, is
[JEE Advanced 2014, 3M]
116. The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with
vertices (0, 0), (0, 41) and (41, 0) is
[JEE Advanced 2015, 4M]
(a) 820
(c) 901
(b) 780
(d) 861
117. Two sides of a rhombus are along the lines, x − y + 1 = 0
and 7 x − y − 5 = 0. If its diagonals intersect at ( −1, − 2),
then which one of the following is a vertex of this
rhombus?
[JEE Main 2016, 4M]
 1 8
(a)  , − 
 3 3
 10 7
(b)  − , − 
 3
3
(c) ( −3, − 9 )
(d) ( −3, − 8 )
166
Textbook of Coordinate Geometry
Answers
Exercise for Session 6
Exercise for Session 1
1. (c)
6. (d)
11. (d)
2. (b)
7. (c)
12. y = 9
132
14. PQ =
12 3 + 5
3. (b)
4. (d)
5. (b)
8. (d)
9. (a,d) 10. (c)
13. (2 + 2 3 , 4) and (2 − 2 3 , 0)
15. 4 2 units
16. 83x − 35 y + 92 = 0
17. x + y − 11 = 0
Exercise for Session 2
1. (b)
2. (c)
3. (a)
4. (c)
5. (b)
6. (d)
7. (c,d) 8. (d)
9. (d)
10. (d)
11. (b)
12. (d)
13. The two points are on the opposite side of the given line.
15. 3x − 4 y = 0 and 3x − 4 y − 10 = 0
17. 7x + y − 31 = 0
18. 2x + 2 y +
2=0
Exercise for Session 3
1. (c)
2. (b)
3. (c)
4. (a)
5. (d)
6. (a)
7. (a)
8. (c)
9. (a)
10. (a,b) 11. (c)
12. (c)
13. 2a + b2 + b = 0 16. (i) y = 3 (ii) x = 4, (iii) 3x + 4 y = 24
−5 −5 
18.  ,

 3 3
Exercise for Session 4
1. (c)
7. (c)
2. (d)
8. (a)
3. (b)
4. (b)
5. (c)
6. (b)
9. x = 7 and x + 3 y = 7 + 9 3
10. x (4 3 + 3) + y(4 − 3 3 ) = 11 − 2 3 and
4 3
y (4 + 3 3 ) − x (4 3 − 3) = 11 + 2 3,
15
1 73 
1 77 


11.  − ,  and  , 
 14 28 
 16 32 
5
12.  0,  and (0, 0) 13. x + 2 y − 6 = 0
 2
14. 3x = 19
15. 10x − 10 y − 3 = 0
Exercise for Session 5
1. (d)
6. (d)
11. (b)
2. (a)
7. (d)
12. (c)
3. (b)
4. (d)
5. (c)
8. (c)
9. (d)
10. (b)
13. (3, − 2) 14. 14x + 23 y − 40 = 0
4 14

16. (2)
15. 4x − y + 6 = 0,  − , 
 5 5
1. (c)
2. (c)
3. (a,b)
4. (a)
5. (a)
1

6. (c)
7. m ∈  −1, 

5
5π π
5π
π

−1 
9. θ ∈  0,  ∪ 
8. θ ∈  0,
, 
− tan 3

 12   12 2 

6
π
5π
10. θ ∈  ∪  2nπ, 2nπ +   ∪  ∪  2mπ +
, 2mπ  

6 
6
n = z 
m = z 
11. Outside
12. 29x − 2 y + 33 = 0
Chapter Exercises
1. (b)
2. (a)
3. (c)
4. (d)
5. (c)
6. (b)
7. (b)
8. (b)
9. (b)
10. (c)
11. (c)
12. (b)
13. (c)
14. (d)
15. (a)
16. (b)
17. (b)
18. (c)
19. (b) 20. (b)
21. (b)
22. (b)
23. (c)
24. (a)
25. (a)
26. (b)
27. (a)
28. (a)
29. (b)
30. (a)
31. (a,b,c,d)
32. (a,b,c,d)
33. (a,c) 34. (a,d)
35. (b,d) 36. (a,b,c,d) 37. (a,c,d) 38. (a,d) 39. (a,b) 40. (a,b)
41. (a,b,c,d)
42. (b,d) 43. (a,b,c) 44. (a,b) 45. (a,b,c)
46. (d) 47. (d)
48. (a)
49. (a)
50. (d)
51. (a)
52. (b) 53. (c)
54. (a)
55. (c)
56. (d)
57. (b)
58. (a)
59. (a)
60. (b)
61. (3)
62. (6)
63. (5)
64. (8) 65. (3)
66. (9)
67. (8)
68. (2)
69. (2)
70. (4) 71. (A) → (p); (B) → (p, q); (C) → (p, r) (D) → (p, r, s)
72. (A) → (p, r); (B) → (q); (C) → (q, s) (D) → (p)
73. (A) → (p, q); (B) → (p, q, r, s); (C) → (p, q, r, s); (D) → (p, q, r, s)
74. (A) → (t); (B) → (p, q, r); (C) → (s)
75. (A) → (q, s); (B) → (p, t); (C) → (r)
76. (a)
77. (a)
78. (d) 79. (d)
80. (c)
81. (b)
82. (b)
83. (d)
6 −1
2 −13   3   −8 3 
85.  ,  ,  − ,
,
,
,
,
0
 
 

 5 10   5 10   2   5 10 
n
n


1
p
±∑

±∑
2
2

(1 + p )
(1 + p ) 
94. 2x + 3 y + 22 = 0
87.  p = 1
, p =1

c
c






h + mk mh − k 

95. 
96. 3x + 4 y − 18 = 0 and x − 2 = 0
,

 1 + m2 1 + m2 
97. (2 − 3 ) sq units. 99. (a)
100. (c) 101. (c) 102. (c) 103. (c)
103. (c) 104. (A) → (s); (B) → (p, q); (C) → (r); (D) → (p, q, s)
105.(a) 106. (a)
107. (c)
108. (b) 109. (b) 110. (c) 111. (b)
112.(a) 113. (d)
114. (a)
115. (6) 116. (b) 117. (a)
Solutions
1. Equation of line passing through (2, 0) and perpendicular to
2.
4. We have,
ax + by + c = 0
Then, required equation is
b
y − 0 = (x − 2)
a
ay = bx − 2b
⇒
ay − bx + 2b = 0
2
2
1
2m
Q
+
+
=
1 !9 ! 3 !7 ! 5 !5 ! n !
1 2 × 10 ! 2 × 10 ! 10 !  2m
+
+

=
10 !  1 !9 !
3 !7 !
5 !5 !  n !
1 10
2m
⇒
{2 C1 + 210C 3 + 10C 5 } =
10 !
n!
1 10
2m
10
10
10
{ C1 + C 3 + C 5 + C 7 + 10C 9 } =
n!
10 !
1 10 − 1 2m
⇒
=
(2 )
n!
10 !
∴
m = 9 and n = 10
Hence, x − y + 1 = 0 and x + y + 3 = 0 are perpendicular to
each other, then orthocentre is the point of intersection which
is ( −2, − 1 )
∴
−2 = 2m − 2n and −1 = m − n
∴ Point is (2m − 2n, m − n ).
5.
y = cos x cos( x + 2 ) − cos2( x + 1 )
1
y = {2 cos x cos( x + 2 ) − 2 cos2( x + 1 )}
2
1
= {cos(2 x + 2 ) + cos2 − 1 − cos(2 x + 2 )}
2
1
= (cos2 − 1 )
2
1
= (1 − 2 sin 2 1 − 1 )
2
= − sin 2 1
which is a straight line passing through ( λ , − sin 2 1 ); ∀ λ ∈ R
and parallel to the X -axis.
x y
Let line
…(i)
+ =1
a b
Its passes through (2, 2), then
2 2
+ =1
a b
…(ii)
⇒
2(a + b ) = ab
Y
(0, b) B
3. ∴ Required area
6
1 8
 64 2
= 4 ×  × 4 =
=
 3
2 3
3
…(i)
X¢
O
(a, 0) A
f ( x + y ) = f ( x ) f (y )
Q
Y¢
Y
1
Q Area of ∆AOB = ab = | λ |
2
∴
ab = 2| λ |
from Eq. (ii), a + b = | λ |
Hence, required equation is
x 2 − (a + b ) x + ab = 0
(0, 4)
(–8/3, 0)
X¢
(8/3, 0)
O
X
or
6.
(0, –4)
Y¢
∴
f (2 ) = f (1 ) f (1 ) = 2
∴
f (3 ) = f (1 + 2 ) = f (1 ) f (2 ) = 2 3
…………………………………………
…………………………………………
f (n ) = 2n
∴
Area =
2
x 2 − | λ | x + 2| λ | = 0
Value of (a 2x + aby + 1 ) at (1, 1 )
>0
Value of (a 2x + aby + 1 ) at ( 0, 0 )
or
2 6 f (6 )
sq units
=
3
3
X
or
∴
⇒
⇒
∴
i.e.
a 2 + ab + 1
> 0; ∀ a ∈ R
1
a 2 + ab + 1 > 0; ∀ a ∈ R
D<0
b −4<0
2
−2 < b < 2 but b > 0
0 <b <2
b ∈( 0, 2 )
(given)
168
Textbook of Coordinate Geometry
x
y
= 1 and let the axis be rotated through
b
an angle θ and let ( X , Y ) be the new coordinates of any point
P ( x, y ) in the plane, then
x = X cosθ − Y sin θ, y = X sin θ + Y cosθ, the equation of the
line with reference to original coordinates is
x y
+ =1
a b
X cosθ − Y sin θ X sin θ + Y cosθ
i.e.
…(i)
+
=1
a
b
and with reference to new coordinates is
X Y
…(ii)
+ =1
p
q
7. Equation of L is a +
Comparing Eqs. (i) and (ii), we get
cosθ sin θ 1
+
=
a
b
p
sin θ cosθ 1
and
−
+
=
a
b
q
…(iii)
…(iv)
Squaring and adding Eqs. (iii) and (iv), we get
1
1
1
1
+ 2 = 2 + 2
2
a
b
p
q
8. d (x, y ) = max{| x |, | y | }
∴
and if
9.
E
75°
30°
2
X
(0, 0) A
∴ Equation of BD is
y − 1 = ( 3 − 2 )( x − 3 )
⇒
(2 − 3 ) x + y = 2( 3 − 1 )
and equation of the circumcircle of the triangle ABE
(Apply diametric form as AB is diameter)
( x − 0 )( x − 3 ) + (y − 0 )(y − 1 ) = 0
x2 + y 2 − x 3 − y = 0
11. If (α, β) be the image of (4, 1) w.r.t y = x − 1, then (α, β) = (2, 3),
say point Q
Y
Q
(2, 3)
R
(3, 3)
45°
| y | > | x |, then a = | y |
=
B
90°
2
x = ±a
p 3 = | m′ 2 cosα + 2m′ sin α +
C
2
D
R¢

sin 2 α

p 2 = mm′ cos α + (m + m′ )sin α +
cosα 

|(m cosα + sin α )| | m′ cosα + sin α |
=
| cosα |
Q
Y
…(ii)
∴
y =±a
Therefore locus represents a straight line.
sin 2 α
P1 = | m 2 cosα + 2m sin α +
|
cosα
(m cosα + sin α ) 2
=
| cosα |
and
3 − 1 ( 3 − 1 )( 3 − 1 )
=
= 3 −2
−1 − 3
−2
⇒
…(i)
but
d ( x, y ) = a
From Eqs. (i) and (ii), we get
a = max{| x |, | y | }
if
| x | > | y |, then a = | x |
Slope of BD =
sin 2 α
|
cosα
(m′ cosα + sin α ) 2
| cosα |
p 22 = p1 p 3
Hence, p1, p 2, p 3 are in GP.
10. Side of the square = 2 unit
Coordinates of B, C and D are ( 3, 1 ), ( 3 − 1, 3 + 1 ) and
( −1, 3 ) respectively.
O
45°
(1, 0)
P
(4, 1)
X
After translation through a distance 1 unit along the positive
direction of X -axis at the point whose coordinate are R ≡ (3, 3 ).
π
After rotation through are angle about the origin in the
4
anticlockwise direction, then R goes toR ′ such that
OR = OR′ = 3 2
∴The coordinates of the final point are ( 0, 3 2 ).
A ≡ ( 0, 0 ); B ≡ (2, 0 ); C ≡ (2, 2 ); D ≡ ( 0, 2 )
(i) f1( x, y ) → (y , x ), then
A ≡ ( 0, 0 ); B ≡ ( 0, 2 ); C ≡ (2, 2 ), D ≡ (2, 0 )
(ii) f 2( x, y ) → ( x + 3y , y ), then
A ≡ ( 0, 0 ); B ≡ (6, 2 ); C ≡ (8, 2 ), D ≡ (2, 0 )
x −y x + y
(iii) f 3( x, y ) → 
,
 , then
 2
2 
12. Q
Now,
A ≡ ( 0, 0 ); B ≡ (2, 4 ); C ≡ (3, 5 ), D ≡ (1, 1 )
AB = DC = 2 5, AD = BC = 2
and
AC = 34, BD = 10
i.e.
AC ≠ BD
∴ Final figure is a parallelogram.
Chap 02 The Straight Lines
13. Let
AN
=λ
BN
169
It is clear from the figure
−3 π 

π 
α ∈  −π,
 ∪  , π .

4 
4 
 a
aλ 
Then, coordinate of N are 
,

1 + λ 1 + λ
16. Q Equation of AB is x + y = 1, then coordinates of A and B are
Q Slope of AB = − 1
(1, 0) and (0, 1) respectively.
 1 1
∴ Coordinates of P are  , 
 2 2
Y
B(0, a)
Q PP1 is perpendicular to OA
N
Y
x+y= a
M
B(0, 1)
O
X
A(a, 0)
∴ Slope of MN = 1
∴ Equation on MN is
1−λ
aλ
a
y −
=x−
⇒ x − y = a

 λ + 1
1+λ
1+λ
P
Then,
⇒
We have,
Also,
a λ
3 1
= ⋅ a2
2
8 2
(1 + λ )
1
λ = 3 or λ =
3
2
⇒
⇒
∴
( x + 1 ) 2 + y 2 + ( x − 2 ) 2 + y 2 = 2[( x − 1 2 ) + y 2 ]
2 x 2 + 2y 2 − 2 x + 5 = 2 ( x 2 + y 2 − 2 x + 1 )
3
⇒
2x + 3 = 0 ⇒ x = −
2
A straight line parallel to Y -axis.
Value of (3 x − 2y + 1 ) at A
>0
Value of (3 x − 2y + 1 ) at B
(sin α − 3 ) − (cosα − 2 ) + 1
⇒
>0
(3 − 2 + 1 )
sin α − cosα > 0 ⇒ sin α > cosα
–p –3p/4
–p
2
(say)
(OPn − 1 ) = (OMn − 1 ) 2 + ( Pn − 1Mn − 1 ) 2
2
1 2
α n − 1 = 2α n2
2
1
αn = αn − 1
2
1
OPn = α n = α n − 1
2
1
1
= 2 αn − 2 = 3 αn − 3
2
2
……………………………………………
……………………………………………
……………………………………………
1
= n − 1 α1
2
1  1 1
= n −1   = n.
 2 2
2
17. Let O ≡ (0, 0), A ≡ (1, 1) and B ≡ (9, 1)
1
×8 ×1 = 4
2
It is clear that 1 < c < 9
Y
X¢
1
2
(OMn − 1 ) 2 = (OPn ) 2 + ( Pn Mn − 1 ) 2
1
α n2 − 1 = 2α n2 + α n2 − 1
2
⇒
⇒
Area of ∆OAB =
O
Y¢
p
4
p
2
X
OP1 = PP1 =
= 2α n2
14. Let S = (x, y ), given (SQ )2 + (SR )2 = 2(SP )2
15.
P1
= 2(OPn ) 2
1
For λ = , then M lies outside the segment OB and hence the
3
required value of λ = 3.
⇒
P3 P2
Equation of OP is y = x
3
area of ∆OAB
8
1
3 1
⋅ AN ⋅ MN = ⋅ a ⋅ a
2
8 2
1
2
2
3
a
λ
a

= ⋅ 1 a ⋅ a
⋅
⋅
2 1 + λ 1 + λ 8 2
∴
1
A(1, 0)
O
Therefore, area of ∆AMN =
⇒
y=
M2
  λ − 1 
So, the coordinates of M are  0,a 

  λ + 1 
⇒
x+
M1
p
X
and
 c
M ≡ (c, 1 ) and N ≡ c, 
 9
{Qy = x}
170
Textbook of Coordinate Geometry
Y
20. Let A ≡ (3, 4), B ≡ (0, y ), C ≡ (x, 0), D ≡ (8, 2)
∴ Slope of AB = − Slope of BC
0 −y
y −4
⇒
=−

 x − 0
0 −3
x=c
A(1, 1)
(a, 1)
B
M
x
y= 9
N
X
O
∴ Area of ∆BMN = 2
1
c

⇒
× (9 − c ) × 1 −  = 2

2
9
or
(given)
(9 − c ) = 36
2
9 − c = ± 6 ⇒ c = 3 or 15
1 <c <9
c =3
or
but
∴
or
4 x − xy = 3y
and slope of BC = − slope of CD
0 −y
 2 − 0
⇒

 = −

 x − 0
8 − x
…(i)
or
2 x + xy = 8y
adding Eqs. (i) and (ii), we get
6 x = 11y
from Eqs. (ii) and (iii), we get
13
1
x= =4
3
3
…(ii)
21.
y=x
Y
18. The three lines are concurrent if
(–n, m) C
B (n, m)
1 2 −9
3 −5 −5 = 0
a
b
…(iii)
A (m, n)
−1
X¢
or
5a + 2b = 1
which is three of the line 5 x + 2y = 1 passes through (a, b ).
X
O
19. Q BC = AC
⇒
2 2 + ( λ − 1 ) 2 = λ2
⇒
4 = λ2 − ( λ − 1 ) 2
∴
D (–n, –m)
= (2 λ − 1 )(1 )
5
λ=
2
C
Area of rectangle BCDE = (2n )(2m )
= 4mn
1
and area of ∆ABE = × 2m × (m − n )
2
= m(m − n )
∴ Area of pentagon = 4mn + m(m − n )
= m (m + 3n )
(2, l)
B
(0, 1)
22. The equation of the line L, be y − 2 = m(x − 8), m < 0
D
O
Q Equation. of AB is
E (n,–m)
Y¢
2 

coordinates of P and Q are P 8 − , 0 and Q( 0, 2 − 8m ).

m 
A (2, 0)
x=2
x y
 1
+ = 1, D ≡ 1, 
 2
2 1
So,
2
+ 2 − 8m
m
2
= 10 +
+ 8( −m ) ≥
( −m )
OP + OQ = 8 −
(mid-point of AB)
∴ Equation of CD is 2x − y = µ
QCD pass through D, thus
1
3
2 − = µ or µ =
2
2
3
2x − y =
∴ Equation of CD is
2
and Eq. (i) of line ⊥ to AC and pass through B is y = 1
from Eqs. (i) and (ii), we get
5 
Orthocentre ≡  , 1
4 
10 + 2
2
× 8( −m ) ≥ 18
( −m )
So, absolute minimum value of OP + OQ = 18
…(i)
…(ii)
23. Let the two perpendiculars through the origin intersect
2x + y = a at A and B so that the triangle OAB is isosceles.
OM = length of perpendicular from O to
a
.
AB, OM =
5
Chap 02 The Straight Lines
27. Q B is the reflection of A(5, 7) w.r.t the line x + y = 0
Y
∴ B ≡ ( −7, − 5 )
and C is the reflection of A(5, 7 ) w.r.t the line x − y = 0
B
A (5, 7)
M
O
2x+y=a
X
A
M
AM = MB = OM
2a
AB =
⇒
5
1
Area of ∆OAB = ⋅ AB ⋅ OM
2
1 2a a
a2
sq units
= ⋅
⋅
=
2 5 5
5
=
x–y
N
=0
B
∴ Equation of BC is y + 5 =
5+5
( x + 7 ) or 7y = 5 x
7+7
28. Let P ≡ (2, − 1)
5
3 + 4m
x is an integer, if 3 + 4m = 1, − 1, 5, −5
−2 −4 2 −8
or
m=
,
, ,
4 4 4 4
1
1
or
m = − , − 1, , − 2
2
2
Hence, m has two integral values.
25. Let the coordinates of A be (a, 0). Then the slope of the
(say) …(i)
Then the slope of the incident ray
2−0
=
= tan( π − θ )
1 −a
From Eqs. (i) and (ii), we get
tan θ + tan( π − θ ) = 0
3
2
+
=0
⇒
5 −a 1 −a
P(2, − 1 ) goes 2 units along x + y = 1 upto A and 5 units along
x − 2y = 4 upto B.
Now,
slope of x + y = − 1 is −1 = tanθ
(say)
∴
θ = 135 °
1
and
slope x − 2y = 4 is = tan φ
(say)
2
1
2
, cos φ =
sin φ =
∴
5
5
The coordinates of A
i.e.
(2 + 2 cos135 °, − 1 + 2 sin 135 ° )
or
(2 − 2, 2 − 1 )
The coordinates of B
i.e. (2 + 5 cos φ, − 1 + 5 sin φ ) or (2 + 2 5, 5 − 1 )
29. Q P ≡ (5, 3)
Let P′ and P′′ be the images of P w.r.t y = x and y = 0 (X -axis)
respectively, then P′ ≡ (3, 5 ) and P′′ ≡ (5, − 3 )
Q PQ + QR + RP is minimum
∴ P ′ , R, Q, P ′′ are collinear.
P¢(3, 5)
Y
x
y=
R
⇒
3 − 3a + 10 − 2a = 0
13
a=
5
 13 
Thus, the coordinate of A is  , 0
5 
X¢
or
∴
5 x − 2y − 7 = 0
a = 5, b = − 2 ⇒a + b = 3
P(5, 3)
X
Q
O
26. Lines 5x + 3y − 2 + λ(3x − y − 4) = 0 are concurrent at (1, −1)
and lines
x − y + 1 + µ(2 x − y − 2 ) = 0 are concurrent at (3, 4).
Thus equation of line common to both family is
4+1
y +1=
(x − 1)
3 −1
0
C
∴ C ≡ (7, 5 )
24. Solving given equations, we get
reflected ray is
3−0
= tanθ
5 −a
x+
y
(0, 0)
Also,
x=
171
P¢¢(5, –3)
Y¢
∴ Equation of P ′ P ′′ is
 5 + 3
y +3=
 (x − 5)
 3 − 5
or
∴
4 x + y = 17
 17 
Q ≡  , 0
4 
(QQ on Y -axis)
172
Textbook of Coordinate Geometry
30. Equation of incident ray is
or
or
31.
y − 0 = tan(90 ° + 60 ° )( x − 2 )
1
y =−
(x − 2)
3
(x − 2) + y 3 = 0
and equation of refracted ray is
y − 0 = − tan 60 °( x − 2 )
or
y = − 3(x − 2)
y
or
(x − 2) +
=0
3
∴ Combined equation is
y 

[( x − 2 ) + y 3 ] ( x − 2 ) +
 =0

3
4
i.e.
(x − 2)2 + y 2 +
( x − 2 )y = 0
3
x y
x y
Point of intersection of + = 1 and + = 1 is
a b
b a
 ab
ab 
P
,
 , this point P satisfies alternates (a), (b), (c)
a + b a + b
and (d).
35. Equation of any line through the point of intersection of the
given lines is (3 x + y − 5 ) + λ ( x − y + 1 ) = 0.
Since this line is perpendicular to one of the given lines
3+λ
1
= − 1 or
3
λ −1
⇒ λ = − 1 or −5, therefore the required straight line is
x + y −3 = 0
or
x − 3y + 5 = 0
36. If B lies on Y -axis, then coordinates of B are (0, a ) or (0, − a )
Y
B
60°
a
A
X¢
A¢
O
30°
30°
60°
A¢¢
c − a = 0 or c 2 + a 2 + ca = k
and
a − b = 0 or a 2 + b 2 + ab = k
⇒
or
a = b or b = c or c = a
b 2 + c 2 + bc = c 2 + a 2 + ca
⇒
or
b = c or c = a
a = b or a + b + c = 0
33. As the third vertex lies on the line y = x + 3, its coordinates
are of the form ( x, x + 3 ). The area of the triangle with vertices
(2, 1), (3, −2) and ( x, x + 3 ) is given by
x x + 3 1
1 
(given)
|2
1
1
| = |2 x − 2| = 5
2
3
−
2
1


−3 7
,
∴
2x − 2 = ± 5 ⇒ x =
2 2
 7 13
Thus, the coordinates of the third vertex are  ,  or
2 2 
 −3 3 
 , .
 2 2
1
34. 3
λ
−2 −6
1 −4
= 0
2
4 λ
⇒
λ + 2λ − 8 = 0
∴
⇒
( λ + 4 )( λ − 2 ) = 0
λ = − 4, 2
30°
30°
X
B¢
k such that
b 3 − c 3 = k(b − c ), c 3 − a 3 = k(c − a ) and a 3 − b 3 = k(a − b )
b − c = 0 or b 2 + c 2 + bc = k
A
x– 3y=0
A¢
32. The two lines will be identical if their exists some real number
⇒
60°
60°
a
Y¢
If third vertex in IV quadrant or in II quadrant, then its
coordinates are (a cos30 °, − a sin 30 ° ) and ( −a cos30 °, a sin 30 ° )
 a 3 a
a 3 a
i.e.
, ,
, −  and  −

 2 2
 2
2
37. Since, ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are
concurrent
∴
a b c 
b c a = 0


c a b
⇒
3abc − a 3 − b 3 − c 3 = 0
⇒ − (a + b + c )(a 2 + b 2 + c 2 − ab − bc − ca) = 0
∴
a+b+c≠0
a 2 + b 2 + c 2 − ab − bc − ca = 0
1
{(a − b ) 2 + (b − c ) 2 + (c − a ) 2 } = 0
2
As a, b, c are real numbers
∴
b − c = 0, c − a = 0, a − b = 0
⇒
a =b =c
38. Q E ≡ (4, 4)
∴ zC = 7 + 5i , z E = 4 + 4i
Now, (in ∆BEC)
π
i
zB − zE
=e 2 =i
zC − z E
2
⇒
or
z B − 4 − 4i = i (7 + 5i − 4 − 4i )
z B = 3 + 7i
Chap 02 The Straight Lines
∴ B ≡ (3, 7 ), then D ≡ (5, 1 )
B
173
After solving, we get
x1 = 1 ± 100 2, 1
C (7, 5)
and
y1 = 7, 7 ± 100 2
Hence,
(1 + 100 2, 7 ), (1 − 100 2, 7 ),
(1, 7 + 100 2 ), (1, 7 − 100 2 )
E
42. Equation of the other diagonal is x + y = λ which pass
A(1, 3)
through (a, b ), then
a+b=λ
∴ Equation of other diagonal is
x + y =a + b
i.e. then centre of the square is the point of intersection of
b b

x − y = a and x + y = a + b is a + ,  , then vertex

2 2
D
Equation of AB is
7 −3
( x − 1 ) or 2 x − y + 1 = 0
3 −1
and equation of AD is
1 −3
y −3 =
( x − 1 ) or x + 2y − 7 = 0
5 −1
Given,
6a 2 − 3b 2 − c 2 + 7ab − ac + 4bc = 0
y −3 =
39.
B
b b
a+ ,
2 2
A(a, b)
⇒ 6a + (7b − c )a − (3b − 4bc + c ) = 0
2
2
2
− (7b − c ) ± (7b − c ) 2 + 24 (3b 2 − 4bc + c 2 )
12
⇒
12a + 7b − c = ± (11b − 5c )
⇒
12a − 4b + 4c = 0
or
12a + 18b − 6c = 0
⇒
3a − b + c = 0
or
−2a − 3b + c = 0
Hence (3, − 1 ) or ( −2, − 3 ) lies on the line ax + by + c = 0,
O
⇒ a=
C
D
∴
If
40. x + 2y + 4 = 0 and 4x + 2y − 1 = 0
⇒
x + 2y + 4 = 0
and
−4 x − 2y + 1 = 0
Here,
(1 )( −4 ) + (2 )( −2 ) = − 8 < 0
∴Bisector of the angle including the acute angle bisectors and
origin is
x + 2y + 4 ( −4 x − 2y + 1 )
=
5
2 5
⇒
6 x + 6y + 7 = 0
Then,
⇒
∴
and
43. (y − y1 ) − m(x − x1 ) = 0 is family of lines
x1 + y1 − 8
= ± 100
2
x1 − y1 + 6
= ± 100
2
∴
Then,
(0, 6)
M
y–x=6
X′
(–6, 0)
P(x, y) x+y=8
O
Y′
y − y1 = 0, x − x1 = 0
y = y1 and x = x1
44. Given lines L1 = 0 and L2 = 0 are perpendicular and given
Y
(0, 8)
b ib 
ib b

 b ib 
z − a + +  = i  − +  = − −

 2
2
2
2
2 2
∴
z =a
∴
B ≡ (a, 0 )
then,
D ≡ (a + b, b )
Hence, other vertices are (a + b, 0 ), (a, 0 ) and (a + b, b ).
41. Let position of bunglow is P (x1, y1 ), then PM = 100 and
PN = 100
C ≡ (2a + b − a, b − b )
C ≡ (a + b, 0 )
B ≡z
b ib 

z − a + + 
π

BO i 2
2
2
=
e = i (Q BO = AO)
b ib  AO

(a + ib ) − a + + 

2
2
(8, 0)
45.
X
bisectors are λ 1L1 − λ 2L2 = 0 and λ 1L1 + λ 2L2 = 0
∴bisectors are perpendicular to each other.
Hence, bisectors of λ 1L1 − λ 2L2 = 0 and λ 1L1 + λ 2L2 = 0 are
L1 = 0 and L2 = 0.
 θ + φ
∴ One bisector makes an angle 
 with X -axis, then
 2 
 θ + φ
other bisector makes an angle 90 ° + 
 with X -axis.
 2 
174
Textbook of Coordinate Geometry
48. QOT = CT
f–q
2
q f
(3, –5)
q+
f+q
f–q
=
2
2
∴ Equations of bisectors are
x −3
y +5
=
 θ + φ
 θ + φ
cos
 sin 

 2 
 2 
and
⇒
x −3
π
θ+

cos +
2
2
x −3
 θ + φ
− sin 

 2 
y +5
π
θ+

sin  +
2
2
y +5
=
 θ + φ
cos

 2 
φ


=
…(i)
φ


…(ii)
x −3 y + 5
=
cosα sin α
x −3 y + 5
θ+φ
[from Eq. (i)]…(iii)
and
=
α=
β
γ
2
But given bisector are
∴
∴
 θ + φ
β = − sin 
 = − sin α
 2 
and
 θ + φ
γ = cos 
 = cosα
 2 
⇒
| x − 0| + | y − 0| = | x − 4| + | y − 3|
Q
x ≥ 0, y ≥ 0
⇒
x + y = | x − 4| + | y − 3|
Case I : If 0 ≤ x ≤ 4 and 0 ≤ y ≤ 3
x + y = 4 − x + 3 −y
7
x+y =
⇒
2
Case II : If 0 ≤ x ≤ 4 and y ≥ 3
x + y = 4− x + y − 3
1
x=
⇒
2
Case III : If x ≥ 4 and 0 ≤ y ≤ 3
x + y = x − 4 + 3 −y
(impossible)
y = −1 / 2
Case IV : If x ≥ 4 and y ≥ 3
x + y = x − 4 + y −3
(impossible)
⇒
0 = −7
Combining all cases, we get
7
x + y = , ∀ 0 ≤ x ≤ 4 and 0 ≤ y ≤ 3
2
1
and
x = , ∀ 0 ≤ x ≤ 4 and y ≥ 3
2
Y
[from Eq. (ii)]
3
46. QOR = AR
⇒
⇒
Q
∴
⇒
| x − 0| + | y − 0| = | x −1| + | y − 2|
| x| + | y | = | x − 1| + | y − 2|
0 ≤ x < 1 and 0 ≤ y < 2
x + y = − ( x − 1 ) − (y − 2 )
2 x + 2y = 3
O
1/2
X
3.5
Sol. (Q. Nos. 49 to 51)
AB : 2 x − y + 4 = 0,
BC : x − 2y − 1 = 0
and
CA : x + 3y − 3 = 0
47. OS = BS
⇒ | x − 0| + | y − 0| = | x − 2| + | y − 3|
Y
Y
A
O
C
O
X¢
infinte ray
X
X
1
2
⇒ | x| + | y | = | x − 2| + | y − 3|
Q
x ≥ 2 and 0 ≤ y < 3
∴
x + y = x −2 + 3 −y
⇒
2y = 1
1
∴
y =
2
B
∴
mAB = m1 = 2
1
mBC = m2 =
2
and
mCA = m3 = −
Q
m1 > m2 > m3
Y¢
1
3
Chap 02 The Straight Lines
49. Q ∠A is obtuse
∴
m3 − m1
1 + m3 m1
1
− −2
= 3
= −7
2
1−
3
or
tan A =
or
i.e.
AB : 2 x − y + 4 = 0
BC : − x + 2y + 1 = 0
Q
(2 )( −1 ) + ( −1 )(2 ) = − 4 < 0
∴ External bisector of B is
( − x + 2y + 1 )
 2x − y + 4

 =−


5
5
∴ Equation of AB is
or
(Q B lies on x + y = 2)
52. Equation of BC is
2
− − (2 − λ )
y − (2 − λ ) = 5
(x − λ )
2
− −λ
5
If ( x, y ) be image of the point (t, 3t − 2 ) in the line y = x − 1 or
x − y − 1 = 0, then
x − t y − (3t − 2 )
=
1
−1
2(t − 3t + 2 − 1 )
=−
1+1
x − t y − 3t + 2
=
= 2t − 1
⇒
1
−1
or
x − t = 2t − 1
…(i)
⇒
x + 1 = 3t
and
y − 3t + 2 = − 2t + 1
…(ii)
⇒
y + 1 =t
From Eqs. (i) and (ii), we get
x + 1 = 3 (y + 1 )
⇒
x − 3y = 2
56. Any point on the circle x 2 + y 2 = 4 is (2 cosθ, 2sin θ ),θ being
parameter.
If ( x, y ) be image of the point (2 cosθ, 2 sin θ ), in the line
x + y = 2, then
x − 2 cosθ y − 2 sin θ
=
1
1
−2(2 cosθ + 2 sin θ − 2 )
=
1+1
Let slope of bisector ( x + y = 2 ) = m3 = − 1
m3 − m1
m − m3
Now,
= 2
1 + m3m1 1 + m2m3
1 + λ 12 − 5 λ
−1 −
+1
1−λ
2 + 5λ
⇒
=
1+λ
12 − 5 λ
1−
1−
1−λ
2 + 5λ
−2
14
or
=
−2 λ −10 + 10 λ
∴
9
−3
y −3 = 2
(x − 1)
5
− −1
2
3 x + 7y = 24
55. Any point on the line 3x − y = 2 is (t, 3t − 2), t being parameter.
x + 3y − 3 = 0, then
α + 3 β + 2 −2( −3 − 6 − 3 )
=
=
1
3
1+9
α + 3 β + 2 12
or
=
=
1
3
5
3
26
or
α = − and β =
5
5
 3 26
∴ Required image is  − ,  ,
 5 5
14 λ = − 10 + 10 λ
−5
λ=
2
[from Eq. (i)]
5 9
2 2
51. Let (α, β) be the image of B(−3, − 2) w.r.t. the line
or
 5 9
− , 
 2 2
54. A ≡ (1, 3) and B ≡  − , 


x+y +5=0
Sol. (Q. Nos. 52 to 54)
Let
B ≡ (λ, 2 − λ )
1+λ
Slope of line AB = m1 =
1−λ
5 λ − 12
and Slope of line BC = m2 =
−5 λ − 2
12 − 5 λ
=
2 + 5λ
2 9
− −
5
5

y − 2 − = 5 2 x + 
2
2 −2 + 5 
5 2
7 x + 3y + 4 = 0
53. Coordinates of vertex B are (λ, 2 − λ )
50. For external bisector of B
or
175
x − 2 cosθ = y − 2 sin θ
= − 2 cosθ − 2 sin θ + 2
or
x − 2 cosθ = − 2 cosθ − 2 sin θ + 2
⇒
x − 2 = − 2sinθ
and
y − 2 sin θ = − 2 cosθ − 2 sin θ + 2
⇒
y − 2 = − 2 cosθ
From Eqs. (i) and (ii),
( x − 2 ) 2 + (y − 2 ) 2 = 4
or
…(i)
⇒
x 2 + y 2 − 4 x − 4y + 4 = 0
…(i)
…(ii)
176
Textbook of Coordinate Geometry
57. Any point on the parabola x 2 = 4y is (2t, t 2 ), t being parameter.
If ( x, y ) be image of the point (2t, t ) in the x + y = a, then
2
x − 2t y − t 2
+
1
1
−2 (2t + t 2 − a )
=
1+1
⇒
⇒
⇒
or
= − 2t − t 2 + a
or
x − 2t = − 2t − t 2 + a
⇒
x − a = −t2
60. Q AO = 2R cos A
…(i)
=2 ×
y − t 2 = − 2t − t 2 + a
and
⇒
y − a = −2t
From Eqs. (i) and (ii) we get
(y − a ) 2 = 4t 2 = − 4( x − a )
…(ii)
=
(y − a ) = 4(a − x )
or
Sol. (Q. Nos. 58 to 60)
Given orthocentre O ≡ (1, 2 )
and circumcentre
O′ = (2, 4 )
or
E
O¢
O
A
B
D
A
M 2x–y=3
Q Slope of OO′ = Slope of (2 x − y = 3 )
3
and
OD = O ′ M =
5
Let R be the circumradius
∴
O ′ M = R cos A
3
R cos A =
⇒
5
58. R = AO ′ = (AO ) + (OO ′ )
2
…(i)
2
[from Eq. (i)]
Here,
OA =
or
(2m − 3 ) 2 = ± 24m
62. Q Point of intersection of ax + 3y − 1 = 0 and ax + y + 1 = 0 is
 2 
A  − , 1 and point of intersection of ax + 3y − 1 = 0 and
 a 
 1
1 
x + 3y = 0 is B 
,−

 a − 1 3(a − 1 )
⇒
and
61
5
Q
59. QOD = 2R cos B cosC
3
5
= R cos A
y − 3 = m( x − 2 )
mx − y = 2m − 3
x
y
+
=1
 2m − 3 (3 − 2m )


 m 
Here D > 0, This is a quadratic in m which given two value of
m, and taking negative sign, we get (2m + 3 ) 2 = 0.
−3
This gives one line of m as .
2
Hence, three straight lines are possible.
= (2 R cos A ) 2 + 5
=
6
5
Taking positive sign, we get 4m 2 − 36m + 9 = 0
C
2
 6 
=   +5
 5
[from Eq. (i)]
2m − 3
or OB = 3 − 2m
m
Q The area of ∆OAB = 12
1 × OA × OB = 12
⇒

2
1  2m − 3
or

 (3 − 2m ) = ± 12
2 m 
A
F
3
5
61. The equation of straight line through (2, 3) with slope m is
2
or
∴
cos A = 2 cos B cosC
(QA + B + C = π )
− cos( B + C ) = 2 cos B cosC
− (cos B cosC − sin B sin C ) = 2 cos B cosC
sin B sin C = 3 cos B cosC
3
=3 ×
2R 5

61 
9
=

Q R =
5

2 61
∴
2 R cos B cosC =
[from Eq. (i)] …(ii)
or
∴
a
2
1
=−
3
= −1
Slope of OA is mOA = −
Slope of OB is mOB
mOA × mOB
a
1
− × − = −1
2
3
a = −6
|a | = 6
Chap 02 The Straight Lines
63. Here, B is the image of A w.r.t line y = x
66. Let a be the length of side of square
∴ B ≡ (2, 1 ) and C is the image of A w.r.t line x − 2y + 1 = 0 if
C ≡ (α , β ), then
α − 1 β − 2 −2(1 − 4 + 1 )
=
=
−2
1
1+4
9
2
or
α = and β =
5
5
9
2


C ≡ , 
∴
 5 5
⇒ Equation of BC is
2

 − 1
5

y −1 =
(x − 2)
9

 − 2
5

or
Here,
∴
3x − y − 5 = 0
a = 3, b = − 1
a − 2b = 5
∴
a 2 + a 2 = 22 ⇒ a = 2
i.e. distance between parallel lines is 2
Now, let two lines of family y = x + n are y = x + n, and
y = x + n2, where
n1, n2 ∈ { 0, 1, 2, 3, 4 }
| n1 − n2|
∴
= 2
2
or
| n1 − n2| = 2
⇒ {n1, n2 } are { 0, 2 }, {1, 3 } and {2, 4 }
Hence, both the family have three such pairs. So, the number
of squares possible is 3 × 3 = 9.
67. Let the coordinate of C be (1, c ), then
c −y
1−x
c − m1x
m2 =
1−x
m2 =
(Q Eq. of BC is ax + by − 5 = 0)
or
2007 − 223 
= 198


9
Y
(0, 9) B
2
1
O
A
(223, 0)
(Q slope of AB = m1)
⇒
or
64. On the line y = 1, the number of lattice points is
X
Hence, the total number of points
=
2007 − 223y 

9
y =1
8
∑ 
= 198 + 173 + 148 + 123 + 99 + 74 + 49 + 24 = 888
Hence, tens place digit is 8.
65. A rough sketch of the lines is given.
m2 (1 − x ) = c − m1x
c = (m1 − m2 ) x + m2
1
Now, the area of ∆ABC is | cx − y |
2
1
(Qy = m1x)
= ((m1 − m2 ) x + m2 ) x − m1x|
2
1
[Qm1 > m2 and x ∈( 0, 1 )]
= (m1 − m2 )( x − x 2 )
2
1
Hence,
f (x ) = (x − x 2 )
2
df ( x ) 1
= (1 − 2 x )
∴
dx
2
d 2 f (x )
and
= −1 < 0
dx 2
For maximum of
df ( x )
1
=0 ⇒ x=
f ( x ),
dx
2
1  1 1
f ( x ) max =  − 
∴
2  2 4
=
There are three triangle namely ABC, BCD and ABD
A
1
=λ
8
(given)
1
=8
λ
⇒
68. Equation of AB is 3x − 2y + 6 = 0
Y
B
Y
A (0, 3)
C
D
2x
y= +3
3x+
2
X¢
3
y=
x+
y=
y=
C (6, 1)
x+
3
P(l, l+1)
X¢
X
O
Y¢
177
B
(–2, 0)
X
O
Y¢
178
Textbook of Coordinate Geometry
Equation of BC is x − 8y + 2 = 0,
Equation of CA is x + 3y − 9 = 0
Let
P ≡ (λ, λ + 1)
Q B and P lie on one side of AC, then
λ + 3( λ + 1 ) − 9
>0
−2 + 0 − 9
71. (A)
L1
L2
4λ − 6 < 0
3
or
λ<
2
and C and P lie on one side of AB, then
3 λ − 2( λ + 1 ) + 6
>0
18 − 2 + 6
or
or
λ+4>0
or
λ > −4
Finally, A and P lie on one side of BC, then
λ − 8( λ + 1 ) + 2
>0
0 − 24 + 2
or
−7 λ − 6 < 0
or
λ>−
6
7
L3
(i)
…(ii)
In this case no circle
∴
n = 0 ⇒ n + 1 =1
(B)
In this case no circle
∴
n = 0 ⇒ 2n + 3 = 3
(C)
…(iii)
From Eqs. (i), (ii) and (iii), we get
6
3
− <λ<
7
2
Integral values of λ are 0 and 1.
Hence, number of integral values of λ is 2.
In this case two circle which are touching all three lines
∴
n =2 ⇒ n + 2 = 4
(D)
69. Lines
(2a + b ) x + (a + 3b )y + b − 3a = 0
or
a (2 x + y − 3 ) + b ( x + 3y + 1 ) = 0
are concurrent at the point of intersection of lines
2 x + y − 3 = 0 and x + 3y + 1 = 0 which is (2, − 1 ).
Now, line λx + 2y + 6 = 0 must pass through (2, − 1 ), therefore,
2 λ − 2 + 6 or λ = − 2
∴
| λ| = 2
70. Since, PQ is of fixed length.
1
Area of ∆PQR = | PQ | | RP |sinθ
2
This will be maximum, if sinθ = 1 and RP is maximum.
P (4, 4)
q
(0, 7)R¢
Q 3x+4y+5=0
Since, line y = mx + 7 rotates about ( 0, 7 ), if PR′ is
perpendicular to the line than PR′ is maximum value of PR.
 4 − 0 4
m=−
∴
 =
 4 − 7 3
3m = 4
72. (A) The given lines an concurrent. So,
or
R
Hence,
In this case four circle which are touching all three lines
∴
n = 4⇒ n + 2 =6
1
−2 −6
3
1
−4 = 0
λ
4
λ2
λ2 + 2 λ − 8 = 0
or
λ = 2, − 4
∴
| λ | = 2, 4
(B) Given family is
3 x(a + 1 ) + 4y (a − 1 ) − 3(a − 1 ) = 0
or
a(3 x + 4y − 3 ) + (3 x − 4y + 3 ) = 0
for fixed point=
3 x + 4y − 3 = 0
and
3 x − 4y + 3 = 0
3
∴
x = 0, y =
4
 3
Fixed point is  0,  ,
 4
Chap 02 The Straight Lines
−4
1
< λ < or −8 < 6 λ < 3
3
2
Integral values of 6λ are
−7, −6, −5, −4, −3, −2, −1, 0, 1, 2
∴
|6λ | = 7, 6, 5, 4, 3, 2, 1, 0
(D)Q A ≡ ( −2, 4 )
7
The points on the line y = , whose x-coordinates lies between
2
7
(put y = in 4y + x − 14 = 0)
0
2
7
−11
and
(put y = in y + 3 x + 2 = 0)
2
6
−11
<λ<0
∴
6
or
−11 < 6 λ < 0
Integral value of 6λ are
−10, − 9, − 8, − 7, −6, −3, − 2, − 1
∴
|6 λ | = 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
3
4
4| λ | = 4| p − q | = 3
∴
p = 0, q =
Here
∴
(C) The point of intersection of x − y + 1 = 0 and
3 x + y − 5 = 0 is (1, 2). It lies on the line
λ
x + y − 1 −  = 0
2
λ
1 + 2 − 1 −  = 0
⇒
2
or
| λ | = 4 or λ = − 4, 4
∴
λ + 1 = − 3, 5 or | λ + 1| = 3, 5
(D) The mid-point of (1, − 2 ) and (3, 4) will satisfy
y − x −1 + λ = 0
or
1 −2 −1 + λ = 0
∴
λ = 2 or | λ | = 2
y+
73.
3x+
Y
74. (A)Q max. {| x|,| y | } = 1
0
2=
x=1
A
(0, 7/2)
E
D
B
X¢
0
5=
(0,5/3)
O
C
y=2
4 y +x
–1 4 =
179
If | x| = 1 and if | y | = 1
then x = ± 1 and y = ± 1
0
Y
y=1
X
2x–
3 y–
X¢
x=–1
Y¢
(A) The points on the line x = 0, whose y-coordinate lies
5
7
between and inside the triangle ABC.
3
2
5
7
or 5 < 3 λ < 10.5
∴
<λ<
3
2
∴
|3 λ | = 6, 7, 8, 9, 10
(B)QC ≡ (2, 3 )
The points on the line x = 1, whose y-coordinate lies between
8
( put x = 1 in 3y − 2 x − 5 = 0)
3
13
and
(put x = 1 in 4y + x − 14 = 0)
4
8
13
or 8 < 3 λ < 9.75
<λ<
∴
3
4
∴
|3 λ | = 9
(C)Q B ≡ ( −1, 1 )
The point on the line y = 2, whose x-coordinate lies between
−4
(put y = 2 in y + 3 x + 2 = 0)
3
1
and
(put y = 2 in 3y − 2 x − 5 = 0)
2
X
x=1
O
y=–1
Y¢
∴ Required area = 2 × 2 = 4 sq units
(B) The line y = x cuts the lines | x + y | = 6
i.e.
x + y = ±6
at
x = ± 3, y = ± 3
or ( −3, − 3 ) and (3, 3)
then
−3 < a < 3
∴
0 ≤ | a| < 3
∴
[| a| ] = 0, 1, 2
(C) Since (0, 0) and (1, 1) lie on the same side.
So,
a 2 + ab + 1 > 0
Q Coefficient of a 2 is > 0
∴
D<0
b 2 − 4 < 0 or −2 < b < 2
⇒
b = − 1, 0, 1
∴ Number of values of b is 3.
180
Textbook of Coordinate Geometry
75. (A) Q d (x, y ) = 2| x| + 3| y | = 6
(given)
(C)∴ Slope of AC × slope of BC = − 1
Y
| x| | y |
+
=1
3
2
∴
C(l, m)
Y
(0, 6) B
B (0, 2)
13
13
X′
C
(–3, 0)
A
X
(3, 0)
Q
A(6, 0)
X
13
13
D (0, –2)
Y′
∴Perimeter, λ = 4 13
1
and area,
µ = 4 × × 3 × 2 = 12
2
2
λ
then
−µ =1
16
and
λ2 − µ 2 = 64
 µ − 0  µ − 6 

 ×
 = −1
 λ − 6  λ − 0
⇒
µ 2 − 6µ = − λ2 + 6 λ
or
λ2 + µ 2 − 6 λ − 6µ = 0
76. Q
x(a + 2b ) + y (a + 3b ) = a + b
⇒
a ( x + y − 1 ) + b (2 x + 3y − 1 ) = 0
then
x + y − 1 = 0 and 2 x + 3y − 1 = 0
∴ point of intersection is (2, − 1 )
Hence, both statement are true and statement II is correct
explanation for statement I.
x 2 − y 2 = 64
(B) It is clear that orthocentre is (6, 6)
O′ ≡ (6, 6 ),
Circumcentre is C′ ≡ (3, 3 ) and centroid is G′ ≡ ( 4, 4 )
77. Q Algebraic perpendicular from (3, 2) to the line
3 x − 2y + 1 = 0 is
9−4+1
6
i.e.
= p1
(9 + 4 )
13
(say)
and algebraic perpendicular distance from (1, 4) to the line
3 x − 2y + 1 = 0 is
3 −8 + 1
−4
i.e.
(say)
= p2
9+4
13
Y
(0, 6) C
⇒
Hence, locus of ( λ , µ ) is
x 2 + y 2 − 6 x − 6y = 0
Hence, locus of ( λ , µ ) are
x 2 − 16y = 16
and
O
B (6, 6)
6
−4
−24
×
=
<0
13
13
13
Hence, both statements are true and statement II is a correct
explanation for statement I.
∴
O
A (6, 0)
λ = O′ C ′ = ( 0 − 3 ) 2 + (6 − 3 ) 2
∴
= 9 + 9 =3 2
µ = C ′ G′ = ( 4 − 3 ) 2 + ( 4 − 3 ) 2
and
= 1+1= 2
∴
λ − µ = 16 and λ = 3µ
2
2
Hence, locus of ( λ , µ ) are
x 2 − y 2 = 16 and x = 3y
X
p1p 2 =
78. Sum of algebraic distances from points A(1, 2), B(2, 3), C(6, 1) to
the line ax + by + c = 0 is zero (given), then
a + 2b + c (2a + 3b + c ) (6a + b + c )
+
+
=0
(a 2 + b 2 )
(a 2 + b 2 )
(a 2 + b 2 )
⇒
9a + 6b + 3c = 0
or
3a + 2b + c = 0
∴ Statement I is false.
 1 + 2 + 6 2 + 3 + 1
Also, centroid of ∆ABC is 
,



3
3
i.e. (3, 2)
∴ Statement II is true.
Chap 02 The Straight Lines
79. Equation of AB is
y −1 =
Also, (2a − 5, a 2 ) and ( 0, 0 ) on the same side of x + y − 3 = 0,
then
2a − 5 + a 2 − 3
>0
0 + 0 −3
0 −1
( x − 0 ) ⇒ x + 2y − 2 = 0
2−0
Q
| PA − PB| ≤ | AB|
⇒| PA − PB| to be maximum, then A, B and P must be
collinear.
Solving
x + 2y − 2 = 0
and
4 x + 3y + 9 = 0,
 24 17
we get,
P ≡ , 
5 5
Hence, Statement I is false and Statement II is obviously true.
80. Statement II is false as the point satisfying such a property can
be the excentre of the triangle.
 π
 π
Let
L1 ≡ x cos  + y sin   − π = 0,
 9
 9
 8π 
 8π 
L2 ≡ x cos  + y sin   − π = 0 and
 9
 9
⇒
84.
a 2 + 2a − 8 < 0
or
(a + 4 )(a − 2 ) < 0
∴
a ∈ ( −4, 2 )
⇒ Statement I is false
Hence, statement I is false and statement II is true.
BD
R
In ∆OBD,
=
sin ( π − 2 C ) sin θ
DC
R
In ∆ODC,
=
sin ( π − 2 B )
sin ( π − θ )
BD sin 2 C
From Eqs. (i) and (ii),
=
DC sin 2 B
...(i)
...(ii)
(x1,y1)A
 13 π 
 8π 
L3 ≡ x cos
 + y sin   − π = 0
 9 
 9
and
P ≡ ( 0, 0 )
Length ⊥ from P to L1 = Length of ⊥ from P to L2 = Length of ⊥
from P to L3 = π and P lies inside the triangle.
∴P( 0, 0 ) is incentre of triangle.
Hence, statement I is true and statement II is false.
181
h
R
2C
R
(x2,y2)B
2B
R
O
π–2C π–2B
θ π– θ
D
C(x3,y3)
81. Q Mid-point of (5, 1) and (−1, − 5) i.e. (2, − 2) lies on x + y = 0
and (slope of x + y = 0) × (slope of line joining (5, 1)
−6
and (−1, − 5)) = ( −1 ) ×
−6
∴ Statement I is true.
Statement II is also true.
Hence, both statements are true but statement II is not correct
explanation of statement I.
82. Equation of AC and BC are 3x + 2y = 0 and 2x + 3y + 6 = 0
Q(3 )(2 ) + (2 )(3 ) = 12 > 0
∴Internal angle bisector of C is
 2 x + 3y + 6
 3 x + 2y 

 = −





13
13
or
5 x + 5y + 6 = 0
⇒ Statement I is true.
Also, the image of A about the angle bisectors of angle B and C
lie on the side BC. (by congruence).
∴ Statement II is true.
Both statements are true and statement II is not correct
explanation of statement I.
83. Q Points (x1, y1 ) and (x 2, y 2 ) lie on the same or opposite sides of
∴ Coordinates of D are
 x 2 sin 2 B + x 3 sin 2 C y 2 sin 2 B + y 3 sin 2 C 
,


 sin 2 B + sin 2 C
sin 2 B + sin 2C 
Let ( x, y ) be any point on AD, then equation of AD is



1
x
y

1= 0
x1
y1


 x 2 sin 2 B + x 3 sin 2 C y 2 sin 2 B + y 3 sin 2 C 1
sin 2 B + sin 2 C
 sin 2 B + sin 2 C

y
x

or 
y1
x1

x 2 sin 2 B + x 3 sin 2 C y 2 sin 2 B + y 3 sin 2 C

= 0
1

sin 2 B + sin 2 C 
1
1 
y
x


or  x1
1 
y1

x 2 sin 2 B y 2 sin 2 B sin 2 B
1 
y
x
1 
y1
x1
= 0
sin
sin
sin
2
y
C
C
2
2
x
C

3
3

+

the line
ax + by + c = 0, as
ax1 + by1 + c
> 0 or < 0
ax 2 + by 2 + c
∴ Statement II is true.
x y
or (sin 2 B ) 
x1 y1
x 2 y 2
1
 x y 1


1 + (sin 2 C ) 
x1 y1 1 = 0
1
x 3 y 3 1
182
Textbook of Coordinate Geometry
85. Let (x1, y1 ) be the coordinates of a point at unit distance from
each of the given lines.
|3 x1 − 4y1 + 1|
|8 x1 + 6y1 + 1|
⇒
= 1 and
=1
32 + 42
82 + 62
∴
⇒
∴
⇒
⇒
y
cos α =
and
Substituting the values of sin α and cos α from Eq. (i) in (i)
then we get the required locus of P
x
y
∴
+
=k
y / (x 2 + y 2 ) x / (x 2 + y 2 )
⇒
( x 2 + y 2 ) ( x 2 + y 2 ) = kxy
Squaring both sides, we get
( x 2 + y 2 ) 2 ( x 2 + y 2 ) = k 2x 2y 2
(1) ∩ (4)
x1 / − 20 = y1 / − 65 = 1 / 50,
 2 13
( x1, y1 ) =  − , − 
 5 10
(2) ∩ (3)
x1 / 0 = y1 / 75 = 1 / 50, ∴ ( x1, y1 ) = ( 0, 3 / 2 )
(2) ∩ (4)
 8 3
x1 / − 80 = y1 / 15 = 1 / 50, ∴ ( x1, y1 ) =  − , 
 5 10
or
 x2
y2 
(x 2 + y 2 )2  2 2 + 2 2  = k 2
xy 
x y
or
( x 2 + y 2 ) 2 ( x −2 + y −2 ) = k 2
87. Let the equation of variable line be
ax + by = 1. Then the coordinates of Ap will be
 1
p 
Ap ≡ 
,

 a + bp a + bp 
nx
Hence, the required four points have the coordinates
1   2 13  3  8 3 
6
 , −  ,  − , −  ,  0,  ,  − ,  .
5
10  5 10  2  5 10
Y
86. Let ∠OAB = α
∴
Q
...(ii)
(x 2 + y 2 )
3x
⇒
...(i)
...(ii)
...(iii)
...(iv)
y=
or
3 x1 − 4y1 + 1 = ± 5 and 8 x1 + 6y1 + 1 = ± 10
3 x1 − 4y1 − 4 = 0
3 x1 − 4y1 + 6 = 0
8 x1 + 6y1 − 9 = 0
8 x1 + 6y1 + 11 = 0
(1) ∩ (3)
x1 / 60 = y1 / − 5 = 1 / 50,
1
6
( x1, y1 ) =  , − 
 5 10
y=
⇒
⇒
or
∴ Equation of OP is y = x tan (90 ° − α )
y
or
cot α =
x
x
∴
sin α =
2
(x + y 2 )
y=
2x
y=
An
OA = AB cos α and OB = AB sin α
(OA ) 2 + (OB ) 2 = k 2
A3
A2
A1
by
+
ax
Y
x
=
X
1
X′
B
O
P
Y′
X′
90°– α
α
A
O
X
OAp =
∴
Y′
i.e.
n
( AB ) 2 (cos2 α + sin 2 α ) = k 2
AB = k
OA = k cos α and OB = k sin α
x
y
+
=1
∴ Equation of AB is
k cos α k sin α
x
y
or
+
=k
cos α sin α
or
then
Let P be the foot of perpendicular from O on AB.
(x 2+y 2)
x
α
y
Σ
Given,
p =1
...(i)

⇒ a ±





or a  ±



| a + bp |
n
(1 + p 2 )
p =1
n
Σ
p =1
n
Σ
p =1
...(i)
1
=c
OAp
Σ
⇒
(1 + p 2 )
| a + bp |

 +b
(1 + p ) 
1
2
=c

±




2
(1 + p ) 
 +b
c



1
[from Eq. (1)]
n



±




 =c
(1 + p ) 
p
Σ
p =1
n
Σ
p =1
2


(1 + p ) 
 =1
c



p
2
Chap 02 The Straight Lines
So, line always passes through a fixed point whose coordinates
are


n
n


p
1
± Σ
 ± p Σ= 1
2
2 
p
=
1
(1 + p )
(1 + p ) 

,


c
c


88. Let the equation of given n lines be
...(i)
...(ii)
Y′
y = mx
y = mnx+cn
R3
y = m2x+c2
y = m1x+c1
X′
Solving Eqs. (i) and (ii) , we get
 cr
mcr 
Rr ≡ 
,

 m − mr m − mr 
2
∴
y1 = mx1 ⇒ m =
Given,
⇒
α − 1 β − 1 − 2 (1 + 1 − 2 )
=
=
= 2 −2
1
1
12 + 12
∴
α = 2 − 1, β = 2 − 1
i.e.
Q ≡ ( 2 − 1, 2 − 1 )
2
90. Let R (h, k ) be the foot of perpendicular from Q on OP.
...(iii)
Let equation of OP be
Y
y1
x1
...(iv)
P(a,ma)
mx
y=
n
n
1
= Σ
OR r = 1 OR r
n
( x12
+
y12 )
=
R
m − mr
1

cr  (1 + m 2 )
Σ 
n
r = 1
[from Eq . (iii)]
n
  n  1 
 m 
m  Σ ±  + Σ m r  

2
r
=
1
r
=
1


 cr  
c


r
(1 + m ) 
1
=
(ma + b )
(1 + m 2 )
=
...(i)
Hence, equation of required family is
(2 cos θ + 3 sin θ ) x + (3 cos θ − 5 sin θ ) y
= ( 2 − 1 ) (5 cos θ − 2 sin θ ).
 c 
=  r  (1 + m 2 )
m − mr
R ≡ ( x1, y1 )
(2 x + 3y − 5 ) cos θ + (3 x − 5y + 2 ) sin θ = 0
⇒
(2 x + 3y − 5 ) + (3 x − 5y + 2 ) tan θ = 0
It is clear that these lines will pass through the point of
intersection of the lines
2 x + 3y − 5 = 0 

3 x − 5y + 2 = 0 
λ = ( 2 − 1 ) (5 cos θ − 2 sin θ )
 cr 
 m cr 
OR r = 
 +

 m − mr 
 m − mr 
Let
[from Eq. (iv)]
2
If the required family of lines is
(2 cos θ + 3 sin θ ) x + (3 cos θ − 5 sin θ ) y = λ
in order that each member of the family pass through Q, we
have
λ = ( 2 − 1 ) (2 cos θ + 3 sin θ + 3 cos θ − 5 sin θ )
X
O
Y′
∴
y 
1 +  1
 x1 
⇒
n = ay1 + bx1
Hence, locus of point R is bx + ay = n.
Then
y = m3x+c3
R
=
for all values of θ.
Solving the system of Eq. (i), we get (1, 1).
Hence, the fixed point is P (1, 1 ). Let Q (α , β ) be the reflection of
P (1, 1 ) in the line x + y = 2.
Rn
R1
( x12 + y12 )
y1
a+b
x1
89. First equation can be expressed as
y = m r x + cr ,
where
r = 1, 2, 3, .... ,n
Let equation of line through origin O is
y = mx
R2
then
n
183
X′

 mr 
 1
where a = Σ  ±  and b = Σ  m 
r = 1  cr 
r =1 
cr 

n
x–a=0
θ
O
Q
A(a,0)
X
Y′
1
n
x x
y = mx
k = mh
k
or
m=
h
and coordinates of P ≡ (a , ma )
then
...(i)
184
Textbook of Coordinate Geometry

3β 
Similarly, the coordinates of C are  0,

 3 − α
Q PQ is the bisector of OPA
∴
∠APQ = ∠RPQ
and
∠PAQ = ∠QRP = 90 °
∴
PA = PR
then
| ma | = (h − a ) 2 + (k − ma ) 2
From Eq. (i),
ak = (h − a ) 2 + k − ak 

h
h 
⇒
a | k | = |(h − a )| (h 2 + k 2 )
Now, the equation of AD is
x (6 − α )
+
y =1
3
6β
and the equation of OU is
βx = αy
Solving Eqs. (i) and (ii), we get
6α
6β
x=
,y =
6+α
6+α
2
Hence, required locus is
( x − a ) 2 ( x 2 + y 2 ) = a 2y 2
91. Let the coordinates of the vertex be (h, k ) and equations of the
bases be
x cos α r + y sin α r − pr = 0 where r = 1 , 2 , 3 , ...., n
and their lengths be respectively l1, l 2, l 3, ...., ln .
Q Length of perpendicular from (h, k ) on
x cos α r + y sin α r − pr = 0 is
| h cos α r + k sin α r − pr |
,
(cos2 α + sin 2 α )
i.e.
| h cos α r + k sin α r − pr |
Given, sum of areas of all triangles = constant
then
n
1
Σ lr ⋅ | h cos αr + k sin αr − pr | = C ′
r =1 2
n
1
Σ . lr . (± (h cos αr + k sin αr − pr )) = C ′
⇒
r =1 2

 n

 n
1
1
⇒ h  Σ ± lr cos α r  + k  Σ ± lr . sin α r 

r = 1 2

r = 1 2
=
n
...(ii)
 6α
6β 
Hence, coordinates of V are 
,

6 + α 6 + α
Then, the equation of CV is
⇒
y −
3β
3 −α
y −
3β
3 −α
⇒
6β
3β
−
6 + α 3 −α
(x − 0)
=
6α
−0
6+α
− 9α β
=
x
6 α (3 − α )
y =
3β 
1 −
(3 − α ) 
x

2
which pass through the point (2, 0) for all values of (α , β ).
93. Let the equation of the variable line through `O' be
x
y
=
cos θ sin θ
and let OR = r1, OS = r2 and OP = r3
L2
Y
1
2
Σ ± lr . pr + C ′
r =1
S
⇒ Ah + Bk = − C
∴ Required locus is
Ax + By + C = 0
where A, B, C are constants.
R
by
=
1
C
U(α,β)
V
Y′
X
+
ax
Y
(2,0) A(3,0)
L1
Then coordinates of R , S and P are :
R (r1 cos θ, r1 sin θ ), S (r2 cos θ, r2 sin θ ), P (r3 cos θ, r3 sin θ )
R lies on L1 and S lies on L2.
Let
L1 ≡ y − c = 0
and
L2 ≡ ax + by − 1 = 0
∴
r1 sin θ = c and ar2 cos θ + br2 sin θ = 1
c
1
and r2 =
∴
r1 =
sin θ
a cos θ + b sin θ

6β 
So that the coordinates of D are  0,

 6 − α
O
O
Y′
0 −β
y −β =
(x − α )
6 −α
D
y=c
P(h,k)
X′
92. The equation of BU is
X′
...(i)
B(6,0)
X
From the given condition
m+n m n
= +
r3
r1 r2
m + n m sin θ
⇒
=
+ n (a cos θ + b sin θ )
r3
c
...(i)
Chap 02 The Straight Lines
Let the coordinates of P be (h, k ), then
h = r3 cos θ, k = r3 sin θ
mr sin θ
From Eq. (i), m + n = 3
+ n (ar3 cos θ + br3 sin θ )
c
mk
m+n=
+ n (ah + bk )
⇒
c
my
Locus of P is n (ax + by ) +
= (m + n )
c
m
n (ax + by − 1 ) + (y − c ) = 0
⇒
c
m
(ax + by − 1 ) +
(y − c ) = 0
⇒
nc
m

⇒
L2 + λL1 = 0  where, λ = 

nc 
Hence, locus of P is a point of intersection of L1 and L2.
94. The given lines are
...(i)
x + 3y + 2 = 0
...(ii)
2x + y + 4 = 0
...(iii)
x −y −5 = 0
Equation of the line passing through A ( − 5 , − 4 ) and making
an angle θ with the positive direction of X-axis is
x+5 y +4
...(iv)
=
= r ( AB, AC , AD )
cos θ sin θ
∴ Points ( − 5 + AB cos θ, − 4 + AB sin θ ),
( − 5 + AC cos θ, − 4 + AC sin θ ) and
( − 5 + AD cos θ, − 4 + AD sin θ ) lie on Eqs. (i), (ii) and (iii)
respectively.
( − 5 + AB cos θ ) + 3 ( − 4 + AB sin θ ) + 2 = 0
⇒
AB (cos θ + 3 sin θ ) = 15
15
= cos θ + 3 sin θ
⇒
AB
10
Similarly,
= 2 cos θ + sin θ
AC
6
and
= cos θ − sin θ
AD
From given condition
2
2
 15 
 10 
 6 

 +
 =

 AB 
 AC 
 AD 
⇒
2
4 cos θ + 9 sin θ + 12 sin θ cos θ = 0
2
(2 cos θ + 3 sin θ ) 2 = 0
2
tan θ = −
∴
3
Hence the equation of the line from Eq. (iv) is
2
y + 4 = − ( x + 5 ) ⇒ 2 x + 3y + 22 = 0
3
∴
y=
P
mx
B(b,mb)
R(h,k)
X′
O
(b,0)Q A(a,0)
X
Y′
Q P be the foot of perpendicular from A on y = mx, then
α − a β − 0 − ( 0 − ma )
=
=
−m
1
(1 + m 2 )
a
am
∴
α=
, β=
2
1+m
1 + m2
i.e.
 a
am 
P ≡
,

2
 1 + m 1 + m2 
∴ Equation of PQ is
am
−0
1 + m2
(x − b )
y −0=
a
−
b
1 + m2
a
...(ii)
⇒
(y − mx ) + am − (1 + m 2 ) y = 0
b
Adding Eqs. (i) and (ii), then
a
(y − mx + k ) + (mh − k − (1 + m 2 ) y ) = 0
b
⇒
(mh − k − (1 + m 2 ) y ) + λ (y − mx + k ) = 0
 h + mk mh − k 
Hence, fixed point is 
,
.
 1 + m2 1 + m2 
96. Given lines are parallel and distance between them < 2
95. Q A, R and B are collinear
k − 0 mb − 0
=
h −a
b −a
a
k − am + mh − k = 0
b
Y
Hence PQ pass through a fixed point.
For fixed point
mh − k − (1 + m 2 ) y = 0, y − mx + k = 0
h + mk
mh − k
,x=
y =
(1 + m 2 )
(1 + m 2 )
⇒
then,
Let P ≡ (α , β )
a

 where, λ = 

b
we get (cos θ + 3 sin θ ) 2 + (2 cos θ + sin θ ) 2 = (cos θ − sin θ ) 2
2
185
...(i)
Given lines are
...(i)
2x + y = 3
and
...(ii)
2x + y = 5
Equation of any line through Eqs. (ii) and (iii) is
y − 3 = m (x − 2)
or
...(iii)
y = mx − 2m + 3
Let line (iii) cut lines (i) and (ii) at A and B respectively.
Solving Eqs. (i) and (iii), we get
 2m 6 − m 
A≡
,

 m + 2 m + 2
186
Textbook of Coordinate Geometry
and solving Eqs. (ii) and (iii), we get
 2m + 2 m + 6
B≡
,

 m + 2 m + 2
Also
tanθ = − 2
Now, equation of required lines are
y − 3 = tan (θ ± α ) ( x − 2 )
According to question AB = 2
⇒
( AB ) 2 = 4
⇒
 tan θ ± tan α 
y −3 = 
 (x − 2)
 1 m tan θ tan α 
⇒
1
2 (x −2)
y −3 =
 1
1 m ( −2 )  
 2
2
2
⇒
 2 
 2m 

 +
 =4
 m + 2
 m + 2
⇒
1 + m 2 = m 2 + 4m + 4
...(iv)
Case I : When m is finite (line is not perpendicular to X-axis)
then from Eq. (iv).
1 = 4m + 4
3
m=−
∴
4
Case II : When m is infinite (line is perpendicular to X-axis)
then from Eq. (iv),
1
4
4
+ 1 =1 +
+
m m2
m2
0 + 1 =1 + 0 + 0
1 = 1 which is true
Hence m → ∞ acceptable.
Hence, equation of the required lines are
3
y − 3 = − (x − 2)
4
y −3
and
= x −2 ⇒x −2 = 0
∞
i.e.
3 x + 4y = 18 and x − 2 = 0
Aliter I :
Q 2 x + y = 3 cuts Y-axis at ( 0, 3 ) and line 2 x + y = 5 cuts
Y-axis at ( 0, 5 )
X′
Y′
α
B
then
AM 1
=
MB 2
i.e.
3
(x − 2)
2
x − 2 = 0 and 3 x + 4y − 18 = 0
x − 2 = 0 and 2y − 6 = −
 θ
 θ
 θ
2 − 2 tan 2   + 2 tan   = − 1 − tan 2  
 2
 2
 2
⇒
 θ
 θ
tan 2   − 2 tan   − 3 = 0
 2
 2
∴
 θ
tan   = − 1 or 3
 2
2
X
2x+y=5
2x+y=3
MB = ( AB ) 2 − ( AM ) 2 = 4 −
tanα =
⇒
⇒
Therefore intercept on Y-axis is 2.
Also,
AM = distance between parallel lines
| −5 + 3|
2
=
=
2
2
5
2 +1
∴
1

⇒ (1 m ( −1 )) (y − 3 ) =  −2 ±  ( x − 2 )

2
A
M
O
1

 −2 ± 

2
y −3 =
(x − 2)
1 m ( −1 )
P(2,3)
2
α
D
( −2 ) ±
Aliter II : Any line through (2 , 3 ) is
x −2 y −3
=
=r
cosθ sin θ
Suppose this line cuts 2 x + y = 5 and 2 x + y = 3 at D and C
respectively but given DC = 2
then
D ≡ (2 + r cosθ, 3 + r sin θ )
and
C ≡ (2 + (r + 2 ) cosθ, 3 + (r + 2 ) sin θ )
Q D and C lies on
2 x + y = 5 and 2 x + y = 3
then
... (v)
2 (2 + r cosθ ) + (3 + r sin θ ) = 5
and
2 (2 + (r + 2 ) cosθ ) + (3 + (r + 2 ) sin θ ) = 3 ... (vi)
Subtracting Eq. (v) from Eq. (vi), then
4 cosθ + 2 sin θ = − 2
or
2 cosθ + sin θ = − 1


 θ 
2  θ 
 1 − tan     2 tan   


 2
2
 +
 = −1
2
⇒
θ





2  θ 
2
 1 + tan     1 + tan   
2
2

Y
C
⇒
(slope of 2 x + y = 5)
4
4
=
5
5
∴
tanθ = ∞ or −
∴ Required lines are
y − 3 = ∞ (x − 2)
3
and
y − 3 = − (x − 2)
4
i.e.
x −2 = 0
and 3 x + 4y − 18 = 0
3
4
Chap 02 The Straight Lines
97. If I be the incentre of ∆OAB.
99. The line passing through the intersection of lines
ax + 2by + 3b = 0 and bx − 2ay − 3a = 0 is
ax + 2by + 3b + λ(bx − 2ay − 3a ) = 0
⇒(a + bλ ) x + (2b − 2aλ )y + 3b − 3 λa = 0
As this line is parallel to X-axis.
a
∴
a + bλ = 0 ⇒ λ = −
b
a
⇒ ax + 2by + 3a − (bx − 2ay − 3a ) = 0
b
2a 2
3a 2
⇒ ax + 2by + 3b − ax +
y +
=0
b
b

2a 2 
3a 2
y 2b +
=0
 + 3b +
b 
b

If inradius =r
then
ID = IE = IF = r
Y
1
B 1,
√3
F r r E
I
X′
15°
15°
O
r
D
15°
15°
A(2,0)
X
Y′
If P at I,then
d ( P , OA ) = d ( P , OB ) = d ( P , AB ) = r
But
d ( P , OA ) ≤ min{d ( P , OB ), d ( P , AB )}
which is possible only when P lies in the ∆OIA.
ID r
∴
tan15 ° =
=
OD 1
r = (2 − 3 )
⇒
1
∴ Required area = ⋅ 2 ⋅ r = r = (2 − 3 ) sq units.
2
 2b 2 + 2a 2 
 3b 2 + 3a 2 
y
 = −

b
b




−3(a 2 + b 2 ) −3
=
2
2(b 2 + a 2 )
y =
So, it is
3
units below X-axis.
2
100.
Y
98. Let A ≡ (x1, y1 ), B ≡ (x 2, y 2 ) and C ≡ (x 3, y 3 ) are the vertices of
P(0, b)
a triangle ABC and P ≡ (a1, b1 ), Q ≡ (a 2, b2 ) and R ≡ (a 3, b3 ) are
the vertices of triangle PQR.
Equation of perpendicular from A to QR is
(a − a 3 )
y − y1 = − 2
( x − x1 )
(b2 − b3 )
or (a 2 − a 3 ) x + (b2 − b3 ) y − x1 (a 2 − a 3 ) − y1(b2 − b3 ) = 0 …(i)
Similarly, equations of perpendiculars from B to RP and C to
PQ are respectively,
(a 3 − a1 ) x + (b3 − b1 ) y − x 2 (a 3 − a1 ) − y 2 (b3 − b1 ) = 0 ...(ii)
and (a1 − a 2 ) x + (b1 − b2 ) y − x 3 (a1 − a 2 ) − y 3 (b1 − b2 ) = 0 ...(iii)
Given that lines (i), (ii) and (iii) are concurrent, then adding,
we get
( x 2 − x 3 ) a1 + ( x 3 − x1 ) a 2 + ( x1 − x 2 ) a 3 + (y 2 − y 3 ) b1 +
...(iv)
(y 3 − y1 )b2 + (y1 − y 2 ) b3 = 0
Now, equation of perpendicular from P to BC is
(x − x 3 )
y − b1 = − 2
( x − a1 )
(y 2 − y 3 )
( x 2 − x 3 ) x + (y 2 − y 3 ) y − a1
( x 2 − x 3 ) − b1(y 2 − y 3 ) = 0 ...(v)
Similarly, equations of perpendiculars from Q to CA and R to
AB are respectively,
( x 3 − x1 ) x + (y 3 − y1 ) y − a 2
...(vi)
( x 3 − x1 ) − b2 (y 3 − y1 ) = 0
and
( x1 − x 2 ) x + (y1 − y 2 ) y − a 3
...(vii)
( x1 − x 2 ) − b3 (y1 − y 2 ) = 0
Adding Eqs. (v), (vi) and (vii), we get
LHS = 0 (identically)
[ from Eq. (iv)]
Hence perpendiculars from P to BC, Q to CA and R to AB are
concurrent.
A(3, 4)
Q(a, 0)
O
X
Q A is the mid-point of PQ, therefore
a+0
0+b
= 3,
=4
2
2
⇒
a = 6, b = 8
x y
∴Equation of line is + = 1
6 8
or
4 x + 3y = 24
101. Clearly for point P,
y
y=3x
or
P(a, a2)
y= x
2
O
a 2 − 3a < 0 and a 2 −
⇒
1
<a <3
2
x
a
>0
2
187
188
Textbook of Coordinate Geometry
102. Point of intersection of L1 and L2 is A(0, 0).
Also P( −2, − 2 ), Q(1, − 2 )
A
(0, 0)
:2
:y
–x
=0
L2
L
1
0
y=
x–
R
L3 : y+2=0
P
(– 2, –2)
Q
(1, –2)
Q AR is the bisector of ∠PAQ, therefore R divides PQ in the
same ratio as AP : AQ.
Thus PR : RQ = AP : AQ = 2 2 : 5
∴ Statement I is true.
Statement II is clearly false.
103. Given : The coordinates of points P , Q, R are (−1, 0), (0, 0),
(3, 3 3 ) respectively.
Y
(C) Lines are not concurrent or not parallel, then
6
k ≠ 5, k ≠ − 9, k ≠ −
5
5
∴
k=
6
(D) The given lines do not form a triangle if they are
concurrent or any two of them are parallel.
6
∴
k = 5, k = − 9, k = −
5
3−4
−1
105. Slope of PQ =
=
k −1 k −1
∴ Slope of perpendicular bisector of PQ = (k − 1 )
 k + 1 7
Also mid-point of PQ 
, 
 2 2
∴ Equation of perpendicular bisector is
7
k + 1

y − = (k − 1 )  x −


2
2 
⇒
2y − 7 = 2(k − 1 ) x − (k 2 − 1 )
⇒
2 (k − 1 ) x − 2y + (8 − k 2 ) = 0
R (3,3Ö3)
8 −k2
= −4
−2
8 − k 2 = −8 or k 2 = 16 ⇒ k = ± 4
∴ Y -intercept = −
M
⇒
106. If the line p(p 2 + 1)x − y + q = 0
X¢
2p/3
P (–1, 0)
p/3
Q (0, 0)
X
Y¢
y 2 − y1 3 3
=
x 2 − x1
3
π
tanθ = 3 ⇒ θ =
⇒
3
π
⇒
∠RQX =
3
π 2π
∠RQP = π − =
∴
3
3
Let QM bisects the ∠PQR,
2π
=− 3
∴ Slope of the line QM = tan
3
∴ Equation of line QM is (y − 0 ) = − 3( x − 0 )
Slope of equation QR =
⇒
y = − 3x ⇒
3x + y = 0
104. (A)Q L1, L2, L3 are concurrent, then
1
3
3 −k
5 2
−5
−1 = 0 ⇒k = 5
−12
(B) slope of ( L1 ) = slope of ( L2 )
1 3
− = ∴ k = −9
⇒
3 k
and slope of ( L3 ) = slope of ( L2 )
5 3
6
⇒
− =
∴ k=−
2 k
5
and
( p 2 + 1 ) 2 x + ( p 2 + 1 )y + 2q = 0
are perpendicular to a common line, then these lines must be
parallel to each other,
p( p 2 + 1 )
(p 2 + 1)2
∴
m1 = m2 ⇒ −
=− 2
−1
p +1
⇒
( p 2 + 1 )( p + 1 ) = 0
⇒
p = −1
∴p can have exactly one value.
b
107. Slope of line L = −
5
3
Slope of line K = −
c
Line L is parallel to line K .
b 3
⇒
= ⇒ bc = 15
5 c
(13, 32) is a point on L.
13 32
32
8
∴
+
=1 ⇒
=−
b
5
b
5
3
⇒
b = − 20 ⇒ c = −
4
Equation of K : y − 4 x = 3
⇒
4x − y + 3 = 0
|52 − 32 + 3|
Distance between L and K =
17
23
=
17
Chap 02 The Straight Lines
108. Let the slope of line L be m.
111. Suppose B(0, 1) be any point on given line and coordinate of A

= 3
m + 3 
1 − 3m
Then
189
is ( 3, 0 ). So, equation of
Y
B
(0, 1)
Ö3x+y=1
L
(0, 1)
X¢
A (3, 0)
X
O
60°
(3, –2)
Y¢
B¢ (0, –1)
⇒
m + 3 = ± ( 3 − 3m )
⇒
4m = 0 or 2m = 2 3
⇒
m = 0 or m = 3
Reflected ray is
⇒
Q L intersects X-axis,
∴
m= 3
3y = x − 3
 −c
−c 
,

a + b a + b
112. The intersection point of two lines is 
∴Equation of L is y + 2 = 3( x − 3 )
or
−1 − 0
y −0
=
0− 3 x− 3
 −c
−c 
Distance between (1, 1) and 
,
 <2 2
a + b a + b
3 x − y − (2 + 3 3 ) = 0
109.
L3
=0
y– x
L1
P (–2, –2)
R (–1, –2)
O
(0, 0)
2x+
2
⇒

c 
21 +
 <8

a + b
c
<2
a+b
⇒
1+
⇒
a + b −c > 0
113. Let P , Q, R, be the vertices of ∆PQR
y=
0
P (2, 2)
L2
L1 : y − x = 0, L2 : 2 x + y = 0, L3 : y + 2 = 0
On solving the equation of lines L1 and L2, we get their point of
intersection (0, 0) i.e. origin O.
On solving the equation of lines L1 and L3,
we get
P = ( −2, − 2 )
Similarly, we get
Q = ( −1, − 2 )
We know that bisector of an angle of a triangle, divide the
opposite side the triangle in the ratio of the sides including the
angle [Angle Bisector Theorem of a Triangle]
∴
( −2 ) + ( −2 )
2 2
PR OP
=
=
=
5
RQ OQ
( −1 ) 2 + ( −2 ) 2
2
Let point C divides line AB in the ratio 3 : 2. So, by section
formula we have
 3 × 2 + 2 × 1 3 × 4 + 2 × 1  8 14
C =
,
= , 

 5 5 
3+2
3+2
 8 14
Since Line 2x + y = k passes through C  , 
5 5 
S
R (7, 3)
Since, PS is the median, S is mid-point of QR
2
110. Let the joining points be A(1, 1) and B(2, 4).
∴C satisfies the equation 2x + y = k.
2 + 8 14
⇒
+
= k ⇒k = 6
5
5
Q (6, –1)
So,
 7 + 6 3 − 1  13 
S =
,
 =  , 1
 2
2  2 
2 −1
2
=−
13
9
2−
2
Since, required line is parallel to PS therefore slope of required
line = slope of PS Now, eqn of line passing through (1, − 1 ) and
2
having slope − is
9
2
y − ( −1 ) = − ( x − 1 )
9
9y + 9 = − 2 x + 2
Now, slope of PS =
⇒
2 x + 9y + 7 = 0
Textbook of Coordinate Geometry
116. Total number of integral points inside the square OABC
4ax + 2ay + c = 0
5bx + 2by + d = 0
The point of intersection will be
x
1
−y
=
=
2ad − 2bc 4ad − 5bc 8ab − 10ab
2(ad − bc ) bc − ad
x=
⇒
=
ab
−2ab
5bc − 4ad 4ad − 5bc
⇒
y =
=
−2ab
2ab
Q Point of intersection is in fourth quadrant so x is positive
and y is negative.
Also distance from axes is same
So
x = − y (Q distance from X-axis is −y as y is
negative)
bc − ad 5bc − 4ad
=
⇒ 3bc − 2ad = 0
ab
2ab
= 40 × 40 = 1600
Number of integral points on AC
= Number of integral points on OB
= 40 [namely (1, 1), (2, 2) … (40, 40)]
O
(0, 0)
If x > y , then
2≤
x −y + x + y
≤ 4 or 2 ≤ x ≤ 2 2
2
If x < y , then
y −x+x+y
≤ 4 or 2 ≤ y ≤ 2 2
2
The required region is the shaded region in the figure given
below.
2≤
Y
y=x
y=2 2
y= 2
O
x= 2
x=2 2
=
1600 − 40
= 780
2
117.
D
m=0
For P lying in first quadrant x > 0, y > 0.
Also
2 ≤ d 1( P ) + d 2( P ) ≤ 4
x −y
x+y
2≤
+
≤4
⇒
2
2
A
(41, 0)
X
∴ Required area = (2 2 ) 2 − ( 2 ) 2 = 8 − 2 = 6 sq units
A
x–y+1=0
C
=0
x − y
x + y
and d 2( P ) = 
d 1( P ) = 
 2 
 2 
B
∴ Number of integral points inside the ∆OAC
115. Let the point P be (x, y )
Then
(41, 41)
(0, 41) C
O (–1, –2)
x–y+l=0
7x–y–5
114. Given lines are
7x–y+
190
B
Let other two sides of rhombus are
x −y + λ = 0
and
7x − y + µ = 0
then O is equidistant from AB and DC and from AD and BC
∴
| −1 + 2 + 1 | = | −1 + 2 + λ | ⇒ λ = −3
and
| −7 + 2 − 5 | = | −7 + 2 + µ| ⇒ µ = 15
∴Other two sides are
x −y −3 = 0
and
7 x − y + 15 = 0
On solving the equation of sides pairwise, we get the vertices
 −7 −4 
 1 −8 
as  ,  , (1, 2),  ,  , ( −3, − 6 )
 3 3
3 3 
CHAPTER
03
Pair of
Straight Lines
Learning Part
Session 1
● Introduction
Session 2
2
● Angle between the Pair of Lines ax
+ 2hxy + by 2
●
Homogeneous Equation in Two Variables
Session 3
● Bisectors of the Angle between the Lines Given by a Homogeneous Equation
Session 4
● General Equation of Second Degree
● Important Theorems
Session 5
● To Find the Point of Intersection of Lines Represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 with the Help of Partial Differentiation
●
●
Removal of First Degree Term
Equation of the Lines Joining the Origin to the Points of Intersection of a
Given Line and a Given Curve
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
Session 1
Introduction, Homogeneous Equation in
Two Variables
Introduction
Let the equation of two lines be
...(i)
ax + by + c = 0
and
...(ii)
a 1 x + b 1y + c 1 = 0
Hence, (ax + by + c ) (a 1 x + b 1 y + c 1 ) = 0 is called the joint
equation of lines Eqs. (i) and (ii) and conversely, if joint
equation of two lines be
(ax + by + c ) (a 1 x + b 1 y + c 1 ) = 0,
then their separate equations will be
ax + by + c = 0 and a 1 x + b 1 y + c 1 = 0
Remark
In order to find the joint equation of two lines, make RHS of two
lines equal to zero and then multiply the two equations.
y Example 1 Find the joint equation of lines y = x and
y = − x.
Sol. The given lines can be rewritten as
x − y = 0 and x + y = 0
∴ Joint equation of lines is ( x − y ) ( x + y ) = 0
or
x2 − y2 = 0
Wrong process : Since, the lines are
y = x and y = − x
Then joint equation is
y2 = − x2
⇒
x2 + y2 = 0
This process is wrong, since RHS of two equations are not
equal to zero.
Remark
In order to find the separate equations of two lines when their joint
equation is given, first of all make RHS equal to zero and then
resolve LHS into two linear factors or use Shri Dharacharya method.
The two factors equated to zero will give the separate equations of
lines.
y Example 2 Find the separate equation of lines
represented by the equation x 2 − 6 xy + 8 y 2 = 0.
Sol. Separate equation of lines represented by the equation
x 2 − 6xy + 8y 2 = 0
i.e.
( x − 4y ) ( x − 2y ) = 0
⇒
x − 4y = 0 and x − 2y = 0
Aliter :
We have,
x 2 − 6xy + 8y 2 = 0
6y ± ( − 6y )2 − 4 ⋅ 8y 2
2
x = 3y ± y (9 − 8)
By Shri Dharacharya method, x =
⇒
x = 3y ± y
∴
x = 4y and x = 2y
Hence, the lines are
x − 4y = 0 and x − 2y = 0.
Homogeneous Equation in
Two Variables
An equation of the form
a 0 y n + a 1 y n − 1 x + a 2 y n − 2 x 2 + .... + a n x n = 0
…(i)
in which the sum of the powers of x and y in every term
is the same (here n), is called a homogeneous equation
(of degree n).
We will prove that Eq. (i) represents n straight lines
passing through the origin.
a 0 y n + a 1 y n − 1 x + a 2 y n − 2 x 2 + .... + a n x n = 0
Dividing each term by x n , we get
n
y 
y 
a 0   + a1  
x
x
n −1
y 
+ a2  
x
Above is an equation of nth degree in
n −2
+ ... + a n = 0
y
. Let the roots of
x
this equation be m 1 , m 2 , m 3 , ..., m n .
Then, the above equation will be identical with
y

a 0  − m1 
x

y

 − m2 
x

y
 y

 − m 3  ...  − m n  = 0
x
 x

⇒ a 0 (y − m 1 x ) (y − m 2 x ) (y − m 3 x ) .... (y − m n x ) = 0
Chap 03 Pair of Straight Lines
Hence, Eq. (i) represents n straight lines
y − m 1 x = 0, y − m 2 x = 0,
y − m 3 x = 0,..., y − m n x = 0
all of which clearly pass through the origin.
Corollary : Since, ax 2 + 2hxy + by 2 = 0 is a homogeneous
equation of second degree, it represents two straight lines
through origin. The given equation is
ax 2 + 2hxy + by 2 = 0
...(i)
Dividing by x 2 , we get
2
y 
y 
b   + 2h   + a = 0
x
x
...(ii)
y
=m
x
Putting
...(iii)
[Making coefficient of y 2 equal on both sides]
Now, comparing both sides, we get
2h = − b ( m1 + m2 ) and a = bm1m2
2h
a
and m1m2 =
m1 + m2 = −
∴
b
b
Sol. Let the lines represented by
ax 2 + 2hxy + by 2 = 0 are y = m1x
m1 + m 2 = −
∴
m1 = −
2h
b (1 + n )
…(iii)
2
⇒
− 2h ± 2 (h − ab )
 − 2h 
a
n
 =
b
+
n
b
1
(
)


4nh 2
⇒
b (1 + n )2
2
2
⇒
2b
b
...(ii)
a
b
a
2
nm1 =
b
∴
Thus, y = m 1 x and y = m 2 x are two straight lines which
are given by Eq. (i). Also, from Eq. (iii),
=
...(i)
and from Eq. (ii), m1(nm1 ) =
2
=
(h 2 − ab )
| b|
− h ± (h 2 − ab )
2h
b
a
b
m 2 = nm1
2h
m1 + nm1 = −
b
From Eq. (i),
|m 1 − m 2 | = {(m 1 + m 2 ) 2 − 4m 1m 2 }
m=
and y = m 2 x .
m1m 2 =
and
Given,
If m 1 and m 2 be two roots, then
coefficient of xy
2h
m1 + m2 = −
=−
b
coefficient of y 2
a coefficient of x 2
and
m 1m 2 = =
b coefficient of y 2
∴
ax 2 + 2hxy + by 2 = b( y − m1x ) ( y − m2x )
then
Therefore,
bm 2 + 2hm + a = 0
then,
pass through the origin, let their equations be
y = m1x and y = m2x
then, ( y − m1x ) and ( y − m2x )
must be factors of ax 2 + 2hxy + by 2 = 0
y Example 3 Find the condition that the slope of one
of the lines represented by ax 2 + 2hxy + by 2 = 0 should
be n times the slope of the other.
2
y 
y 
a + 2h   + b   = 0
x
x
⇒
193
y
=
x
∴
by = {− h + (h 2 − ab ) }x
and
by = {− h − (h 2 − ab ) } x
y

Q m = x 


are two lines represented by Eq. (i).
(i) The lines are real and distinct, if h 2 − ab > 0.
(ii) The lines are coincident, if h 2 − ab = 0.
(iii) The lines are imaginary, if h 2 − ab < 0.
Remarks
1. In further discussions, we will consider only real cases.
2. Two very useful identities When lines represented by
ax 2 + 2hxy + by 2 = 0
=
[from Eq. (iii)]
a
b
4nh 2 = ab(1 + n )2
...(iv)
This is the required condition.
Corollary : If slope of one line is double of the other, then
put n = 2 in Eq. (iv), we have
8h 2 = 9ab.
y Example 4 If the slope of one of the lines
represented by ax 2 + 2hxy + by 2 = 0 be the nth power
1
n n +1
of the other, prove that, (ab )
+ (a b )
n
1
n +1
+ 2h = 0.
n
Sol. Let m and m be the slopes of the lines represented by
ax 2 + 2hxy + by 2 = 0
then
m + mn = −
2h
b
...(i)
194
Textbook of Coordinate Geometry
m ⋅ mn =
a
a
⇒ mn + 1 =
b
b
1
 a  (n + 1)
m= 
b
⇒
...(ii)
Substituting the value of m from Eq. (ii) in Eq. (i), then
 a
 
b
a
1
n +1
⋅b
+ 
b
n
n +1
+a
n
n +1
⋅b
2h
b
=−
1
n +1
+ 2h = 0
Corollary : If slope of one line is square of the other, then
put n = 2, then
1
1
(ab 2 ) 3 + (a 2b ) 3 = − 2h
On cubing both sides, we get
1
1 
1
1
ab 2 + a 2b + 3 (ab 2 ) 3 (a 2b ) 3 (ab 2 ) 3 + (a 2b ) 3  = − 8h 3




⇒
ab (a + b ) + 3ab ( − 2h ) = − 8h 3
(a + b ) 8h 2
+
= 6.
h
ab
∴
y Example 5 Find the product of the perpendiculars
drawn from the point ( x 1 , y 1 ) on the lines
ax 2 + 2hxy + by 2 = 0.
Sol. Let the lines represented by ax 2 + 2hxy + by 2 = 0 be
y = m1x and y = m 2 x .
a
2h
Therefore, m1 + m 2 = −
and m1m 2 =
b
b
The lengths of the perpendiculars from ( x 1, y1 ) on these
lines are
| y1 − m1x 1|
| y1 − m 2 x 1|
and
2
1 + m1
1 + m 22
Their product =
=
=
=
×
| y1 − m 2 x 1|
1 + m 22
|(y1 − m1x 1 ) (y1 − m 2 x 1 )|
(1 + m12 ) (1 + m 22 )
| y12 − (m1 + m 2 ) x 1y1 + m1m 2 x 12 |
1 + m12 + m 22 + m12 m 22
| y12 − (m1 + m 2 ) x 1y1 + m1m 2 x 12 |
⇒
bm 2 + 2hm + a = 0
1


Qy = − m x is perpendicular to 


y = mx and passing through origin
2
 1 
 1 
then a ′ x 2 + 2h ′ x  −
x + b′ −
x = 0
 m 
 m 
⇒
a ′ m 2 − 2h ′ m + b ′ = 0
a′
− 2h ′
m
m
1
=
=
2hb ′ + 2h ′ a aa ′ − bb ′ − 2h ′ b − 2a ′ h
∴
m2 = −
(bb ′ − aa ′ )
(hb ′ + h ′ a )
,m =
(h ′ b + a ′ h )
2 (a ′ h + h ′ b )
On eliminating m, we obtain
4 (ha ′ + h ′ b ) (h ′ a + hb ′ ) + (bb ′ − aa ′ )2 = 0.
y Example 7 Show that the centroid ( x ′ , y ′ ) of the
triangle with sides ax 2 + 2hxy + by 2 = 0 and
lx + my = 1, is given by
y′
2
x′
.
=
=
2
bl − hm am − hl 3 (am − 2hlm + bl 2 )
Sol. Let the lines represented by
ax 2 + 2hxy + by 2 = 0
be
y = m1x
and
y = m2x .
Y
B
G(x′,y′ )
1
y=
{(a − b ) + 4h }
a′
⇒
a
2h
x 1y1 + x 12 |
+
b
b
2

4h
2a a 2 
+ 
1 + 2 −
b b2 
b

2
− 2h ′ b ′
2
| y12
2
...(ii)
On solving Eqs. (i) and (ii), by cross-multiplication rule i.e.
1 m2
m2
m
m
2h
2h
b
a b
1 + (m1 + m 2 )2 − 2m1m 2 + (m1m 2 )2
| ax 12 + 2hx 1y1 + by12 |
...(i)
1
then, y = − x be one of the line represented by
m
a ′ x 2 + 2h ′ xy + b ′ y 2 = 0
m
=
1+
m12
then, ax 2 + 2hx (mx ) + b (mx )2 = 0
lx+
=
| y1 − m1x 1|
Sol. Since, both pair are passing through origin, let y = mx be
one of the lines represented by
ax 2 + 2hxy + by 2 = 0
2x
⇒
 n 


 a   n + 1
1
(n + 1)
y Example 6 Find the condition that one of the lines
given by ax 2 + 2hxy + by 2 = 0
may be perpendicular to one of the lines given by
a ′ x 2 + 2h ′ xy + b ′ y 2 = 0.
y=m
and
x A
y=m 1
O
X
Chap 03 Pair of Straight Lines
2h
b
B
1

m2 
1
,


 l + mm 2 l + mm 2 
x A
y=m 1
O
Coordinates of A and B are
 −n
− nm1 
,


 l + mm1 l + mm1 
[Q if coordinates are (0, 0), ( x 1, y1 )and ,
1
( x 2 , y 2 ) , then area = | x 1y 2 − x 2y1 |]
2

n 2 (m 2 − m1 )
1

= 2
2
2l + lm (m1 + m 2 ) + m m1m 2
...(i)
1 n 2 (m1 + m 2 )2 − 4m1m 2 
= 2

2l + lm (m1 + m 2 ) + m 2m1m 2
m2

 m1
+
+ 0

l
+
mm
l
+
mm
1
2

and y ′ = 
3








l (m1 + m 2 ) + 2mm1m 2
= 2

2
 3 {l + lm (m1 + m 2 ) + m m1m 2 ) }
=


2hl 2ma


−
+


b
b
=
 2 2hlm m 2 a  
 3 l −
+

 
b
b  

2
(am − hl )
= ⋅
3 (am 2 − 2hlm + bl 2 )
 −n
− nm 2 
and 
,
,
 l + mm 2 l + mm 2 
1 − n   − nm 2   − n   − nm1 
= 
 

−
 
2 l + mm1   l + mm 2   l + mm 2   l + mm1 


2hm


2l −


b
=
 2 2hml m 2a  
 3 l −
+

 
b
b  

(bl − hm )
2
= ⋅
3 (am 2 − 2hlm + bl 2 )
=
...(ii)
From Eqs. (i) and (ii), we get
2
x′
y′
=
=
2
bl − hm am − hl 3(am − 2hlm + bl 2 )
y Example 8 Show that the area of the triangle formed
by the lines ax 2 + 2hxy + by 2 = 0 and lx + my + n = 0
1
2
n2
 4h 2 4a 
 2 − 
b
 b
l2 −
2hlm m 2a
+
b
b
n 2 (h 2 − ab )
|(am 2 − 2hlm + bl 2 )|
y Example 9 Show that the two straight lines
x 2 (tan 2 θ + cos 2 θ ) − 2xy tan θ + y 2 sin 2 θ = 0
move with the axis of x angles such that the difference
of their tangents is 2.
Sol. Given equation is
x 2 (tan 2 θ + cos 2 θ ) − 2xy tan θ + y 2 sin 2 θ = 0
and homogeneous equation of second degree
ax 2 + 2hxy + by 2 = 0
On comparing Eqs. (i) and (ii), we get
a = tan 2 θ + cos 2 θ
n 2 (h 2 − ab )
.
|(am 2 − 2hlm + bl 2 )|
Sol. Let equation of lines represented by
ax 2 + 2hxy + by 2 = 0 be y = m1x and y = m 2 x
a
and m1m 2 = .
b
X
respectively.
Then, required area


2l + m(m1 + m 2 )
= 2

2
 3 {l + ml (m1 + m 2 ) + m m1 m 2 ) }
2h
therefore, m1 + m 2 = −
b
y=
and
Since, centroid = ( x ′ , y ′ ),
1
1


+
+ 0

l
+
mm
l
+
mm
1
2

then, x ′ = 
3






is
m
y=m
m1m 2 =
lx+
a
b
Coordinates of A and B are

m1 
1
,


 l + mm1 l + mm1 
and
Y
2x
Therefore, m1 + m 2 = −
195
and
h = − tan θ
b = sin 2 θ
Let separate lines of Eq. (ii) are
y = m1x
and
y = m2x
...(i)
...(ii)
196
Textbook of Coordinate Geometry
m1 = tan θ1 and m 2 = tan θ 2
2h 2 tan θ
therefore, m1 + m 2 = −
=
b
sin 2 θ
where,
and
∴
=
=
a tan 2 θ + cos 2 θ
m1 ⋅ m 2 = =
b
sin 2 θ
=
m1 ~ m 2 = (m1 + m 2 ) − 4m1m 2
2
⇒ tan θ1 ~ tan θ 2 =
4 tan 2 θ
sin 4 θ
−
=
4 (tan 2 θ + cos 2 θ )
sin 2 θ
2
sin 2 θ
2 sin θ
sin 2 θ
2 sin θ
sin 2 θ
tan 2 θ − sin 2 θ (tan 2 θ + cos 2 θ )
(sec 2 θ − tan 2 θ − cos 2 θ )
(1 − cos 2 θ )
2
sin θ = 2
sin θ
Exercise for Session 1
1.
The lines given by the equation (2y 2 + 3xy − 2x 2 ) ( x + y − 1) = 0 form a triangle which is
(a) equilateral
(c) right angled
2.
Area of the triangle formed by the lines y 2 − 9xy + 18x 2 = 0 and y = 9 is
(a) 27/4
(c) 9 / 4
3.
4.
(b) 0
(d) 27
The equation 3x 2 + 2hxy + 3y 2 = 0 represents a pair of straight lines passing through the origin. The two lines
are
(a) real and distinct, if h 2 > 3
(b) real and distinct, if h 2 > 9
(c) real and coincident, if h 2 = 3
(d) real and coincident, if h 2 > 3
If one of the lines of the pair ax 2 + 2hxy + by 2 = 0 bisects the angle between positive directions of the axes,
then a, b , h satisfy the relation
(a) a + b = 2 |h |
(c) a − b = 2 |h |
5.
(b) isosceles
(d) obtuse angled
(b) a + b = − 2h
(d) (a − b )2 = 4h 2
If the slope of one of the lines given by a 2 x 2 + 2hxy + b 2 y 2 = 0 be three times of the other, then h is equal to
(a) 2 3ab
2
(c)
ab
3
(b) − 2 3ab
2
(d) −
ab
3
6.
Find the separate equations of two straight lines whose joint equation isab ( x 2 − y 2 ) + (a 2 − b 2 ) xy = 0.
7.
Find the coordinates of the centroid of the triangle whose sides are 12x 2 − 20xy + 7y 2 = 0 and 2x − 3y + 4 = 0.
8.
If the lines ax 2 + 2hxy + by 2 = 0 be two sides of a parallelogram and the line lx + my = 1be one of its diagonal,
show that the equation of the other diagonal is y (bl − hm ) = x (am − hl ).
9.
Find the condition that one of the lines given by ax 2 + 2hxy + by 2 = 0 may coincide with one of the lines given
by a′ x 2 + 2h′ xy + b ′ y 2 = 0.
Session 2
Angle between the Pair of Lines ax2 + 2hxy + by2
Angle between the Pair of
Lines ax 2 + 2 hxy + by 2
Theorem The angle θ between the pair of lines
represented by ax 2 + 2hxy + by 2 = 0
is given by θ = tan
−1
2 (h 2 − ab ) 

.
 | a + b | 
Proof Let y = m 1 x and y = m 2 x be the lines represented by
ax 2 + 2hxy + by 2 = 0.
a
2h
, m 1m 2 =
b
b
Since, θ be the angle between the lines
y = m 1 x and y = m 2 x .
Then, m 1 + m 2 = −
i.e. Coefficient of x 2 + Coefficient of y 2 = 0
Hence, the lines represented by ax 2 + 2hxy + by 2 = 0 are
perpendicular, iff a + b = 0 i.e. coefficient of x 2 +
coefficient of y 2 = 0.
2x
Y
Remark
y=
m
θ
Pair of any two perpendicular lines through the origin.
Q Lines represented by ax 2 + 2hxy + by 2 = 0
x
m1
y=
X′
X
Y′
(m 1 + m 2 ) 2 − 4m 1m 2
m − m 2 
=
Then, tan θ = 1
|1 + m 1m 2 |
1 + m 1m 2
2
=
∴
 −2h 
a 

 −4 
 b 
b 
θ = tan − 1
Corollary 1 Condition for the lines to be perpendicular.
The lines are perpendicular if the angle between them is
π
.
2
π
i.e.
θ=
2
π
⇒
cot θ = cot
2
⇒
cot θ = 0
| a + b|
=0
⇒
⇒ a +b =0
2 (h 2 − ab )


1 + a 
 b
=
2 (h 2 − ab )
be perpendicular, then a + b = 0 or b = − a
Hence, the equation becomes ax 2 + 2hxy − ay 2 = 0
2h
x 2 +   xy − y 2 = 0
⇒
 a
⇒
(Remember)
where, p is any constant.
Corollary 2 Pair of lines perpendicular to the lines
represented by
|a + b |
ax 2 + 2hxy + by 2 = 0
and through origin.
Let lines represented by
2 (h 2 − ab ) 


 | a + b| 
ax 2 + 2hxy + by 2 = 0 be y = m 1 x and y = m 2 x
then
Remark
2
 2 h − ab
θ = sin−1 
 ( a − b) 2 + 4 h2
x 2 + pxy − y 2 = 0 ,



and
m1 + m2 = −
m 1m 2 =
a
b
2h
b
198
Textbook of Coordinate Geometry
lines perpendiculars to y = m 1 x
and
y = m2 x
and passing through origin are
1
y =−
x
m1
y =−
and
then, pair is
x 


y +

m1 
On comparing Eqs. (i) and (ii), we get
a = sin 2 α − cos 2 β, h = sin β cos β,
b = sin 2 α − sin 2 β
Let the angle between the lines representing by Eq. (i) is θ.
1
x
m2
{sin 2 β cos 2 β − sin 4 α + sin 2 α sin 2 β
=2
⇒
2hxy a 2
+ y =0
x2 −
b
b
=2
∴
bx 2 − 2hxy + ay 2 = 0
Aid to memory For perpendicular pairs interchange the
coefficients of x 2 and y 2 and change the sign of xy.
Corollary 3 Condition for the lines to be coincident.
The lines are coincident, if the angle between them is
0° (or π )
i.e.
θ = 0 (or π )
∴
tan θ = 0 °
2 (h 2 − ab )
∴
=0
2 sin α cos α
= tan 2α
| − cos 2α|
θ = 2α
...(i)
The homogeneous equation of second degree
Ax 2 + 2Hxy + By 2 = 0
...(ii)
On comparing Eqs. (i) and (ii), we get
A = a + 2hm + bm 2 , H = (b − a )m − (m 2 − 1)h ,
Hence, the lines represented by ax 2 + 2hxy + by 2 = 0
are coincident, iff h 2 = ab, then ax 2 + 2hxy + by 2
is a perfect square.
B = am 2 − 2hm + b
Let the angle between the lines representing by Eq. (i) is θ.
Remark
The parallel lines will be coincident only as both pass through a
point.
∴ tan θ =
2 ( H 2 − AB )
| A + B|
2
y Example 10 Find the angle between the lines
( x 2 + y 2 ) sin 2 α = ( x cos β − y sin β) 2 .
=
Sol. Given equation is
( x 2 + y 2 ) sin 2 α = ( x cos β − y sin β )2
=
x 2 (sin 2 α − cos 2 β ) + 2xy sin β cos β
+y 2 (sin 2 α − sin 2 β ) = 0 …(i)
The homogeneous equation of second degree is
ax 2 + 2hxy + by 2 = 0
sin 2 α (1 − sin 2 α )
| − cos 2α|
(am 2 − 2hm + b )y 2 = 0
h 2 = ab
⇒
|(2 sin 2 α − 1)|
Sol. Given equation is
(a + 2hm + bm 2 )x 2 + 2 { (b − a )m − (m 2 − 1)h }xy +
2
⇒
=
+ sin 2 α cos 2 β − sin 2 β cos 2 β}
y Example 11 Show that the angle between the lines
given by (a + 2hm + bm 2 ) x 2 + 2 {(b − a ) m − (m 2 − 1) h }
xy + (am 2 − 2hm + b )y 2 = 0 is the same whatever be
the value of m.
h − ab = 0
⇒
|sin 2 α − cos 2 β + sin 2 α − sin 2 β |

x 
 =0
y +

m2 
x 2 + xy (m 1 + m 2 ) + m 1m 2 y 2 = 0
| a + b|
sin 2 β cos 2 β − (sin 2 α − cos 2 β ) (sin 2 α − sin 2 β )
=2
⇒
⇒
h 2 − ab
|a + b |
∴ tan θ = 2
…(ii)
{(b − a ) m − (m 2 − 1)h }2 − (a + 2hm + bm 2 )
(am 2 − 2hm + b )
| a + 2hm + bm 2 + am 2 − 2hm + b |
2 (m 2 + 1)2 (h 2 − ab )
| a + b | ( m + 1)
2
=
2 (h 2 − ab )
|a + b |
 2 (h 2 − ab ) 
θ = tan − 1 

 | a + b | 
which is independent of m. Hence, the angle between the
lines representing by Eq. (i) is same for all values of m.
⇒
Chap 03 Pair of Straight Lines
y Example 12 Show that the straight lines
x 2 + 4 xy + y 2 = 0 and the line x − y = 4 form an
equilateral triangle.
x 2 + 4 xy + y 2 = 0
Sol. Equation
y Example 13 Show that the condition that two of the
three lines represented by ax 3 + bx 2 y + cxy 2 + dy 3 = 0
may be at right angles is a 2 + ac + bd + d 2 = 0.
...(i)
is a homogeneous equation of second degree in x and y.
Therefore, it represents two lines OP and OQ through the
origin.
Equation,
...(ii)
x −y = 4
represent the line PQ .
Y
X′
15°
O
60°
y+(2–√3)x=0
X
15 °
P
+√3)
x–
y=
4
y+(2
x=0
Q
tan θ =
∴
From Eq. (i),
2 [(2) − 1 ⋅ 1]
= 3
| 1 + 1|
x
2
2
θ = 60°
+ 4 xy + y 2 = 0
2
⇒
y 
y 
  + 4   +1=0
x
x
⇒
y − 4 ± (42 − 4)
=
= −2± 3
2
x
y = (− 2 ± 3) x
i.e.
OP : y + (2 − 3 ) x = 0
and
OQ : y + (2 + 3 ) x = 0
⇒
Q
If
then
Comparing the coefficients of similar terms, we get
...(i)
b = ap − d
...(ii)
c = − pd − a
Multiplying Eq. (i) by d and Eq. (ii) by a and adding, we get
bd + ac = − d 2 − a 2
a 2 + ac + bd + d 2 = 0
Aliter :
Let y = m1x , y = m 2 x and y = m 3 x be the lines represented
by the equation
ax 3 + bx 2y + cxy 2 + dy 3 = 0.
Then
Let ∠POQ = θ.
then,
Sol. The given equation being homogeneous of third degree
represents three straight lines through the origin. Since,
two of these lines are to be at right angles.
Let pair of these lines be ( x 2 + pxy − y 2 ), p is constant and
the other factor is (ax − dy ).
Hence, ax 3 + bx 2y + cxy 2 + dy 3 = ( x 2 + pxy − y 2 ) (ax − dy )
⇒
Y′
199
Slope of PQ = 1 and Slope of OP = − (2 − 3 ).
∠OPQ = α
1 − ( − 2 + 3 )
 =
= 3
3− 3
tan α =
1
2
3
−
+
 − 1 + 3

∴
α = 60°
Hence, ∠OQP = 180° − (60° + 60°) = 60°
Hence, ∆OPQ is an equilateral triangle.
ax 3 + bx 2y + cxy 2 + dy 3 = d (y − m1x )
...(i)
(y − m 2 x ) (y − m 3 x )
On equating the coefficients of x 3 , x 2y and xy 2 on both
sides, we get
c
m1 + m 2 + m 3 = −
d
b
m1m 2 + m 2m 3 + m 3m1 =
d
a
...(ii)
and
m1m 2m 3 = −
d
Let the perpendicular lines be y = m1x and y = m 2 x , then
m1m 2 = − 1
a
From Eq. (ii),
...(iii)
m3 =
d
On putting y = m 3 x in Eq. (i), we get
ax 3 + bm 3 x 3 + cm 32 x 3 + dm 33 x 3 = 0
⇒
dm 33 + cm 32 + bm 3 + a = 0
⇒
a
a
a
d   +c   +b  +a=0
d 
d 
d 
3
Hence,
2
a 2 + ac + bd + d 2 = 0
[from Eq. (iii)]
200
Textbook of Coordinate Geometry
Exercise for Session 2
1.
The angle between the pair of straight lines y 2 sin2 θ − xy sin2 θ + x 2(cos 2 θ − 1) = 0 is
(a)
2.
3.
π
4
(b)
π
2
(c)
(a) 5x 2 + 2xy − 3y 2 = 0
(b) x 2 − 2xy − 3y 2 = 0
(c) x 2 − y 2 = 100
(d) xy = 0
2π
3
Which of the following pair of straight lines intersect at right angles ?
(b) (x + y )2 = x (y + 3x )
(c) 2y (x + y ) = xy
5.
(d)
The angle between the lines given by the equation ay 2 − (1 + λ 2 ) xy − ax 2 = 0 is same as the angle between
the lines
(a) 2x 2 = y (x + 2y )
4.
π
3
(d) y = m 2x
If h = ab , then the lines represented by ax + 2hxy + by 2 = 0 are
2
2
(a) parallel
(b) perpendicular
(c) coincident
(d) None of these
Equation ax 3 − 9x 2y − xy 2 + 4y 3 = 0 represents three straight lines. If the two of the lines are perpendicular,
then a is equal to
(a) −5
(c) −4
(b) 5
(d) 4
6.
Find the angle between the lines whose joint equation is2x 2 − 3xy + y 2 = 0.
7.
Show that the lines (1 − cos θ tan α ) y 2 − (2 cos θ + sin2 θ tan α ) xy + cos θ (cos θ + tan α ) x 2 = 0
include an angle α between them.
8.
Find the angle between the lines represented by the equation x 2 − 2pxy + y 2 = 0.
9.
Show that the lines x 2 − 4xy + y 2 = 0 and x + y = 1form an equilateral triangle and find its area.
10.
Prove that the triangle formed by the lines ax 2 + 2hxy + by 2 = 0 and lx + my = 1is isosceles,
if h (l 2 − m 2 ) = (a − b ) m .
Session 3
Bisectors of the Angle between the Lines
Given by a Homogeneous Equation
Bisectors of the Angle
between the Lines Given by a
Homogeneous Equation
Theorem The joint equation of the bisectors of the angles
between the lines represented by the equation
ax + 2hxy + by = 0 is
2
2
x 2 − y 2 xy
= .
a −b
h
∴ The pair of bisectors is
 (y − m x ) (y − m x )   (y − m x ) (y − m x ) 
1
1
2
2


 =0
+
−
 (1 + m 2 )
2
2  
2 
(1 + m 2 )   (1 + m 1 )
(1 + m 2 ) 

1
⇒
⇒
(y − m 1 x ) 2
(1 + m 12 )
−
(y − m 2 x ) 2
(1 + m 22 )
=0
(1 + m 22 ) (y 2 + m 12 x 2 − 2m 1 xy )
− (1 + m 12 ) (y 2 + m 22 x 2 − 2m 2 xy ) = 0
y=
m
2x
Session2
Proof Let the lines represented by ax 2 + 2hxy + by
=0
⇒ (m 22 − m 12 )y 2 − (m 22 − m 12 ) x 2
be y − m 1 x = 0 and y − m 2 x = 0, then
+2 xy (m 2 − m 1 ) − 2m 1m 2 (m 2 − m 1 ) xy = 0
a
2h
2
⇒ (m 2 + m 1 ) (y − x 2 ) + 2 xy − 2m 1m 2 xy = 0
and m 1m 2 =
m1 + m2 = −
b
b
[Qm 1 − m 2 ≠ 0 ]
Since, the bisectors of the angles between the lines are the
2h 

locus of a point which is equidistant from the two given
Qm 1 + m 2 = − 

 2h 
 a
b
lines.
⇒ ( x 2 − y 2 )  −  = 2 xy  1 −  





b
b
a
Y
 mm =

1 2

b

,k) x
P(h m 1
y=
M
N
X¢
∴
Y¢
Let P (h, k ) be a point on a bisector of the angle between
the given lines. Then, PM = PN
⇒
⇒
| k − m 1 h|
(1 + m 12 )
(k − m 1 h )
(1 + m 12 )
=
| k − m 2 h|
=±
(1 + m 22 )
(k − m 2 h )
(1 + m 22 )
Hence, the locus of a P (h, k ) is
(y − m 1 x )
(1 + m 12 )
=±
(y − m 2 x )
(1 + m 22 )
[b ≠ 0 ]
Aliter :
X
O
x 2 − y 2 xy
=
a −b
h
Let the equation ax 2 + 2hxy + by 2 = 0 represent two lines
L1 OM 1 and L2 OM 2 making angles θ 1 and θ 2 with the
positive direction of X-axis.
If slopes of L1 OM 1 and L2 OM 2 are m 1 and m 2 , then
m 1 = tan θ 1 and m 2 = tan θ 2
2h
a
...(i)
and
m 1 + m 2 = − , m 1m 2 =
b
b
Let NON 1 and KOK 1 are the required bisectors,
θ − θ1
Since ∠ NOL1 = ∠ NOL2 = 2
2
θ − θ1 θ1 + θ2
∠ NOX = θ 1 + 2
=
2
2
π
Since, ∠ NOK =
2
202
Textbook of Coordinate Geometry
Y
Remark
L2
The joint equation of the bisectors is
N
L1
K
x 2 − y 2 xy
=
a− b
h
hx 2 − ( a − b) xy − hy 2 = 0
or
i.e. coefficient of x 2 + coefficient of y 2 = 0.
X′
M1
N1
∴
θ1 θ2
Hence, the bisectors of the angle between the lines are always
perpendicular to each other.
X
O
Corollaries
K1
M2 Y′
1. If a = b, the bisectors are x 2 − y 2 = 0
π
π  θ + θ2 
∠ KOX = + ∠ NOX = +  1

2
2  2 
i.e.
x − y = 0, x + y = 0
2. If h = 0, the bisectors are xy = 0
i.e.
x = 0, y = 0
 θ + θ2 
Equation of bisectors arey = x tan  1

 2 
⇒
 θ + θ2 
y − x tan  1
 =0
 2 
and
 π θ + θ2 
y = x tan  + 1

2
2 
⇒
 θ + θ2 
y = − x cot  1

 2 
⇒
 θ + θ2 
y + x cot  1
 =0
 2 
…(ii)
Sol. Given equation is
3x 2 − 5xy + 4y 2 = 0
Comparing it with the equation
ax 2 + 2hxy + by 2 = 0
...(iii)
∴ Pair of bisectors

 θ1 + θ2  
y − x tan  2  



 θ1 + θ2  
y + x cot  2   = 0



 θ + θ2 
 θ1 + θ2  
⇒ y 2 − x 2 + xy  cot  1
 − tan 
  =0
 2 
 2 

⇒

2  θ1 + θ2  
 1 − tan 

 2 
x 2 − y 2 = xy 

 θ1 + θ2  
 
 tan 
2  

⇒


2
x 2 − y 2 = xy 

 tan (θ 1 + θ 2 ) 
⇒
 1 − tan θ 1 tan θ 2 
x 2 − y 2 = 2 xy 

 tan θ 1 + tan θ 2 
⇒
 1 − m 1m 2 
x 2 − y 2 = 2 xy 

 m1 + m2 
⇒
1 −a / b 
x 2 − y 2 = 2 xy 

 − 2h / b 
∴
( x 2 − y 2 ) xy
=
(a − b )
h
y Example 14 Find the equation of the bisectors of the
angle between the lines represented by
3 x 2 − 5 xy + 4y 2 = 0.
...(i)
...(ii)
5
a = 3, h = − , b = 4
2
Hence, the equation of bisectors of the angle between the
pair of the lines (i) is
x2 − y2
xy
=
3− 4
−5 / 2
then
⇒
∴
x 2 − y 2 2xy
=
−1
−5
2
2
5x − 2xy − 5y = 0
y Example 15 Show that the line y = mx bisects the
angle between the lines
ax 2 − 2hxy + by 2 = 0,
if
h (1 − m 2 ) + m (a − b ) = 0.
Sol. Equation of pair of bisectors of angles between lines
ax 2 − 2hxy + by 2 = 0 is
x 2 − y 2 xy
=
a−b
−h
⇒
− h ( x 2 − y 2 ) = (a − b ) xy
...(i)
But y = mx is one of these lines, then it will satisfy
it. Substituting y = mx in Eq. (i),
− h ( x 2 − m 2 x 2 ) = (a − b )x ⋅ mx
Dividing by x 2 , h (1 − m 2 ) + m (a − b ) = 0
Chap 03 Pair of Straight Lines
y Example 16 If pairs of straight lines
x 2 − 2pxy − y 2 = 0 and x 2 − 2qxy − y 2 = 0 be such
that each pair bisects the angle between the other
pair, then prove that pq = − 1.
x 2 − 2qxy − y 2 = 0
⇒
− px 2 − 2xy + py 2 = 0
…(iii)
Since, Eqs. (ii) and (iii) are identical, comparing Eqs. (ii) and
(iii), we get
−2q −1
1
⇒ pq = − 1
=
=
p
−2
−p
Remark
By taking the bisectors of the angles between the pair of lines (ii),
we will get the same result.
y Example 17 Prove that the angle between one of the
lines given by ax 2 + 2hxy + by 2 = 0 and one of the
lines ax 2 + 2hxy + by 2 + λ ( x 2 + y 2 ) = 0 is equal to
angle between other two lines of the system.
Sol. The equation of the bisectors of the angle between the lines
...(i)
ax 2 + 2hxy + by 2 = 0
is
x 2 − y 2 xy
=
a−b
h
and the equation of bisectors of the angle between the lines
…(ii)
ax 2 + 2hxy + by 2 + λ ( x 2 + y 2 ) = 0
⇒
is
(a + λ) x 2 + 2hxy + (b + λ )y 2 = 0
x
R
x
X′
R′
X
O
P′
A′
S′
B'
Y′
Q′
which are also the bisectors of the second pair.
Let P ′ OP , Q ′ OQ , R ′ OR, S ′ OS be the lines of the Eqs. (i) and
(ii) pairs respectively and A ′ OA and B ′ OB be their
bisectors. We have,
∠ ROA = ∠ SOA
and
∠ POA = ∠QOA
⇒ ∠ ROA − ∠ POA = ∠SOA − ∠QOA
⇒
∠ ROP = ∠SOQ
Hence the result.
y Example 19 If the lines represented by
x 2 − 2pxy − y 2 = 0 are rotated about the origin
through an angle θ , one in clockwise direction and the
other in anti-clockwise direction, then find the
equation of the bisectors of the angle between the
lines in the new position.
Sol. Since, lines represented by x 2 − 2pxy − y 2 = 0
are perpendicular to each other. The bisectors of the angles
between the lines in new position are same as the bisectors
of the angles between their old positions. i.e.
OC , OD ; OA , OB be the old and new pairs respectively and
OE and OF be their bisectors, we have
∠COE = ∠ DOE
and
[given]
∠COA = ∠ DOB = θ
x2 − y2
xy
x 2 − y 2 xy
=
⇒
=
(a + λ ) − (b + λ ) h
a−b
h
D
Y
B
E
A
F
y Example 18 Show that the pair of lines given by
a 2 x 2 + 2h(a + b )xy + b 2 y 2 = 0 is equally inclined to the
pair given by ax 2 + 2hxy + by 2 = 0.
Sol. Given pair of lines are
a 2 x 2 + 2h (a + b )xy + b 2y 2 = 0
...(i)
ax 2 + 2hxy + by 2 = 0
...(ii)
Equation of bisectors of first pair is
x2 − y2
xy
x 2 − y 2 xy
=
=
⇒
2
2
h (a + b )
a−b
h
a −b
P
B
∴ Bisectors of angles between lines given by Eqs. (i) and
(ii) are the same. Hence the result.
and
Q
A
…(ii)
∴ The equation of bisectors of the angle between the lines
(i) is
x 2 − y 2 xy
=
1 − ( − 1) − p
S
Y
Sol. According to the question, the equation of the bisectors of
the angle between the lines
...(i)
x 2 − 2pxy − y 2 = 0
is
203
θ
C
θ
X′
X
O
Y′
⇒ ∠COE − ∠COA = ∠DOE − ∠ DOB
⇒
∠ AOE = ∠ BOE
Therefore, the required equation is
i.e.
px 2 + 2xy − py 2 = 0.
x 2 − y 2 xy
=
1 − ( − 1) − p
204
Textbook of Coordinate Geometry
Exercise for Session 3
1.
If coordinate axes are the angle bisectors of the pair of lines ax 2 + 2hxy + by 2 = 0, then
(a) a = b
(b) h = 0
(c) a + b = 0
(d) a + b 2 = 0
2
2.
If the line y = mx is one of the bisector of the lines x 2 + 4xy − y 2 = 0, then the value of m is
(a)
5−1
2
 5 + 1
(c) − 

 2 
3.
(b)
5+ 1
2
 5 − 1
(d) − 

 2 
If one of the lines of my 2 + (1 − m 2 ) xy − mx 2 = 0 is a bisector of the angle between the lines xy = 0, then
cos −1(m ) is
4.
(a) 0
(b) π/2
(c) π
(d) 3 π / 2
The bisectors of the angles between the lines (ax + by )2 = c (bx − ay )2, c > 0 are respectively parallel and
perpendicular to the line
(a) bx − ay + µ = 0
(b) ax + by + λ = 0
(c) ax − by + ν = 0
(d) bx + ay + τ = 0
5.
If the pairs of straight lines ax 2 + 2hxy − ay 2 = 0 and bx 2 + 2gxy − by 2 = 0 be such that each bisects the angles
between the other, then prove that hg + ab = 0.
6.
Prove that the lines 2x 2 + 6xy + y 2 = 0 are equally inclined to the lines 4x 2 + 18xy + y 2 = 0.
7.
Show that the lines bisecting the angle between the bisectors of the angles between the lines
ax 2 + 2hxy + by 2 = 0 are given by (a − b ) ( x 2 − y 2 ) + 4hxy = 0.
8.
Prove that the bisectors of the angle between the lines ax 2 + acxy + cy 2 = 0 and
1 2
1 2


3 +  x + xy + 3 +  y = 0 are always the same.


c
a
9.
The lines represented by x 2 + 2λxy + 2y 2 = 0 and the lines represented by (1 + λ )x 2 − 8xy + y 2 = 0 are equally
inclined, find the values of λ.
Session 4
General Equation of Second Degree,
Important Theorems
General Equation of Second
Degree
Now, ax 12 + 2hx 1 y 1 + by 12 + 2 gx 1 + 2 fy 1 + c = 0
i.e. a = coefficient of x 2 , b = coefficient of y 2 ,
⇒ x 1 (ax 1 + hy 1 + g ) + y 1 (hx 1 + by 1 + f )
+ ( gx 1 + fy 1 + c ) = 0
⇒ x 1 ⋅ 0 + y 1 ⋅ 0 + gx 1 + fy 1 + c = 0
[from Eqs. (iii) and (iv)]
...(vi)
⇒
gx 1 + fy 1 + c = 0
On eliminating x 1 , y 1 from Eqs. (iii), (iv) and (vi), we get
the determinant
a h g
c = constant term,
g = half the coefficient of x,
f = half the coefficient of y,
h = half the coefficient of xy.
Theorem The necessary and sufficient condition for
h b f =0
g f c
2
∴ abc + 2 fgh − af − bg 2 − ch 2 = 0,
as the required condition.
The equation ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
is the general equation of second degree and represents a
conics (pair of straight lines, circle, parabola, ellipse,
hyperbola). It contains six constants a, b, c , f , g, h.
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
Remarks
to represent a pair of straight lines is that
a
h
g
abc + 2 fgh − af − bg − ch = 0 or h
g
b
f
f =0.
c
2
2
2
Proof Necessary condition : Let the equation be
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
…(i)
represent a pair of lines. Assuming that these lines are not
parallel, we suppose further that their point of intersection
is ( x 1 , y 1 ). Shifting the origin at ( x 1 , y 1 ) without rotating
the coordinate axes, we have the Eq. (i) transforms to
a( X + x 1 ) 2 + 2h( X + x 1 )(Y + y 1 ) + b(Y + y 1 ) 2
+ 2 g( X + x 1 ) + 2 f (Y + y 1 ) + c = 0 ...(ii)
Now this Eq. (ii) represents a pair of lines through the new
origin and consequently, it is homogeneous in X and Y.
Hence, the coefficients of X and Y and the constant term
in Eq. (ii) must vanish separately.
i.e. coefficient of X = coefficient of Y = constant term = 0
⇒
and
ax 12
+ 2hx 1 y 1 + by 12
ax 1 + hy 1 + g = 0
...(iii)
hx 1 + by 1 + f = 0
...(iv)
+ 2 gx 1 + 2 fy 1 + c = 0
...(v)
1. Without using determinant On solving Eqs. (iii) and (iv),
we get
 hf − bg gh − af 

( x1, y1 ) = 
,
2
2
 ab − h ab − h 
and then substituting the values of x1 and y1 in Eqs. (vi), we
obtain
 hf − bg 
 gh − af 
 + f 
 + c =0
g 
2
2
 ab − h 
 ab − h 
⇒
2. By making
abc + 2fgh − af 2 − bg 2 − ch2 = 0
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0
homogeneous with the help of a new variable z, i.e.
ax 2 + 2hxy + by 2 + 2gxz + 2fyz + cz 2 = 0
Let f ( x, y, z ) ≡ ax 2 + 2hxy + by 2 + 2gxz + 2fyz + cz 2 = 0
∂f
= 2ax + 2hy + 2gz = 0
∴
∂x
∂f
= 2hx + 2by + 2fz = 0
∂y
∂f
= 2gx + 2fy + 2cz = 0
∂z
and finally putting z = 1, we obtain equations
ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0
which are same as Eqs. (iii), (iv) and (vi), respectively.
3. If ab − h2 = 0, the lines given by Eq. (i) are parallel. In this
case, the method followed in the above proof fails and we
follow the following method.
206
Textbook of Coordinate Geometry
Aliter I : (Proof)
Let the lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
be
then
Case I If a ≠ 0, then writing Eq. (i) as a quadratic
equation in x, we get
lx + my + n = 0 and l ′ x + m ′ y + n ′ = 0
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c
Solving, we have
≡ (lx + my + n ) (l ′ x + m ′ y + n ′ ) ...(ii)
Comparing the coefficients of similar terms in both sides
of Eq. (ii), we get
ll ′ = a, mm ′ = b, nn ′ = c



…(iii)
lm ′ + l ′ m = 2h, ln′ + l ′ n = 2 g,
mn ′ + m ′ n = 2 f



We now eliminate l, m, n, l ′ , m ′ and n′ from these equations,
l l′ 0
l′ l 0
we have
m m′ 0 × m′ m 0 =0
n n′ 0
n′ n 0
[Qeach determinant = 0]
2ll ′
lm ′ + l ′ m
ln ′ + l ′ n
⇒ ml ′ + m ′ l
2mm ′
mn ′ + m ′ n = 0
2nn′
nl ′ + n ′ l nm ′ + n ′ m
2a
⇒
2h 2b
2g 2 f
⇒
∴
2h
2a
− (hy + g ) ± {(h 2 − ab )y 2 + 2( gh − af )y + ( g 2 − ac ) }
a
Eq. (i), will represent two straight lines, if LHS of Eq. (i),
can be resolved into two linear factors, therefore the
expression under the square root should be a perfect
square.
[QAx 2 + Bx + C = 0 is a perfect
square ⇔ B 2 − 4 AC = 0 ]
Hence, 4( gh − af ) 2 − 4(h 2 − ab )( g 2 − ac ) = 0
or
abc + 2 fgh − af 2 − bg 2 − ch 2 = 0
by 2 + 2y (hx + f ) + 2 gx + c = 0
and proceeding above we get the condition
a
h
g
∆= h
g
b
f
f =0
c
2 fgh − bg 2 − ch 2 = 0
Without using determinant
Now, ( lm′ + l ′ m) (ln′ + l ′ n) ( mn′ + m′ n) = 2h ⋅ 2g ⋅ 2f
⇒
2ll ′ mm′ nn′ + ll ′ ( m2n′2 + m′2 n2 )
+ mm′( n2l ′2 + n′2 l 2 ) + nn′ ( l 2m′2 + l ′2 m2 ) = 8fgh
2ll ′ mm′ nn′ + ll ′{( mn′ + m′ n) 2 − 2mm′ nn′}
+ mm′{( nl ′ + n′ l ) 2 − 2nn′ ll ′} + nn′ {( lm′ + l ′ m) 2 − 2ll ′ mm′}
= 8fgh
[from Eq. (iii)]
⇒ 2abc + a( 4 f − 4 bc ) + b( 4 g − 4 ca) + c( 4 h − 4 ab) = 8 fgh
2
...(ii)
Ths is called discriminant of the Eq. (i).
Case II If a = 0, b ≠ 0, then writing Eq. (i) as a quadratic
equation in y
2f =0
2c
Remark
∴
∴x =
−2(hy + g ) ± 4(hy + g ) 2 − 4a(by 2 + 2 fy + c )
i.e.
which is the required necessary condition.
2
x=
2g
∆ = abc + 2 fgh − af 2 − bg 2 − ch 2 = 0,
⇒
ax 2 + 2 x (hy + g ) + by 2 + 2 fy + c = 0
...(i)
2
abc + 2fgh − af 2 − bg 2 − ch2 = 0,
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
which is the required necessary condition.
represents two straight lines is
Aliter II : (Proof)
Given equation is
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
which is condition obtained by putting a = 0 in Eq. (ii).
Case III If a = 0, b = 0 but h ≠0, then Eq. (i) becomes
2hxy + 2 gx + 2 fy + c = 0
h
Multiplying by
[Qh ≠ 0 ]
2
ch
h 2 xy + hgx + hfy + = 0
⇒
2
ch
⇒
(hx + f ) (hy + g ) = fg −
2
Above equation represents two straight lines, if
ch
fg − = 0 ⇒ 2 fgh − ch 2 = 0
2
which is condition obtained by putting a = 0, b = 0 in
Eq. (ii).
Hence in each case, the condition that
abc + 2 fgh − af 2 − bg 2 − ch 2 = 0
...(i)
which is the required necessary condition.
Chap 03 Pair of Straight Lines
b
f
f =0
c
i.e. lines ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0
are concurrent.
Let the point of concurrency be ( x 1 , y 1 ).
...(iii)
Then,
ax 1 + hy 1 + g = 0
...(iv)
hx 1 + by 1 + f = 0
and
...(v)
gx 1 + fy 1 + c = 0
Now, shifting the origin at ( x 1 , y 1 ) without rotating the
coordinate axes the equation
Theorem 1 The angle between the lines represented
by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
is given by
2 (h 2 − ab ) 
θ = tan −1 
.
 | a + b | 
Proof Let y = m 1 x + c 1
and
y = m2 x + c 2
be the lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
Y
P
ax + 2hxy + by + 2 gx + 2 fy + c = 0 reduces to
+
c
2
2
2x
2
a( X + x 1 ) + 2h( X + x 1 ) (Y + y 1 ) + b
y=
(Y + y 1 ) 2 + 2 g( X + x 1 ) + 2 f (Y + y 1 ) + c = 0
X′
⇒ aX + 2hXY + bY + 2 X (ax 1 + hy 1 + g )
2
2
+ 2Y (hx 1 + by 1 + f ) + x 1 (ax 1 + hy 1 + g )
+ y 1 (hx 1 + by 1 + f ) + ( gx 1 + fy 1 + c ) = 0
⇒ aX 2 + 2hXY + bY 2 + 0 + 0 + 0 + 0 + 0 = 0
[from Eqs. (iii), (iv) and (v)]
i.e.
aX 2 + 2hXY + bY 2 = 0
It is homogeneous equation of second degree. So, it
represents a pair of straight lines through the new origin.
Hence, the equation ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
represents a pair of straight lines, if
α
β
R
O
X
Q
Y′
where, m 1 = tan α, m 2 = tan β
Then, ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c
≡ (y − m 1 x − c 1 ) (y − m 2 x − c 2 )
Comparing coefficients of like powers, we obtain
a
2h
m 1 + m 2 = − , m 1m 2 =
b
b
Now, if θ be the acute angle between the lines
y = m 1 x + c 1 and y = m 2 x + c 2 , then
(m 1 + m 2 ) 2 − 4m 1m 2
m − m 2 
=
tan θ = 1
| 1 + m 1m 2 |
1 + m 1m 2
abc + 2 fgh − af 2 − bg 2 − ch 2 = 0.
Some useful identities If y = m 1 x + c 1 , y = m 2 x + c 2 be
lines represented by Eq. (i). Then,
2
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c
=
= b (y − m 1 x − c 1 ) (y − m 2 x − c 2 )
= b (y 2 − (m 1 + m 2 ) xy + m 1m 2 x 2
+ (m 1 c 2 + m 2 c 1 ) x − (c 1 + c 2 )y + c 1 c 2 )
On equating coefficients, we get
2g
a
2h
m 1 + m 2 = − ,m 1m 2 = ,m 1 c 2 + m 2 c 1 = ,
b
b
b
2f
c
c1 + c2 = −
, c 1c 1 =
b
b
These five relations are very useful to solve many problems.
θ
m
2
1 x+c
1
h
g
Important Theorems
y=m
Sufficient condition (Conversely)
Here, we have to show that the equation
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents a pair of
straight lines.
a h g
207
∴
 2h 
a 
−  − 4  
 b
b 


1 + a 
 b
=
2 (h 2 − ab )
|a +b |
2 (h 2 − ab ) 
θ = tan −1 

 | a + b | 
Corollary 1. The angle between the lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
is the same as the angle between the lines represented by
ax 2 + 2hxy + by 2 = 0
208
Textbook of Coordinate Geometry
Corollary 2. The lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
are perpendicular iff a + b = 0 and parallel iff h = ab.
2
Theorem 2 The lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
will be coincident, if h − ab = 0, g − ac = 0
2
2
f 2 − bc = 0.
and
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c
then
Proof Let the lines represented by
≡ (lx + my + n ) (l ′ x + m ′ y + n ′ )
Comparing the coefficients of similar terms in both sides
then
ll ′ = a, mm ′ = b, nn ′ = c
lm ′ + l ′ m = 2h, ln′ + l ′ n = 2 g, mn ′ + m ′ n = 2 f
⇒
(lm ′ − l ′ m ) = (lm ′ + l ′ m ) 2 − 4ll ′ mm ′ = 2 (h 2 − ab )
⇒
(nl ′ − n ′ l ) = (ln ′ + l ′ n ) 2 − 4ll ′ nn ′ = 2 ( g 2 − ac )
and (mn ′ − m ′ n ) = (mn ′ + m ′ n ) 2 − 4 mm ′ nn ′
= 2 ( f 2 − bc )
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
be
lx + my + n = 0 and l ′ x + m ′ y + n ′ = 0
then
ax + 2hxy + by + 2 gx + 2 fy + c
2
2
≡ (lx + my + n ) (l ′ x + m ′ y + n ′ )
Comparing the coefficients of similar terms in both sides,
we get
ll ′ = a, mm ′ = b, nn ′ = c


lm ′ + l ′ m = 2h, ln′ + l ′ n = 2 g 

mn ′ + m ′ n = 2 f

Q Lines lx + my + n = 0 and l ′ x + m ′ y + n ′ =0 are
l m n
coincident, then =
= .
l ′ m ′ n′
Taking the ratios in pairs, then
lm ′ − l ′ m = 0,mn ′ − m ′ n = 0, ln′ − l ′ n = 0
⇒
Now, solving lx + my + n = 0 and l ′ x + m ′ y + n ′ = 0
y
x
1
then
=
=
(mn ′ − m ′ n ) (nl ′ − n ′ l ) (lm ′ − l ′ m )
⇒
x
2 ( f 2 − bc )
=
y
2 ( g 2 − ac )
=
(lm ′ + l ′ m ) 2 − 4ll ′ mm ′ = 0,
=
f 2 h 2 − abf 2 − bch 2 + b (abc )
(h 2 − ab )
f 2 h 2 − abf 2 − bch 2 + b (af 2 + bg 2 + ch 2 − 2 fgh )
(h 2 − ab )
[Qabc + 2 fgh − af 2 − bg 2 − ch 2 = 0 ]
(ln ′ + l ′ n ) 2 − 4ll ′ nn ′ = 0
( f 2 h 2 + b 2 g 2 − 2bfgh )
( 4h 2 − 4ab ) = 0, ( 4 f 2 − 4bc ) = 0
=
and
( 4 g − 4ac ) = 0
 bg − hf 

= 2
 h − ab 
2
h − ab = 0, f 2 − bc = 0, g 2 − ac = 0
2
Theorem 3 The point of intersection of the lines
represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 is
  f 2 − bc   g 2 − ca  


 
 or  bg − hf , af − gh  .



,
  h 2 − ab   h 2 − ab  
 h 2 − ab h 2 − ab 


Proof Let the lines represented by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
be
lx + my + n = 0 and l ′ x + m ′ y + n ′ = 0
2 (h 2 − ab )
 f 2 − bc 
( f 2 − bc ) (h 2 − ab )
 =
Also, x =  2
(h 2 − ab )
 h − ab 
i.e.
i.e.
1
  f 2 − bc   g 2 − ac  

, 
(x , y ) =   2
  h − ab   h 2 − ab  


∴
(mn ′ + m ′ n ) 2 − 4 mm ′ nn ′ = 0
and
=
Similarly,
Hence,
(h 2 − ab )
=
(bg − hf ) 2
(h 2 − ab )
 af − gh 

y = 2
 h − ab 
 bg − hf af − gh 
.
(x , y ) =  2
,
 h − ab h 2 − ab 
Remembering Method (For second point)
a h g
Since,
∆= h b f
g
f
c
Chap 03 Pair of Straight Lines
taking first two rows (repeat first column)
a
h
g
a
h
b
f
ab − h 2 , hf − bg, gh − af
⇒
h 2 − ab, bg − hf , af − gh
⇒
1,
 bg − hf af − gh 
.
Hence, point of intersection is  2
,
 h − ab h 2 − ab 
OR
Cofactors of third column are C 13 , C 23 , C 33
h b
Q
C 13 =
= hf − bg
g f
and
C 33 =
a
g
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
or
C 22 =
a
g
g
c
C 33 =
a h
h b
…(i)
Shifting the origin at (α, β) without rotating the coordinate
axes, the Eq. (i) reduces to
a( X + α ) 2 + 2h( X + α ) (Y + β)
+ b(Y + β) 2 + 2 g( X + α ) + 2 f (Y + β) + c = 0
⇒ (aX + 2hXY + bY ) + 2 X (aα + hβ + g )
2
2
+ 2Y (hα + bβ + f )
+ aα + 2hαβ + bβ + 2 gα + 2 f β + c = 0…(ii)
2
 bg − hf af − hg 
,
 2
.
 h − ab h 2 − ab 
Remembering Mehod (For first point)
Cofactors of leading diagonal are
C 11 , C 22 , C 33
b f
C 11 =
= bc − f 2 ,
Q
f c
= ac − g 2
= ab − h 2
 C 11 C 22 
∴ Point of intersection are 
,

 C 33 C 33 
or
where (α, β) be the point of intersection of the pair of
straight lines represented by Eq. (i).
Proof Since (α, β) be the point of intersection of the lines
represented by
a h
= ab − h 2
h b
 hf − bg hg − af 
,


 ab − h 2 ab − h 2 
i.e.
( x − α ) 2 − ( x − β) 2 ( x − α ) (y − β)
=
(a − b )
h
[Qx = X + α and y =Y + β]
h
= hg − af
f
C C 
Point of intersection is  13 , 23  i.e.
 C 33 C 33 
and
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0,
is
bg − hf af − gh
,
h 2 − ab h 2 − ab
C 23 = −
Theorem 4 The pair of bisectors of the lines
represented by
h
⇒
209
  bc − f 2   ac − g 2  
 

, 
  ab − h 2   ab − h 2  


  f 2 − bc   g 2 − ac  
 

, 
  h 2 − ab   h 2 − ab  


2
This equation represents a pair of straight lines passing
through the new origin. So, it must be homogeneous
equation of second degree in X and Y.
...(iii)
∴
aα + hβ + g = 0
....(iv)
hα + bβ + f = 0
and aα 2 + 2hαβ + bβ2 + 2 gα + 2 f β + c = 0
Now, from Eq. (ii),
aX 2 + 2hXY + bY 2 = 0
…(v)
...(vi)
The equation of the bisectors of the angles between the
lines given by Eq. (vi) is
X 2 − Y 2 XY
...(vii)
=
a −b
h
[with reference to new origin]
Replacing X by x − α and Y by y − β in Eq. (vii), then
( x − α ) 2 − (y − β) 2 ( x − α ) (y − β)
=
(a − b )
h
[with reference to old origin]
which is the required equation of the bisectors of the
angles between the lines given by Eq. (i).
Remark
If
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0
represents two straight lines, then the equation of lines through
the origin and parallel to them is ax 2 + 2hxy + by 2 = 0.
210
Textbook of Coordinate Geometry
y Example 20 For what value of λ does the
equation 12x 2 − 10xy + 2y 2 + 11x − 5y + λ = 0
represent a pair of straight lines? Find their equations
and the angle between them.
Sol. Comparing the given equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0,
11
5
we get a = 12, h = − 5, b = 2, g = , f = − and c = λ
2
2
If the given equation represents a pair of straight lines, then
abc + 2 fgh − af 2 − bg 2 − ch 2 = 0
25
5 11
× ( − 5) − 12 ×
 ×

4
2
2
121
−2×
− λ × 25 = 0
4
λ = 2, also h 2 − ab = 25 − 24 = 1 > 0

⇒ 12 × 2 × λ + 2 ×  −

∴
∴ The given equation will represent a pair of straight lines,
if λ = 2.
To find the two lines
First method
Substituting λ = 2 in the given equation, we get
…(i)
12x 2 − 10xy + 2y 2 + 11x − 5y + 2 = 0
Since,
12x 2 − 10xy + 2y 2 = 2 (3x − y ) (2x − y )
factors of Eq. (i) can be taken as
2 ( 3x − y + l ) ( 2x − y + m )
= 12x 2 − 10xy + 2y 2 + 2 (2l + 3m )x + 2 ( −l − m )y + 2lm
11
5
, l + m = , lm = 1
2
2
1
Solving, we get l = 2, m = .
2
Thus, the factors of Eq. (i) are
1

2(3x − y + 2) 2x − y +  = 0

2
On comparing, 2l + 3m =
or
(3x − y + 2) ( 4 x − 2y + 1) = 0.
∴ The two straight lines represented by the given equation
are
3x − y + 2 = 0 and 4 x − 2y + 1 = 0.
Second Method
Writing Eq. (i) as quadratic equation in x, we get
12x 2 + ( − 10y + 11)x + 2y 2 − 5y + 2 = 0
∴
x=
i.e.
− ( − 10y + 11 ) ±
( −10y + 11)2 − 48 (2y 2 − 5y + 2)
24
24 x = (10y − 11) ± ( 4y 2 + 20y + 25)
= (10y − 11) ± (2y + 5)
∴
24 x = 12y − 6, i.e. 4 x − 2y + 1 = 0
and
24 x = 8y − 16, i.e. 3x − y + 2 = 0
are the required lines.
To find the angle between the lines
If θ be the angle between the lines, then
tan θ =
=
2 h 2 − ab
|a + b |
2 25 − 24 1
=
| 12 + 2|
7
θ = tan
∴
− 1  1
 .
7
y Example 21 Prove that the equation
8 x 2 + 8 xy + 2y 2 + 26 x + 13y + 15 = 0 represents a pair
of parallel straight lines. Also, find the
perpendicular distance between them.
Sol. Given equation is
8x 2 + 8xy + 2y 2 + 26x + 13y + 15 = 0
…(i)
Writing Eq. (i) as quadratic equation in x, we get
8x 2 + 2x ( 4y + 13) + 2y 2 + 13y + 15 = 0
∴ x=
− 2 ( 4y + 13) ± 4 ( 4y + 13)2 − 32 ( 2y 2 + 13y + 15)
16
− ( 4y + 13) ± ( 4y + 13)2 − 8 (2y 2 + 13y + 15)
8
− ( 4y + 13) ± 7
⇒ x=
8
⇒
8x = − 4y − 13 + 7, i.e. 4 x + 2y + 3 = 0
and
8x = − 4y − 13 − 7, i.e. 2x + y + 5 = 0
i.e. the given Eq. (i) represents two straight lines
2x + y + 5 = 0
and
4 x + 2y + 3 = 0
3
i.e.
2x + y + = 0
2
both lines are parallel.
3
5−
7
2
=
∴ Distance between them =
2
2
2
5
2 +1
⇒
x=
Aliter : Here, ∆ = 8 × 2 × 15 + 2 ×
13
× 13 × 4
2
2
 13
− 8 ×   − 2 × (13)2 − 15 × ( 4 )2 = 0
 2
and
h 2 = ( 4 )2 = 16 = 8 × 2 = ab
∴ Given equation
8x 2 + 8xy + 2y 2 + 26x + 13y + 15 = 0
…(i)
represents two parallel straight lines.
Since,
8x 2 + 8xy + 2y 2 = 2(2x + y )2
factors of Eq. (i) can be taken as
2(2x + y + l ) (2x + y + m )
2
= 8x + 8xy + 2y 2 + 2(2m + 2l )x + 2 (m + l ) y + 2lm
Chap 03 Pair of Straight Lines
On comparing, we get l + m =
13
and
2
lm =
15
2
Sol. Given
 g 2 − ca 
 f 2 − bc 
α=  2

 ,β =  2
 h − ab 
 h − ab 
l −m
(l + m )2 − 4lm
=
=
5
( 22 + 12 )
Y
r
For comparing coefficients write equation in form
+
y
+
l
2x
+
y
+
m
X¢
y Example 22 Find the combined equation of the
straight lines passing through the point (1, 1) and
parallel to the lines represented by the equation
x 2 − 5xy + 4 y 2 + x + 2y − 2 = 0.
…(i)
Since, x − 5xy + 4y = ( x − 4y )( x − y )
2
2
Factors of Eq. (i) taken as ( x − 4y + l ) ( x − y + m ) .
Now, equation of line through (1, 1) and parallel to
x − 4y + l = 0 is x − 4y + λ = 0
i.e.
1− 4 + λ =0
∴
λ =3
then line is x − 4y + 3 = 0
and equation of line through (1, 1) and parallel to
x − y + m = 0 is x − y + µ = 0
i.e.
1−1+µ =0
∴
µ =0
then line is
x −y =0
Hence, equation of lines Eqs. (ii) and (iii) is
( x − 4y + 3) ( x − y ) = 0
i.e.
x 2 − 5xy + 4y 2 + 3x − 3y = 0
2h
ca − g 2
ab − h 2
.
C(x2,0)
X
( x − α )2 − (y − β) 2 ( x − α) (y − β)
=
a −b
h
For X-axis, y = 0.
∴
or
− β( x − α )
( x − α) 2 − β 2
=
a−b
h
h ( x − α) 2 + β ( x − α) (a − b ) − h β 2 = 0
…(ii)
Eq. (ii) is a quadratic in ( x − α ) and let two values of x be x 1
and x 2 , so that its roots are
x 1 − α and x 2 − α
− β( a − b )
( x 1 − α) + ( x 2 − α) = Sum of roots =
∴
h
( x 1 − α) ( x 2 − α) = Products of roots = − β 2
x 2 − x 1 = |( x 2 − α) − ( x 1 − α) |
= [( x 2 − α) + ( x 1 − α) ]2 − 4 ( x 2 − α) ( x 1 − α)
…(ii)

 β 2 (a − b )2
∴ | x 2 − x 1| = 
+ 4 β2
2
h


=
…(iii)
represents a pair of lines, prove that the area of
the triangle formed by their bisectors and axis of x
.
M
Hence, equation of the bisectors of the lines given by Eq. (i)
is
∴
y Example 23 If ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
(a − b ) 2 + 4h 2
O B(x1,0)
Y'
coefficient of x is 2m + 2l , cofficient of y is l + m and coefficients
of constant term is lm.
13
15
i.e.
l + m = , lm =
2
2
Sol. Given equation of lines is
x 2 − 5xy + 4y 2 + x + 2y − 2 = 0.
r
Bis
to
ec
ec
to
Bis
Remark
is
A(a,b)
169 60
−
4
2 = 7
5
2 5
2x
…(i)
The point of intersection of the lines given by Eq. (i) are
∴ Distance between them
=
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
211
β
h
( a − b ) 2 + 4h 2
1
2
1
=
2
1
=
2
∴ Area of ∆ABC =
BC AM
x 2 − x1 β
β
h
( a − b ) 2 + 4h 2 × | β |
=
( a − b ) 2 + 4h 2 2
.β
2h
=
(a − b )2 + 4h 2 ca − g 2
.
ab − h 2
2h 
212
Textbook of Coordinate Geometry
Exercise for Session 4
1.
If λx 2 + 10xy + 3y 2 − 15x − 21y + 18 = 0 represents a pair of straight lines. Then, the value of λ is
(a) – 3
2.
(c) 4
(d) – 4
The point of intersection of the straight lines given by the equation 3y 2 − 8xy − 3x 2 − 29x + 3y − 18 = 0 is
1
(a)  1, 
 2
3.
(b) 3
1
(b)  1, − 

2
3 5
(c)  − , 
 2 2
3
5
(d)  − , − 
 2
2
If the equation 12x 2 + 7xy − py 2 − 18x + qy + 6 = 0 represents two perpendicular lines, then the value of p and
q are
(a) 12, 1
4.
(c) 12,
23
2
(d) 12, −
23
2
If the angle between the two lines represented by 2x 2 + 5xy + 3y 2 + 7y + 4 = 0 is tan−1(m), then m is equal to
(a) −
5.
(b) 12, – 1
1
5
(b)
1
5
(c) −
3
5
(d)
3
5
The equation of second degree x 2 + 2 2xy + 2y 2 + 4x + 4 2 y + 1 = 0 represents a pair of straight lines, the
distance between them is
(a) 2
(b) 2 3
(c) 4
(d) 4 3
6.
Find the area of the parallelogram formed by the lines
2x 2 + 5xy + 3y 2 = 0 and 2x 2 + 5xy + 3y 2 + 3x + 4y + 1 = 0.
7.
Find the locus of the incentre of the triangle formed by
xy − 4x − 4y + 16 = 0 and x + y = a (a > 4, a ≠ 4 2 and a is the parameter).
8.
If the equation 2hxy + 2gx + 2fy + c = 0 represents two straight lines, then show that they form a rectangle of
area
fg
with the coordinate axes.
h2
9.
Find the area of the triangle formed by the lines represented by ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 and axis
of x.
10.
Find the equations of the straight lines passing through the point (1, 1) and parallel to the lines represented by
the equation x 2 − 5xy + 4y 2 + x + 2y − 2 = 0.
Session 5
To Find the Point of Intersection of Lines
Represented by ax2 + 2hxy + by2 + 2gx+ 2fy + c = 0
with the Help of Partial Differentiation, Removal
of First Degree Terms, Equation of the Lines
Joining the Origin to the Points of Intersection
of a Given Line and a Given Curve
To Find the Point of Intersection
of Lines Represented by
from first two rows
a h g ⇒ ax + hy + g = 0
and
h b f ⇒ hx + by + f = 0 and then solve.
ax 2 + 2 hxy + by 2 + 2 gx
+ 2 fy + c = 0
y Example 24 Find the point of intersection of lines
represented by 2x 2 − 7 xy − 4 y 2 − x + 22y − 10 = 0.
Let φ ( x , y ) ≡ ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
Sol. Let
∴
and
∂φ
= 2ax + 2hy + 2 g
∂x
∂φ
= 2hx + 2by + 2 f
∂y
[treating y as constant]
and
ax + hy + g = 0 and hx + by + f = 0
y
1
x
Solving them
=
=
f h − bg gh − af ab − h 2
we obtain
 bg − fh af − gh 

(x , y ) =  2
,
 h − ab h 2 − ab 
Working rule In practice, therefore, the general equation of
second degree φ = 0, represents a pair of straight lines, we
∂φ
∂φ
solve its partial derivatives
= 0,
= 0 for their intersecting
∂x
∂y
point ( x , y ) = (α, β).
Remembering Method (without use of partial derivatives)
a h g
Since,
∆= h b f
f
∴
[treating x as constant]
∂φ
∂φ
For point of intersection
= 0 and
= 0,
∂x
∂y
g
φ ≡ 2x 2 − 7 xy − 4y 2 − x + 22y − 10 = 0
c
∂φ
≡ 4 x − 7y − 1 = 0
∂x
∂φ
≡ − 7 x − 8y + 22 = 0
∂y
then, the point of intersection is ( x , y ) = (2, 1).
Removal of First Degree Terms
Let point of intersection of lines represented by
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
…(i)
is (α, β).
 bg − fh af − gh 
Here, (α, β) =  2
,

 h − ab h 2 − ab 
For removal of first degree terms, shift the origin to
(α, β).
i.e. Replacing x by (X + α) and y be (Y + β) in Eq. (i).
Aliter : Direct equation after removal of first degree
terms is
aX 2 + 2hXY + bY 2 + ( gα + f β + c ) = 0
where,
α=
and
β=
bg − fh
h 2 − ab
af − gh
h 2 − ab
.
214
Textbook of Coordinate Geometry
y Example 25 Find the new equation of curve
12x 2 + 7 xy − 12y 2 − 17 x − 31y − 7 = 0 after removing
the first degree terms.
Sol. Let
φ ≡ 12x 2 + 7 xy − 12y 2 − 17 x − 31y − 7 = 0
…(i)
∂φ
= 24 x + 7y − 17 = 0
∂x
∂φ
≡ 7 x − 24y − 31 = 0
∂y
∴
and
Their point of intersection is ( x , y ) ≡ (1, − 1)
Here,
α = 1, β = − 1
Shift the origin to (1, − 1) then replacing x = X + 1 and
y = Y − 1 in Eq. (i) the required equation is
12( X + 1)2 + 7( X + 1) (Y − 1) − 12(Y − 1)2 − 17( X + 1)
− 31(Y − 1) − 7 = 0
i.e. 12X 2 + 7 XY − 12Y 2 = 0
Aliter :
Here,
α = 1 and β = − 1
17
31
and
g = − , f = − ,c = − 7
2
2
31
17
gα + f β + c = −
×1−
× −1−7 =0
∴
2
2
∴ Removed equation is
aX 2 + 2hXY + by 2 + ( g α + f β + c ) = 0
3 π
π

we can also take θ = 4 
4
1
1
⇒
cos θ =
, sin θ =
2
2
Now, if ( X ′ , Y ′ ) be the coordinates of the point when the
π
axes are rotated through θ = , we have
4
X ′ −Y ′
X ′ +Y ′
,Y =
X =
2
2
then Eq. (i) becomes
Y ′2 X ′2
Y ′2 X ′2
−
=1 ⇒
− 2 =1
6
2
b2
a
where,
a 2 = 2 and b 2 = 6.
θ=
Taking
Equation of the Lines Joining
the Origin to the Points of
Intersection of a Given Line
and a Given Curve
Theorem The combined equation of the straight lines
joining the origin to the points of intersection of a second
degree curve
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
12X 2 + 7 XY − 12Y 2 + 0 = 0
12X 2 + 7 XY − 12Y 2 = 0.
y Example 26 Transform the equation x 2 + 4 xy + y 2
Y ′2
X ′2
− 2x + 2y + 4 = 0 into the form 2 − 2 = 1.
b
a
Sol. To remove the first degree terms, we shift the origin to the
point (α, β).
bg − fh 1 × ( − 1) − 1 × 2
Then,
= −1
=
α= 2
4 −1×1
h − ab
af − gh 1 × 1 − ( − 1) × (2)
and
=1
=
β= 2
4 −1×1
h − ab
then, the transformed equation is
X 2 + 4 XY + Y 2 + ( g α + f β + c ) = 0
and a straight line lx + my + n = 0 is :
 lx + my 
 lx + my 
ax 2 + 2hxy + by 2 + 2 gx 
 + 2 fy 

 −n 
 −n 
2
 lx + my 
+c 
 =0
 −n 
Proof The equation of the curve ( PAQ ) is
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
…(i)
and the equation of the line PQ be
lx + my + n = 0
…(ii)
Y
P
⇒ X + 4 XY + Y + ( − 1 × ( − 1) + 1 × 1 + 4 ) = 0
2
∴
2
X 2 + 4 XY + Y 2 + 6 = 0
…(i)
Now, to remove the XY term from Eq. (i), we rotate the axes
through an angle θ given by
a−b
[Q here a = b]
cot2θ =
=0
2h
⇒
cot2θ = 0
π
3π
or
2θ =
⇒
2
2
3π
π
or
θ=
⇒
4
4
A
Q
X′
X
O
Y′
From the equation of the line Eq. (ii), find the value of ‘1’
in terms of x and y,
lx + my
…(iii)
i.e.
=1
−n
Chap 03 Pair of Straight Lines
If θ is the acute angle between pair of lines of Eq. (iii), then
Now, the Eq. (i) can be written as
or ax + 2hxy + by + (2 gx + 2 fy )(1) + c (1) = 0
2
or
2
2
tan θ =
 lx + my 
ax 2 + 2hxy + by 2 + (2 gx + 2 fy ) 

 −n 
2
 lx + my 
+c 
 = 0 …(iv)
 −n 
lx + my
from Eq. (iii)]
[replacing 1 by
−n
Hence, the Eq. (iv) is homogeneous equation of second
degree. Above Eq. (iv) on simplification will be of the form
Ax 2 + 2 Hxy + By 2 = 0 and will represent the required
straight lines. If θ be the angle between them, then
 2 ( H 2 − AB ) 

θ = tan −1 
  A + B 


y Example 27 Prove that the angle between the lines
joining the origin to the points of intersection of the
straight line y = 3x + 2 with the curve
2 2
x 2 + 2xy + 3y 2 + 4 x + 8 y − 11 = 0 is tan −1 
.
 3 
Sol. Equation of curve is x 2 + 2xy + 3y 2 + 4 x + 8y − 11 = 0 …(i)
y − 3x
…(ii)
=1
2
Making Eq. (i) homogeneous with the help of Eq. (ii), then
 y − 3x 
 y − 3x 
x 2 + 2xy + 3y 2 + 4 x 

 + 8y 

 2

 2
y = 3x + 2 ⇒
2
 y − 3x 
−11 
 =0

 2
⇒
x + 2xy + 3y + 2xy − 6x + 4y − 12xy
2
2
2
−
2
11
( y − 3x ) 2 = 0
4
⇒
11 2
(y − 6xy + 9 x 2 ) = 0
4
− 20x 2 − 32xy + 28y 2 − 11y 2 + 66xy − 99 x 2 = 0
⇒
119 x 2 − 34 xy − 17y 2 = 0
⇒
7 x 2 − 2xy − y 2 = 0
⇒
− 5x 2 − 8xy + 7y 2 −
a + b
=
2 (1 + 7 )
7 − 1
=
2 8 4 2
2 2
=
=
6
6
3
2 2
θ = tan −1 

 3 
y Example 28 Find the condition that the pair of
straight lines joining the origin to the intersections of
the line y = mx + c and the circle x 2 + y 2 = a 2 may be
at right angles.
Sol. The equations of the line and the circle are
y = mx + c
2
and
x + y 2 = a2
⇒ x 2 (c 2 − a 2m 2 ) + 2ma 2 xy + y 2 (c 2 − a 2 ) = 0
…(i)
…(ii)
…(iii)
…(iii)
The lines given by Eq. (iii), are at right angles, then
coefficient of x 2 + coefficient of y 2 = 0
⇒ c 2 − a 2m 2 + c 2 − a 2 = 0
2c 2 = a 2 (1 + m 2 )
∴
which is the required condition.
y Example 29 Prove that the pair of lines joining the
2
x2 y
origin to the intersection of the curve 2 + 2 = 1 by
a
b
the line lx + my + n = 0 are coincident, if a
a 2l 2 + b 2m 2 = n 2 .
x2
x2
⇒
⇒
n 2x 2
a
2
+
+
+
y2
…(i)
=1
a2 b2
and line
lx + my + n = 0
⇒
lx + my = − n
lx + my
…(ii)
⇒
=1
−n
Making Eq. (i) homogeneous with the help of Eq. (ii), then
Sol. The given curve is
a2
This is the equation of lines joining the origin to the points
of intersection of Eqs. (i) and (ii).
Comparing Eq. (iii) with ax 2 + 2hxy + by 2 = 0
a = 7, h = − 1, b = − 1
∴
2 h 2 − ab
The pair of straight lines joining the origin to the
intersections of Eqs. (i) and (ii), is obtained by making
homogeneous Eq. (ii) with the help of Eq. (i).
y − mx
Q
y = mx + c ⇒
=1
c
2
 y − mx 
∴
x 2 + y 2 = a 2 ( 1) 2 ⇒ x 2 + y 2 = a 2 

 c 
Hence, the equation of pairs of straight lines passing
through the origin and the points of intersection of a
curve and a line is obtained by making the curve
homogeneous with the help of the line.
and line
215
 lx + my 
=

b 2  −n 
y2
n 2y 2
b2
2
= l 2 x 2 + m 2y 2 + 2lmxy
n2
n2
2 2
2 2
 2 − l  x − 2lmxy +  2 − m  y = 0

b

a
…(iii)
216
Textbook of Coordinate Geometry
This is of the form Ax 2 + 2Hxy + By 2 = 0,
A=
then
B=
n
2
2
a
n2
y Example 30 Show that the straight lines joining the
origin to the points of intersection of curves
ax 2 + 2hxy + by 2 + 2gx = 0
and
a ′ x 2 + 2h ′ xy + b ′ y 2 + 2g ′ x = 0
are at right angles, if g ′ (a + b ) = g (a ′ + b ′ ).
− l 2 , H = − lm
− m2
b2
The lines given by Eq. (iii) will be coincident, if
H 2 − AB = 0 ⇒ H 2 = AB
and
n2

l 2m 2 =  2 − l 2 

a
⇒
l m =
⇒
2
2
n4
⇒
2 2
=
n
4
a 2b 2
n 2m 2
2
−
+
Sol. The two curves meet in two points and the required lines
joining the origin to these points will be obtained by making
one equation homogeneous with the help of the other.
…(i)
ax 2 + 2hxy + by 2 + 2gx = 0
n2
2
 2 −m 
b

2
n m
2
a2
−
2 2
l n
b2
a ′ x 2 + 2h ′ xy + b ′ y 2 + 2g ′ x = 0
+l m
2
2
n 2l 2
⇒
ab
a
b
2
2 2
2 2
n =b m +a l
⇒
a 2l 2 + b 2m 2 = n 2
…(ii)
Multiplying Eq. (i) by g ′ and Eq. (ii) by g and subtracting,
we get (ag ′ − a ′ g )x 2 + (2hg ′ − 2h ′ g )xy + (bg ′ − b ′ g )y 2 = 0
If the lines are at right angles, then coefficient of
x 2 + coefficient of y 2 = 0
2
⇒
∴
ag ′ − a ′ g + bg ′ − b ′ g = 0
(a + b )g ′ = (a ′ + b ′ )g .
Exercise for Session 5
1.
If the straight lines joining origin to the points of intersection of the line x + y = 1with the curve
x 2 + y 2 + x − 2y − m = 0 are perpendicular to each other, then the value of m should be
(a) −
2.
1
2
(b) 0
(c)
1
2
(d) 1
The pair of straight lines joining the origin to the common points of x 2 + y 2 = 4 and y = 3x + c are
perpendicular, if c 2 is equal to
(a) –1
3.
(a − b )
2h
(d) 20
(b)
2h
(a + b )
(c)
(a + b )
2h
(d)
2h
(a − b )
The lines joining the origin to the points of intersection of 2x 2 + 3xy − 4x + 1 = 0 and 3x + y = 1 are given by
(a) x 2 − y 2 − 5xy = 0
5.
(c) 13
Mixed term xy is to be removed from the general equation of second degree
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, one should rotate the axes through an angle θ, then tan 2θ is equal to
(a)
4.
(b) 6
(b) x 2 − y 2 + 5xy = 0
(c) x 2 + y 2 − 5xy = 0
(d) x 2 + y 2 + 5xy = 0
The equation of the line joining the origin to the point of intersection of the lines 2x 2 + xy − y 2 + 5x − y + 2 = 0 is
(a) x + y = 0
(b) x − y = 0
(c) x − 2y = 0
(d) 2x + y = 0
6.
Find the equation of the lines joining the origin to the points of intersection of 3x − 2y = 1 with
3x 2 + 5xy − 3y 2 + 2x + 3y = 0 and show that they are at right angles.
7.
If the straight line joining the origin and the points of intersection of y = mx + 1and x 2 + y 2 = 1 be perpendicular
to each other, then find the value of m.
8.
Prove that the straight lines joining the origin to the points of intersection of the straight line kx + hy = 2hk with
the curve ( x − h )2 + ( y − k )2 = c 2 are at right angles, if h 2 + k 2 = c 2.
9.
Show that for all values of λ, the lines joining the origin to the points common to x 2 + 2hxy − y 2 + gx + fy = 0
and fx − gy = λ are at right angles.
10.
Find the equations of the straight lines joining the origin to the points of intersection of x 2 + y 2 − 4x − 2y = 0
and x 2 + y 2 − 2x − 4y = 4.
Shortcuts and Important Results to Remember
1 If slope of one of the lines represented by
ax 2 + 2 hxy + by 2 = 0 should be n times the slope of the
other, then 4nh 2 = ab(1 + n )2 .
2 If the slope of one of the lines represented by
ax 2 + 2 hxy + by 2 = 0 be the nth power of the other, then
(abn )1/ n + 1 + (an b)1/ n + 1 + 2 h = 0
3 If two of the three lines represented by
ax 3 + bx 2 y + cxy 2 + dy 3 = 0 may be at right angles, then
a2 + ac + bd + d 2 = 0
4 If pairs of straight lines x 2 + 2 m1 xy − y 2 = 0 and
x 2 + 2 m2 xy − y 2 = 0 be such that each pair bisects the
angle between the other pair, then m1m2 = − 1
5 If the equation ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0
represents a pair of parallel lines, then
(i) h 2 = ab, bg 2 = af 2 .
(ii) the distance between them = 2
(g 2 − ac )
.
a(a + b)
6 If ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = 0 and
ax 2 + 2 hxy + by 2 − 2 gx − 2 fy + c = 0 each represent a pair
of lines, then the area of the parallelogram enclosed by
2|c|
them is
.
(h 2 − ab )
JEE Type Solved Examples :
Single Option Correct Type Questions
n
This section contains 6 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
ONLY ONE is correct.
Ex. 1 If the pairs of lines x 2 + 2 xy + ly 2 = 0 and
lx 2 + 2 xy + y 2 = 0 have exactly one line in common, then
the joint equation of the other two lines is given by
l
(a) 3 x 2 + 8 xy - 3y 2 = 0
(b) 3 x 2 + 10xy + 3y 2 = 0
(c) x 2 + 2xy - 3y 2 = 0
(d) 3 x 2 + 2xy - y 2 = 0
Ex. 3 The pair of lines 3 x 2 - 4 xy + 3y 2 = 0 are
p
rotated about the origin by in the anti-clockwise sense.
6
The equation of the pair in the new position is
l
Þ
x 2 + 2xy + ly 2 = x 2 + 2xy - 3y 2 = ( x - y ) ( x + 3y )
and lx 2 + 2xy + y 2 = - 3x 2 + 2xy + y 2 = - ( x - y ) (3x + y )
Hence, the joint equation of other two lines is
( x + 3y ) (3x + y ) = 0 or 3x 2 + 10xy + 3y 2 = 0.
Ex. 2 The combined equation of the lines l1 and l 2 is
2 x 2 + 6 xy + y 2 = 0 and that of the lines m1 and m 2 is
4 x 2 + 18 xy + y 2 = 0. If the angle between l1 and m 2 is a ,
then the angle between l 2 and m1 will be
l
p
(a) - a
2
p
(b) + a
4
(c) a
(d) 2a
Sol. (c) The combined equation of the bisectors of the angles
between the lines of the first pair is
x 2 - y 2 xy
1
=
Þ x 2 - y 2 = xy
3
2-1
3
and the combined equation of the bisectors of the angles
between the lines of the second pair is
x 2 - y 2 xy
1
=
Þ x 2 - y 2 = xy
3
4 -1
9
It is clear that the two pairs are equally inclined to each
other.
Hence, the angle between l 2 and m1 is a.
(d) 3y 2 - xy = 0
y = 3x and y =
Þ
1
x
3
Þ
y = x tan 60° and y = x tan 30°
After rotation, the separate equations are
y = x tan 90° and y = x tan 60°
m2
m
1
=
=
2
2( l - 1) (1 - l ) 2( l - 1)
1
m 2 = 1 and m = - ( l + 1)
2
( l + 1) 2 = 4 Þ l + 1 = 2 , - 2
\
l = 1 or - 3
But for l = 1, the two pairs have both the lines common.
So, l = - 3 and the slope m of the line common to both the
pairs is 1. Now,
(c) 3 x 2 - xy = 0
( 3x - y ) ( x - 3y ) = 0
lm 2 + 2m + 1 = 0 and m 2 + 2m + l = 0
Þ
(b) y 2 - 3 xy = 0
Sol. (c) The given equation of pair of straight lines can be written
as
Sol. (b) Let y = mx, be a line common to the given pairs of lines,
then
Þ
(a) x 2 - 3 xy = 0
Þ
x = 0 and y = x 3
Hence, the combined equation in the new position is
x ( 3x - y ) = 0
or
3x 2 - xy = 0.
Ex. 4 If the pair of lines ax 2 - 2 xy + by 2 = 0 and
bx 2 - 2 xy + ay 2 = 0 be such that each pair bisects the angle
between the other pair, then | a - b | equals to
l
(a) 1
(c) 3
(b) 2
(d) 4
Sol. (b) According to the example, the equation of the bisectors of
the angle between the lines
ax 2 - 2xy + by 2 = 0
is
2
2
bx - 2xy + ay = 0
...(i)
...(ii)
\The equation of bisectors of the angle between the lines
Eq. (i) is
x 2 - y 2 xy
=
a-b
-1
Þ
x 2 + (a - b )xy - y 2 = 0
...(iii)
Since, Eqs. (ii) and (iii) are identical, comparing Eqs. (ii) and
(iii), we get
-2
b
a
=
=
1 a-b -1
Þ
and
(a - b )b = - 2
(a - b )a = 2
\
(a - b )2 = 4
or
a-b =2
Chap 03 Pair of Straight Lines
Ex. 5 The equation of line which is parallel to the line
common to the pair of lines given by 3 x 2 + xy - 4y 2 = 0 and
6 x 2 + 11xy + 4y 2 = 0 and at a distance of 2 units from it is
l
(a) 3 x - 4y = - 10
(c) 3 x + 4y = 10
2
(b) x - y = 2
(d) 2x + y = - 2
2
2
2
(c) g - f
=c
(b) g 2 - f
2
=c
= 2c
2
2
= c2
(d) g + f
Sol. (c) Given,
x 2 + y 2 + 2gx + c = 0
and
2
Sol. (c) We have, 3 x + xy - 4y = 0
or
and
(a) g 2 + f
( x - y ) ( 3x + 4y ) = 0
6x 2 + 11xy + 4y 2 = 0
...(i)
or
( 2x + y ) ( 3x + 4y ) = 0
Equation of line common to Eqs. (i) and (ii) is
3x + 4y = 0
Equation of any line parallel to Eq. (iii) is
3x + 4y = l
Since, its distance from Eq. (iii) is 2, we have
l -0
= 2 or l = ± 10
( 32 + 4 2 )
...(ii)
219
2
...(i)
2
...(ii)
x + y + 2 fy - c = 0
On subtracting Eq. (ii) from Eq. (i), we get
2gx - 2 fy + 2c = 0
fy - gx
or
=1
c
On adding Eqs. (i) and (ii), we get
...(iii)
...(iii)
2( x 2 + y 2 + gx + fy ) = 0
or
x 2 + y 2 + gx + fy = 0
...(iv)
Homogenising Eq. (iv) with the help of Eq. (iii), then
æ fy - gx ö
x 2 + y 2 + ( gx + fy ) ç
÷ =0
ø
è c
Hence, required lines are 3x + 4y = ± 10.
\The lines will be at right angles, when
æ
g2 ö æ
f 2ö
ç1 ÷ + ç1 +
÷ =0
c ø è
c ø
è
Ex. 6 The lines joining the origin to the points of
intersection of x 2 + y 2 + 2 gx + c = 0 and
x 2 + y 2 + 2 fy - c = 0 are at right angles, if
l
g2 - f
Þ
2
= 2c
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 3 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
Ex. 7 The lines joining the origin to the point of
intersection of 3 x 2 + mxy - 4 x + 1 = 0 and 2 x + y - 1 = 0 are
at right angles. Which of the following is/are possible value
of m?
l
(a) - 4
(b) 3
(c) 4
(d) 7
Sol. (a,b,c,d) Given line is 2 x + y = 1
and curve is 3x 2 + mxy - 4 x + 1 = 0
...(i)
...(ii)
Ex. 8 The lines (lx + my ) 2 - 3(mx - ly ) 2 = 0 and
lx + my + n = 0 form
l
(a) an isosceles triangle (b) a right angled triangle
(c) an equilateral triangle (d) None of these
Sol. (a,c) (lx + my ) 2 - 3(mx - ly ) 2 = 0
Þ (l 2 - 3m 2 )x 2 + 8mlxy + (m 2 - 3l 2 )y 2 = 0
Þ {(l + m 3 )x + (m - l 3 )y } {(l - 3m )x + (m + 3l )y } = 0
Let slope of (l + m 3 )x + (m - l 3 )y = 0 be m1 and slope of
(l - m 3 )x + (m + l 3 )y = 0 be m 2 and slope of
lx + my + n = 0 is m 3 .
Homogenising Eq. (ii) with the help of Eq. (i), then
3x 2 + mxy - 4 x (2x + y ) + (2x + y )2 = 0
or
Now,
- x 2 + mxy + y 2 = 0
m 3 - m1
tan q 1 =
=
1 + m 3m1
the lines are at right angles as a + b = 0, when h 2 > ab
m2
i.e.
+1>0
4
which is true for all m Î R.
=
Þ
3( l 2 + m 2 )
(l 2 + m 2 )
q 1 = 60°
l
l +m
+
m m -l
l æl + m
1+ ×ç
m èm - l
-
= 3
3
3
3ö
÷
3ø
220
Textbook of Coordinate Geometry
m 2 - m1
= 3
1 + m 2m1
Similarly, tan q 2 =
Þ
q 2 = 60°
Hence, an equilateral triangle is formed which is also an
isosceles one.
Ex. 9 If the equation ax 2 - 6 xy + y 2 + bx + cy + d = 0
represents a pair of lines whose slopes are m and m 2 , then
the value(s) of a is/are
l
(a) - 27
(c) 8
(b) - 8
(d) 27
Sol. (a,c) Q m and m 2 are the roots of the equation
2
æy ö
æy ö
ç ÷ -6ç ÷ +a=0
èx ø
èx ø
m + m 2 = 6 and m × m 2 = a
\
2 3
Now,
( m + m ) = ( 6)
3
6
...(i)
3
2
Þ m + m + 3m × m (m + m 2 ) = 216
Þ
a + a 2 + 3a (6) = 216
Þ
2
a + 19a - 216 = 0
Þ
\
(a + 27 ) (a - 8) = 0
a = - 27, 8
[from Eq. (i)]
JEE Type Solved Examples :
Paragraph Based Questions
n
This section contains one solved paragraph based upon
each paragraph 2 multiple choice questions have to be
answered. Each of these questions has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
Paragraph
(Q. Nos. 10 to11)
Consider the equation of a pair of straight lines as
lxy - 8x + 9 y - 12 = 0
10. The value of l is
(a) 0
(c) 4
(b) 2
(d) 6
11. The point of intersection of lines is (a, b), then the
equation whose roots are a, b, is
(a) 4 x 2 + x - 8 = 0
(b) 6 x 2 + x - 12 = 0
(c) 4 x 2 - x - 8 = 0
(d) 6 x 2 - x - 12 = 0
Sol. Given equation is,
Here,
lxy - 8x + 9y - 12 = 0
a = 0, b = 0, c = - 12,
9
l
f = , g = - 4 and h =
2
2
10. (d) For D = 0,
0+2´
Þ
\
Hence,
9
l
l2
´ - 4 ´ - 0 - 0 + 12 ´
=0
2
2
4
3l2 - 18l = 0
l = 0, 6
l =6
[Q for l = 0, it will give an
equation of first degree]
11. (b) Let f ( x, y ) = 6 xy - 8 x + 9y - 12
\
¶f
¶f
= 6y - 8 and
= 6x + 9
¶x
¶y
¶f
¶f
= 0 and
= 0, we get
¶x
¶y
3
4
x = - ,y =
2
3
3
4
\
a = - and b =
2
3
Hence, required equation is
æ 3 4ö
x 2 - ç- + ÷x - 2 = 0
è 2 3ø
For point of intersection
Þ
6x 2 + x - 12 = 0
[Q l = 6]
Chap 03 Pair of Straight Lines
221
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
This section contains 2 examples. The answer to each
example is a single digit integer, ranging from 0 to 9
(both inclusive).
Ex. 12 If the sum of the slopes of the lines given by
x 2 - 2cxy - 7y 2 = 0 is four times their product, then the
value of c is
l
Sol. (2) Given, m1 + m2 = 4m1m2
Þ
\
Ex. 13 If one of the lines given by 6 x 2 - xy + 4cy 2 = 0 is
3 x + 4y = 0, then the value of | c | is
l
Sol. (3) Q 3 x + 4y = 0 is one of the lines given by
6 x 2 - xy + 4cy 2 = 0, then
2
Þ
2h
a
= 4 × Þ h = - 2a
b
b
-c = -2´1 Þ c =2
3x ù
é
êëQy = - 4 úû
æ 3x ö
æ 3x ö
6x 2 - x ç ÷ =0
÷ + 4c ç è 4 ø
è 4 ø
6+
Hence,
3 9c
+
=0 ; \ c = -3
4
4
|c| = 3
JEE Type Solved Examples :
Matching Type Questions
n
l
This section contains one solved example. This example
has four statements (A, B, C and D) is Column I and four
statements (p, q, r and s) in Column II. Any given
statements in Column I can have correct matching with
one or more statement(s) given in Column II.
Ex. 14 Match the following
Column I
Column II
(A) If the slope of one of the lines represented by
ax 2 + 2hxy + by 2 = 0 is the square of the
a + b 8h 2
other, then
is a
+
h
ab
(p) Odd prime
number
(B) The product of perpendiculars drawn from
the point (1, 2 ) to the pair of lines
x 2 + 4 xy + y 2 = 0 is l units, then [ l ] is a
(where [×] denotes the greatest integer
function).
(q) Composite
number
(C) Distance between two lines represented by
the line pairx 2 - 4 xy + 4y 2 + x - 2y - 6 = 0
is l unit, then [ l ] is a (where [×] denotes the
greatest integer function).
(r) Even prime
number
(D) If the pairs a1x 2 + 2h1xy + b1y 2 = 0 and
a 2x 2 + 2h2xy + b2y 2 = 0 have one line
(a1b2 - a 2b1 ) 2
common, then
is a
(b1h2 - b2h1 ) (a 2h1 - a1h2 )
(s) Perfect
number
m + m2 = -
(A) Let m and m 2 be the slopes of the lines represented by
ax 2 + 2hxy + by 2 = 0, then
...(i)
a
b
...(ii)
æ 2h ö
From Eq. (i), (m + m 2 )3 = ç - ÷
è bø
3
Þ m 3 + m 6 + 3m × m 2 (m + m 2 ) = 2
Þ
Þ
Sol. (A) ® (q, s); (B) ® (p); (C) ® (r); (D) ® (q)
m × m2 =
and
2h
b
8h 3
b3
3
a a
3a æ 2h ö
8h
+ 2 +
ç- ÷ = - 3
ø
è
b b
b
b
b
ab + a 2 - 6ah = -
[from Eqs. (i) and (ii)]
8h 3
b
a + b 8h 2
+
=6
h
ab
(B) Let y = m1x and y = m 2 x lines represented by
x 2 + 4 xy + y 2 = 0, then m1 + m 2 = - 4, m1m 2 = 1. Then,
product of perpendiculars drawn from the point (1, 2)
(2 - m1 ) (2 - m 2 )
=
(1 + m12 ) (1 + m 22 )
\
=
=
Q
\
4 - 2 (m1 + m 2 ) + m1m 2
1 + (m1 + m 2 )2 - 2m1m 2 + (m1m 2 )2
4 +8+1
(1 + 16 - 2 + 1)
13
4
[l] = 3
l=
=
13
units
4
222
Textbook of Coordinate Geometry
(C) Given, line pair is x 2 - 4 xy + 4y 2 + x - 2y - 6 = 0
2
Þ
( x - 2y ) + ( x - 2y ) - 6 = 0
(D) Let y = mx be the common line, then
b1m 2 + 2h1m + a1 = 0 and b 2m 2 + 2h 2m + a 2 = 0.
Þ
( x - 2y + 3) ( x - 2y - 2) = 0
\ Lines are x - 2y + 3 = 0 and x - 2y - 2 = 0
3 - ( - 2)
Hence, distance between lines =
= 5 unit
1+ 4
\
Þ
Þ
\
l= 5
Hence,
b1
2h1
b 2 2h 2
´
2h1
a1
2h 2 a 2
=
a1
b1
2
a2 b2
4(b1h 2 - b 2h1 ) (a 2h1 - a1h 2 ) = (a1b 2 - a 2b1 )2
(a1b 2 - a 2b1 )2
=4
(b1h 2 - b 2h1 ) (a 2h1 - a1h 2 )
[ l] = [ 5 ] = 2
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions (Ex. Nos. 15-16) Each of these examples contains
two statements.
Statement I (Assertion) and
Statement II (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice, as given below :
(a) Statement I is true, Statement II is true and Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true and Statement II is
not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Ex. 15 Statement I The combined equation of l1 , l 2 is
3 x 2 + 6 xy + 2y 2 = 0 and that of m1 , m 2 is
5 x 2 +18 xy + 2y 2 = 0. If angle between l1 , m 2 is q, then angle
between l 2 , m1 is q.
Statement II If the pairs of lines l1l 2 = 0, m1m 2 = 0 are
equally inclinded that angle between l1 and m 2 = angle
between l 2 and m1 .
Sol. (a) The pair of bisectors of 3 x 2 + 6 xy + 2y 2 = 0 is
x 2 - y 2 xy
= , i.e. 3 x 2 - xy - 3y 2 = 0 and pair of bisectors of
3 -2
3
x 2 - y 2 xy
2
5 x + 18 xy + 2y 2 = 0 is
=
5 -2
9
i.e. 3 x 2 - xy - 3y 2 = 0 are coincides.
\ Angle between l1, m2 is same as angle between l 2, m1.
\ Both statements are true and Statement II is a correct
explanation for Statement I.
Ex. 16 Statement I The equation
2 x 2 - 3 xy - 2y 2 + 5 x - 5y + 3 = 0 represents a pair of
perpendicular straight lines.
Statement II A pair of lines given by
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 are perpendicular if
a + b = 0.
Sol. (d) Here, D ¹ 0, h 2 > ab and a + b = 0
\
2x 2 - 3xy - 2y 2 + 5x - 5y + 3 = 0
represents a rectangular hyperbola.
\ Statement I is false and Statement II is true.
Subjective Type Examples
n
In this section, there are 10 subjective solved examples.
Ex. 17 If the lines represented by 2 x 2 - 5 xy + 2y 2 = 0 be
the two sides of a parallelogram and the line 5 x + 2y = 1 be
one of its diagonal. Find the equation of the other diagonal,
and area of the parallelogram.
l
Sol. Given pair of lines, 2 x 2 - 5 xy + 2y 2 = 0
i.e.
\
( x - 2y ) (2x - y ) = 0
x - 2y = 0 and 2x - y = 0
and the diagonal 5x + 2y = 1 does not pass through origin,
hence it is AC.
On solving, x - 2y = 0 and 5x + 2y = 1, we get
æ1 1 ö
A=ç , ÷
è 6 12 ø
and solving 2x - y = 0 and 5x + 2y = 1 , we get
æ1 2ö
C º ç , ÷.
è9 9 ø
223
Chap 03 Pair of Straight Lines
B
Y
Þ {(1 + m1 )y - (m1 - 1)x } {(1 + m 2 )y - (m 2 - 1)x } = 0
Þ (1 + m1 ) (1 + m 2 )y 2 - xy {(1 + m1 )(m 2 - 1)
2x–
+ (m1 - 1) (1 + m 2 )} + (m1 - 1) (m 2 - 1)x 2 = 0
H
y =1
X′
5x+2
y=
0
C
y=0
x–2
Þ (1 + m1 + m 2 + m1m 2 )y 2 - 2xy (m1m 2 - 1)
A
O
+ {m1m 2 - (m1 + m 2 ) + 1} x 2 = 0
X
Þ
Y′
Since, diagonals of parallelogram bisect each other, if bisect
at H .
Then,
ì 1 æ 1 1 ö 1 æ 1 2 öü
H º í ç + ÷ , ç + ÷ý
î 2 è 6 9 ø 2 è 12 9 øþ
i.e.
æ 5 11 ö
H =ç , ÷
è 36 72 ø
\
2h a
æ
+
ç1 è
b b
ö
ö æ a 2h
æa
ö 2
+ 1÷ x 2 = 0
÷y - 2xy ç - 1÷ + ç +
ø
ø èb b
èb
ø
(a + 2h + b )x 2 - 2(a - b )xy + (a - 2h + b )y 2 = 0
Ex. 19 If u º ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
represents a pair of straight lines, prove that the equation of
the third pair of straight lines passing through the points
where these meet the axes is cu + 4( fg - ch ) xy = 0.
l
Hence, the equation of other diagonal which passes through
O and H is
11
-0
( x - 0)
y - 0 = 72
5
-0
36
11
Þ
y= x
10
Þ
11x - 10y = 0
Now, area of parallelogram = 2 ´ Area of D OAC
11 2 1 1
1
sq units
=2´
´ - ´
=
2 6 9 9 12 36
Ex. 18 Prove that the equation
(a + 2h + b ) x 2 - 2(a - b ) xy + (a - 2h + b ) y 2 = 0 represents
a pair of lines each inclined at an angle of 45° to one or
other of the lines given by, ax 2 + 2hxy + by 2 = 0.
Sol. u º ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
Q u represents a pair of straight line, then D = 0
\
abc + 2 fgh - af
2
- bg 2 - ch 2 = 0
…(i)
Combined equation of axes is xy = 0
Now, the curve through the intersection of
u = 0 and xy = 0 is
u + lxy = 0
…(ii)
Y
Third pair
l
Sol. Given equation is
ax 2 + 2hxy + by 2 = 0
…(i)
Let the lines represented by Eq. (i) are
…(ii)
y - m1x = 0
and
…(iii)
y - m2 x = 0
2h
a
therefore, m1 + m 2 = - , m1m 2 =
b
b
If m1 = tanq 1 and m 2 = tanq 2 , the equation of lines through
the origin making 45° with the lines Eqs. (ii) and (iii) will be
y - x tan(q 1 - 45° ) = 0
and
y - x tan(q 2 - 45° ) = 0
Their combined equation is
{y - x tan(q 1 - 45° )} {y - x tan(q 2 - 45° ) } = 0
æ
m -1 ö
x÷
Þ çy - 1
è
1 + m1 ø
æ
m2 - 1 ö
x÷ = 0
çy è
1 + m2 ø
X'
X
O
Y'
2
i.e.
ax + 2hxy + by 2 + 2gx + 2 fy + c + lxy = 0
Þ
ax 2 + xy (2h + l ) + by 2 + 2gx + 2 fy + c = 0
If it represents a pair of lines, then
lö
æ
abc + 2 fg çh + ÷ - af
è
2ø
Þ (abc + 2 fgh - af
2
2
2
lö
æ
- bg 2 - c çh + ÷ = 0
è
2ø
- bg 2 - ch 2 ) + l( fg - ch ) -
cl2
=0
4
cl2
[from Eq. (i)]
=0
4
4( fg - ch )
…(iii)
\
l=
c
Hence, the equation of third pair from Eqs. (ii) and (iii) is
4( fg - ch )
xy = 0
u+
c
\
cu + 4( fg - ch ) xy = 0
Þ
0 + l ( fg - ch ) -
224
Textbook of Coordinate Geometry
we show that by an example that these conditions are not
sufficient because
Ex. 20 If the equation ax 2 + 2hxy + by 2 + 2 gx
+ 2 fy + c = 0 represents a pair of parallel lines, prove that
(i) h = ab and g b = f a or (h = - ab and
g b = - f a ).
æ g 2 - ac ö
÷.
(ii) the distance between them is 2 ç
è a (a + b ) ÷ø
l
h2 = ab Þ
2
and
Þ
bg = af
h=
ab, g
ab
Þ g b=±f a
b = f a or
( h = - ab, g b = - f a)
Here, a = 4, h = 2, b = 1 , g = 2, f = - 1, c = 5 and
h=
ab, g b = - f a
But abc + 2fgh - af 2 - bg 2 - ch2
…(i)
Let the equation of the parallel lines represented
by Eq. (i) be
lx + my + n = 0 and lx + my + n1 = 0.
Then, (lx + my + n ) (lx + my + n1 )
º ax 2 + 2hxy + by 2 + 2gx + 2 fy + c
h= ±
Consider, for example, 4 x 2 + 4 xy + y 2 + 4 x - 2 y + 5 = 0
Sol. Given equation is,
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
2
= 4 ´ 1 ´ 5 + 2 ´ ( -1) ´ 2 ´ 2 - 4 ´ ( -1) 2 - 1 ´ ( 2) 2 - 5 ´ ( 2) 2
= 20 - 8 - 4 - 4 - 20 = - 16 ¹ 0
Hence, the Eq. (i) does not represent a pair of lines at all.
nn1 = c
…(iv)
2lm = 2h
…(v)
Ex. 21 A parallelogram is formed by the lines
ax 2 + 2hxy + by 2 = 0 and the lines through ( p, q ) parallel to
them. Show that the equation of the diagonal of the
parallelogram which does not pass through origin is
( 2 x - p )(ap + hq ) + ( 2y - q ) (hp + bq ) = 0.
Show also that the area of the parallelogram is
|(ap 2 + 2hpq + bq 2 )| / 2 (h 2 - ab ) .
(n + n1 )l = 2g
…(vi)
Sol. The combined equation of AB and AD is
m( n + n1 ) = 2 f
Eq. (i) From Eq. (v)
…(vii)
Equating the coefficients, we get
l2 = a
…(ii)
2
m =b
…(iii)
Þ
h = lm = ± ab
Now,
h = ab
Þ abc + 2 fg ab - af
S1 º ax 2 + 2hxy + by 2 = 0
Now, equation of lines through ( p , q ) and parallel to S1 º 0
is S 2 º a( x - p )2 + 2h ( x - p )(y - q ) + b(y - q )2 = 0
or h = - ab
abc + 2 fgh - af
Q
l
2
2
2
C(p,q)
2
- bg - ch = 0
D
2
- bg - c × ab = 0
[substituting the value of h = ab ]
Þ
2
-( f a - g b ) = 0 Þ
B
f a=g b
Thus, the given equation represents a pair of lines.
A(0,0)
Also, if h = - ab , then g b = - f a
Hence, the equation of diagonal BD is S1 - S 2 = 0
(ii) The distance between parallel lines
=
=
| n - n1|
(l 2 + m 2 )
=
ö
æ 4g 2
ç 2 - 4c ÷
ø
è l
(a + b )
æ g 2 - ac ö
=2 ç
÷
è a(a + b ) ø
a( -2xp + p 2 ) + 2h ( - py - qx + pq ) + b ( -2qy + q 2 ) = 0
(n + n1 )2 - 4nn1
-ap (2x - p ) - hq (2x - p ) - hp (2y - q ) - bq (2y - q ) = 0
l 2 + m2
[from Eqs. (vi), (ii) and (iii)]
[Ql 2 = a ]
Remark
In some books, the conditions for parallel lines are stated as
h2 = ab and bg 2 = af 2
[Q2hpq is written as hpq + hpq]
Hence, diagonal of BD is
(2x - p ) (ap + hq ) + (2y - q ) (bq + hp ) = 0
pp
Area of parallelogram = 1 2
sinq
Þ p 1p 2 = product of perpendiculars from ( p , q) on AB and
AD whose combined equation is
ax 2 + 2hxy + by 2 = 0
\
p 1p 2 =
| ap 2 + 2hpq + bq 2 |
{(a - b )2 + 4h 2 }
Chap 03 Pair of Straight Lines
\
tan q =
\
sin q =
\ Required area =
On squaring both sides, we get
2 (h 2 - ab )
|(a + b )|
( x 12 + y12 ) (h 2 - ab ) = k 2 {(a - b )2 + 4h 2 }
Hence, locus of P ( x 1, y1 ) is
2 (h 2 - ab )
( x 2 + y 2 ) (h 2 - ab ) = k 2 {(a - b )2 + 4h 2 }.
( a - b ) 2 + 4h 2
| ap 2 + 2hpq + bq 2 |
Ex. 23 Show that if two of the lines ax 3 + bx 2 y + cxy 2
+ dy 3 = 0 (a ¹ 0 ) make complementary angles with X-axis in
anti-clockwise sense, then
a (a - c ) + d (b - d ) = 0.
l
2
2 (h - ab )
Ex. 22 A point moves so that the distance between the
feet of perpendiculars drawn from it to the lines ax 2 + 2hxy
l
Sol. The given equation can be written as
dm 3 + cm 2 + bm + a = 0 , where m =
+ by 2 = 0 is a constant 2k. Show that the equation of its
locus is ( x 2 + y 2 ) (h 2 - ab ) = k 2 {(a - b ) 2 + 4h 2 }.
Sol. Q ÐONP = Ð OMP =
a
…(i)
d
If m1 = tana, then m 2 = tan(90° - a )
[Q two lines makes complementary
angles with X-axis]
\
m 2 = cota then m1m 2 = 1
a
From Eq. (i),
m3 = d
Since, m 3 is root of the above cubic, we have
\
\ O, M, P, N are concyclic with diameter OP.
N
P(x1,y1)
k
r
θ
X′
L
θ
O'
r
k
θ
r
M
Y′
Let
P º ( x 1, y1 )
æx y ö
\ Coordinates of centre are O ¢ = ç 1 , 1 ÷
è2 2ø
\
Radius r = O ¢ N = OM = OO ¢
2
2
æx ö
æy ö
= ç 1÷ + ç 1÷ =
è2ø
è2ø
+
2
2 (h 2 - ab )
From Eqs. (i) and (ii), we get
=
- a 3 + ca 2 - abd + ad 2 = 0
\
2k
( x 12 + y12 )
a (a - c ) + d (b - d ) = 0.
Ex. 24 Show that the equation
a ( x 4 + y 4 ) - 4bxy ( x 2 - y 2 ) + 6cx 2 y 2 = 0 represents two
pairs of lines at right angles and that if 2b 2 = a 2 + 3ac , the
two pairs will coincide.
l
…(i)
Sol. Given equation is
a( x 4 + y 4 ) - 4bxy ( x 2 - y 2 ) + 6c x 2y 2 = 0
…(i)
Eq. (i) is a homogeneous equation of fourth degree and
since it represents two pairs at right angles. i.e. sum of the
coefficients of x 2 and y 2 should be zero.
Let a( x 4 + y 4 ) - 4bxy ( x 2 - y 2 ) + 6cx 2y 2
…(ii)
( a - b ) 2 + 4h 2
( a - b ) 2 + 4h 2
Þ
-a 2 + ca - bd + d 2 = 0
2 (h 2 - ab )
.
ax 2 + 2hxy + by 2 = 0 is tan q =
|a + b |
2 (h 2 - ab )
æ a 3 ö æ ca 2 ö æ ab ö
ç- 2 ÷ + ç 2 ÷ - ç ÷ + a = 0
è d ø èd ø èd ø
y12 )
2
But angle between the lines represented by
sin q =
Þ
On dividing each by a, we get
( x 12
Let
Ð MON = q
\
Ð MO ¢ N = 2q Þ Ð NO ¢ L = Ð MO ¢ L = q
and given MN = 2k , then NL = LM = k
Now, in DNO ¢ L,
2k
NL k
k
sin q =
= =
=
2
2
2
O¢N r
( x 1 + y1 )
( x 1 + y12 )
\
m1 m 2 m 3 = -
æ a2 ö
æ a3 ö
æ aö
d ç- 3 ÷ + c ç 2 ÷ + bç- ÷ + a = 0
è dø
èd ø
è d ø
X
O
y
x
Let its roots be m1, m 2 , m 3 .
p
2
Y
225
= (ax 2 + pxy - ay 2 )( x 2 + qxy - y 2 ),
where, p and q are constants.
On comparing similar powers, we get
p + aq = - 4b
-2a + pq = 6c
…(ii)
…(iii)
226
Textbook of Coordinate Geometry
Again, given two pairs coincide, then
p
=q
a
or
p = aq
From Eqs. (ii) and (iv),
2b
q=a
and
p = - 2b
Substituting the values of p and q in Eq. (iii), we get
-2a +
= k 3 (a 2 + b 2 - 2ac + c 2 - 2bd + d 2 )
= k 3 {(a - c )2 + (b - d )2 }
(ay 3 + by 2 x + cyx 2 + dx 3 ) = k 3 {(a - c )2 + (b - d )2 }
Ex. 26 If one of the lines given by the equation
ax 2 + 2hxy + by 2 = 0 coincides with one of the lines given
by a ¢ x 2 + 2h ¢ xy + b ¢ y 2 = 0 and the other lines represented
by them be perpendicular, then .
ha ¢ b ¢ h ¢ ab 1
( -aa ¢ bb ¢ )
=
=
b¢ - a¢ b - a 2
l
4b 2
= 6c
a
-a 2 + 2b 2 = 3ac
Þ
2b 2 = a 2 + 3ac
\
Hence, locus of P is
…(iv)
Sol. Let the two lines represented by
ax 2 + 2hxy + by 2 = 0
Ex. 25 Show that the locus of a point such that the
product of the perpendiculars let fall from it on three lines
represented by ay 3 + by 2 x + cyx 2 + dx 3 = 0 is
constant = k 3 , is
l
ay 3 + by 2 x + cyx 2 + dx 3 = k 3 (a - c ) 2 + (b - d ) 2 .
Sol. Let the three represented lines are
y = m1x , y = m 2 x and y = m 3 x
ay 3 + by 2 x + cyx 2 + dx 3 = a(y - m1x ) (y - m 2 x ) (y - m 3 x )
On comparing similar powers on both sides
b
Þ
m1 + m 2 + m 3 = a
c
m1m 2 + m 2m 3 + m 3m1 =
a
d
m1m 2m 3 = a
Let the point P ( x 1, y1 ), from given condition
(y1 - m1x 1 ) (y1 - m 2 x 1 ) (y1 - m 3 x 1 )
.
.
= k3
2
2
2
(1 + m 3 )
(1 + m 2 )
(1 + m1 )
Þ
y = m1x
y = m2x
2h
\
m1 + m 2 = b
a
and
m1m 2 =
b
and the lines represented by
a ¢ x 2 + 2h ¢ xy + b ¢ y 2 = 0
be
…(ii)
and
a¢
æ 1 ö
- ç ÷m 2 =
è m1 ø
b¢
…(iii)
From Eqs. (iii) and (vi), we get
Þ
+
= ak
3
= ak 3
= ak
3
+
(1 +
Sm12
+
+
dx 13
2
2
-
3
2
- 2m1m 2m 3 Sm1 + (m1m 2m 3 ) }
2c c 2 2d
1+ 2 +
+
a a2
a
a
æ bö æ d ö
.ç- ÷ + ç- ÷
è aø è aø
2
-aba ¢ b ¢
+
a¢b
Þ
and
+ m12m 22m 32 )
{1 + ( Sm1 )2 - 2 Sm1m 2 + ( Sm1m 2 )2
b2
…(iv)
…(v)
…(vi)
( -aba ¢ b ¢ )
aa ¢
Þ m2 =
bb ¢
bb ¢
( -aba ¢ b ¢ )
a¢b
Substituting these values of m1 and m 2 in Eqs. (ii) and (v),
we get
)
Sm12m 22
…(iii)
From Eqs. (iii), we get m1 = -
= a (y - m1x ) (y - m 2 x )(y - m 3 x )]
cy1x 12
1
2h ¢
+ m2 = b¢
m1
- m 22 =
…(ii)
and y = m 2 x
-
[Qay + by x + cyx + dx
by12 x 1
1
x
m1
\
= k 3 (1 + m12 ) (1 + m 22 ) (1 + m 32 )
(ay13
y=-
…(i)
1
(ay13 + by12 x 1 + cy1x 12 + dx 13 )
a
3
…(i)
be
and
a¢b
+
-aba ¢ b ¢
2h
-aba ¢ b ¢
=bb ¢
b
ha ¢ b ¢ 1
=
-aa ¢ bb ¢
b¢ - a¢ 2
2h ¢
-aba ¢ b ¢
=bb ¢
b¢
h ¢ ab 1
=
- aba ¢ b ¢
b-a 2
From Eqs. (vii) and (viii), we get
ha ¢ b ¢
h ¢ ab 1
=
- aba ¢ b ¢
=
b¢ - a¢ b - a 2
…(vii)
…(viii)
#L
Pair of Straight Lines Exercise 1 :
Single Option Correct Type Questions
n
This section contains 12 multiple choice questions.
Each question has four choices (a), (b), (c), (d) out of which
ONLY ONE is correct.
1. If the sum of the slopes of the lines given by
4 x + 2λxy − 7y = 0 is equal to the product of the
slopes, then λ is equal to
2
2
(a) − 4
(b) − 2
(c) 2
(d) 4
2. The equation 3ax 2 + 9 xy + (a 2 − 2)y 2 = 0 represents two
(b) 2 5
(d) 4 5
8. The four straight lines given by the equations
12x 2 + 7 xy − 12y 2 = 0 and
12x 2 + 7 xy − 12y 2 − x + 7y − 1 = 0 lie along the sides of a
(a) square
(c) rectangle
(b) rhombus
(d) parallelogram
9. Distance between the parallel lines
perpendicular straight lines for
4 x 2 + 20xy + 25y 2 + 2x + 5y − 12 = 0 is
(a) only one value of a
(b) for all values of a
(c) for only two values of a (d) for no value of a
(a)
3. The image of the pair of lines represented by
ax 2 + 2hxy + by 2 = 0 by the line mirror y = 0 is
(a) ax 2 − 2hxy − by 2 = 0
(b) bx 2 − 2hxy + ay 2 = 0
(c) bx + 2hxy + ay = 0
(d) ax − 2hxy + by = 0
2
2
2
2
4. Number of points lying on the line 7 x + 4y + 2 = 0 which
is equidistant from the lines 15x 2 + 56xy + 48y 2 = 0 is
(a) 0
(c) 2
(b) 1
(d) 4
5. Orthocentre of the triangle formed by the lines
xy − 3x − 5y + 15 = 0 and 3x + 5y = 15 is
(a) ( − 5, − 3 )
(c) ( − 3 , − 5 )
(b) (5, 3 )
(d) (3, 5 )
6. Two of the straight lines given by
3x + 3x y − 3xy + dy = 0 are at right angles, if d
equal to
3
2
2
(a) − 4
(c) − 2
3
(b) − 3
(d) − 1
7. Two lines are given by ( x − 2y ) 2 + λ( x − 2y ) = 0. The
value of | λ | so that the distance between them is 3, is
#L
(a) 5
(c) 3 5
3
29
7
(c)
29
5
29
9
(d)
29
(b)
10. The point of intersection of the two lines given by
2x 2 − 5xy + 2y 2 + 3x + 3y + 1 = 0 is
(a) ( − 2, 2 )
(c) (3, 3 )
(b) ( − 3, 3 )
(d) (2, 2 )
11. If α, β > 0 and α < β and
αx 2 + 4 γxy + βy 2 + 4 p ( x + y + 1) = 0 represents a pair
of straight lines, then
(a) α ≤ p ≤ β
(c) p ≥ α
(b) p ≤ α
(d) p ≤ α or p ≥ β
12. If the equation of the pair of straight lines passing
through the point (1, 1), one making an angle θ with the
positive direction of the X -axis and the other making the
same angle with the positive direction of the Y -axis, is
x 2 − (a + 2)xy + y 2 + a ( x + y − 1) = 0, a ≠ − 2 , then the
value of sin 2θ is
(a) a − 2
2
(c)
(a + 2 )
(b) a + 2
2
(d)
a
Pair of Straight Lines Exercise 2 :
More than One Option Correct Type Questions
n
This section contains 6 multiple choice questions. Each
questions has four choices (a), (b), (c), (d) out of which
MORE THAN ONE may be correct.
13. The equation of image of pair of lines y = | x − 1 | in
Y -axis is
(a) y = | x + 1 |
(c) x 2 − y 2 + 2 x + 1 = 0
(b) y = | x − 1 | + 3
(d) x 2 − y 2 + 2 x − 1 = 0
14. The equation ax 2 + by 2 + cx + cy = 0 represent a pair of
straight lines, if
(a) a + b = 0
(b) c = 0
(c) a + c = 0
(d) c (a + b ) = 0
228
Textbook of Coordinate Geometry
15. If x 2 + αy 2 + 2βy = a 2 represents a pair of
perpendicular straight lines, then
(a) α = 1, β = a
(b) α = 1, β = − a
(c) α = − 1, β = − a
(d) α = − 1, β = a
y 2 + xy − 12x 2 = 0 and ax 2 + 2hxy + by 2 = 0.
One line will be common among them, if
(a) a = − 3(2h + 3b )
(c) a = 2(b + h )
16. If the pair of lines ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
intersect on Y -axis, then
(a) f 2 = bc
(b) abc = 2 fgh
(c) bg 2 ≠ ch 2
(d) 2 fgh = bg 2 + ch 2
#L
17. Two pair of straight lines have the equations
(b) a = 8(h − 2b )
(d) a = − 3(b + h )
18. The combined equation of three sides of a triangle is
( x 2 − y 2 ) (2x + 3y − 6) = 0. If ( − 2 , a ) is an interior and
(b, 1) is an exterior point of the triangle, then
10
3
9
(c) − 1 < b <
2
(b) − 2 < a <
(a) 2 < a <
10
3
(d) − 1 < b < 1
Pair of Straight Lines Exercise 3 :
Paragraph Based Questions
n
This section contains 3 paragraphs based upon each of
the paragraph, 3 multiple choice questions have to be
answered. Each of these question has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
Paragraph I
(Q. Nos. 19 to 21)
Consider the equation of a pair of straight lines as
x 2 − 3xy + λy 2 + 3x − 5 y + 2 = 0
19. The value of λ is
(a) 1
(c) 3
(c) ax 2 + 2hxy + by 2 = 0
(d) None of the above
23. If f i + 1 ( x , y ) = 0 represents the equation of a pair of
perpendicular lines, thenf 3 ( x , y ) = 0 is same as
(a) f1( x, y ) = 0
(b) f 2( x, y ) = 0
(c) hx 2 − (a − b ) xy − hy 2 = 0
(d) None of the above
24. The value of
(b) 2
(d) 4
20. The point of intersection of lines is (α, β), then the value
5
f n + 2 ( x, y )
n=2
f n ( x, y )
∑
(a) 14
(c) 54
(b) 5
(d) 17
21. The angle between the lines is θ, then the value of cos 2θ
is
1
3
3
(c)
5
2
3
4
(d)
5
(a)
(b) 4
(d) 6
Paragraph III
of α 2 + β 2 is
(a) 2
(c) 10
is
(b)
Paragraph II
(Q. Nos. 22 to 24)
Let f1 ( x, y ) ≡ ax + 2hxy + by 2 = 0 and let f i + 1 ( x, y ) = 0
2
denote the equation of the bisectors of f i ( x, y ) = 0 for all
i = 1, 2, 3, ...
22. f 3 ( x , y ) = 0 is
(a) hx 2 − (a − b ) xy − hy 2 = 0
(b) (a − b ) x 2 + 4hxy − (a − b )y 2 = 0
(Q. Nos. 25 to 27)
Consider a pair of perpendicular straight lines
2x 2 + 3xy + by 2 − 11x + 13 y + c = 0
25. The value of c is
(a) − 2
(c) − 3
(b) 2
(d) 3
26. The value of | b + 2c | is
(a) 4
(c) 8
(b) 6
(d) 10
27. If point of intersection of lines is C and points of
intersection of the lines with the X -axis are A and B, if
distance between the orthocentre and the circumcentre
of ∆ABC is λ, then [λ] is (where [. ] denotes the greatest
integer function
(a) 2
(c) 4
(b) 3
(d) 5
Chap 03 Pair of Straight Lines
#L
229
Pair of Straight Lines Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 5 questions. The answer to each
question is a single digit integer, ranging from 0 to 9
(both inclusive).
28. Equation λx 3 − 10x 2 y − xy 2 + 4y 3 = 0 represented
three straight lines, out of these three, two lines makes
equal angle with y = x and λ > 0, then the value of λ is
29. Area enclosed by curves y − 5xy + 6x + 3x − y = 0 and
2
2
y 2 − 5xy + 6x 2 + 2x − y = 0 is λ sq units, then the value
of λ is
#L
30. The lines represented by x 2 + 2λxy + 2y 2 = 0 and ( λ + 1)
x 2 − 8xy + y 2 = 0 are equally inclined, then the value of
| λ | is
31. If the lines joining the origin to the intersection of the
line y = nx + 2 and the curve x 2 + y 2 = 1 are at right
angles, then the value of n 2 is
32. If area of the triangle formed by the line x + y = 3 and
the angle bisectors of the pair of straight lines
x 2 − y 2 + 2y = 1 is λ sq units, then the value of λ is
Pair of Straight Lines Exercise 5 :
Matching Type Questions
n
This section contains one question. This question has four statements (A,B,C and D) given in Column I and four
statements (p, q, r and s) in Column II. Any given statements in Column I can have correct matching with one or more
statement(s) given in Column II.
33. Match the following
Column I
Column II
(A) The pair of lines joining the origin to the
points of intersection of the curve
9x 2 + 16 y2 = 144 by the line 2x + 2 y + λ = 0 are coincident, then | λ | is divisible by
(p)
2
(B) If the straight lines joining the origin to the points of intersection of the straight line 4 x + 3 y = 24
and the curve (x − 3)2 + ( y − 4 )2 = λ2, are at right angles, then | λ | is divisible by
(q)
3
(C) The two line pairs y2 − 4 y + 3 = 0 and x 2 + 4 xy + 4 y2 − 5x − 10 y + 4 = 0 enclose a 4 sided
convex polygon, if area of polygon is λ sq units, then λ is divisible by
(r)
5
(D) If the pairs of lines 3x 2 − 2 pxy − 3 y2 = 0 and 5x 2 − 2qxy − 5 y2 = 0 are such that each pair bisects
the angle between the other pair. If λ = | pq | , then λ is divisible by
(s)
6
230
#L
Textbook of Coordinate Geometry
Pair of Straight Lines Exercise 6 :
Statement I and II Types Questions
n
35. Statement I Two of the straight lines represented by
Directions (Q. Nos. 34-37) are Assertion-Reason type
questions. Each of these questions contains two
statements:
Statement I (Assertion) and Statement II (Reason)
Each of these questions also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below :
36. Statement I If αβ = − 1, then the pair of straight lines
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is not a
correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
37. Statement I If a + b = − 2h, then one line of the pair of
34. Statement I The four straight lines given by
6x 2 + 5xy − 6y 2 = 0 and 6x 2 + 5xy − 6y 2 − x + 5y − 1 = 0
are the sides of a square.
dx 3 + cx 2 y + bxy 2 + ay 3 = 0 will be at right angles if
d 2 + bd + bc + a 2 = 0.
Statement II Product of the slopes of two perpendicular
lines is − 1.
x 2 − 2αxy − y 2 = 0 and y 2 + 2βxy − x 2 = 0 are the angle
bisector of each other.
Statement II Pair of angle bisector lines of the pair of
lines ax 2 + 2hxy + by 2 = 0 is h( x 2 − y 2 ) = (a − b )xy.
lines ax 2 + 2hxy + by 2 = 0 bisects the angle between
coordinate axes in positive quadrant.
Statement II If ax + y(2h + a ) = 0 is a factor of
ax 2 + 2hxy + by 2 = 0, then b + 2h + a = 0.
Statement II The lines represented by general equation
of second degree ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
are perpendicular if a + b = 0.
#L
Pair of Straight Lines Exercise 7 :
Subjective Type Questions
n
42. If ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 and
In this section, there are 7 subjective questions.
38. Prove that the straight lines represented by
(y − mx ) = a (1 + m ) and (y − nx ) = a (1 + n )
form a rhombus.
2
2
2
2
2
2
39. Prove that the equation m ( x 3 − 3xy 2 ) + y 3 − 3x 2 y = 0
represents three straight lines equally inclined to each
other.
40. Show that the straight lines
( A 2 − 3B 2 )x 2 + 8ABxy + ( B 2 − 3A 2 ) y 2 = 0 form with
the line Ax + By + C = 0 an equilateral triangle whose
C2
area is
3( A 2 + B 2 )
41. Find the equations of the diagonals of the parallelogram
formed by the lines L 2 − aL = 0 and L ′ 2 − aL ′ = 0, where
L = x cos θ + y sin θ − p and L ′ = x cos θ′ + y sin θ′ − p ′.
ax 2 + 2hxy + by 2 − 2gx − 2 fy + c = 0 each represents a
pair of lines, then prove that the area of the
2| c |
parallelogram enclosed by them is
.
(h 2 − ab )
43. Prove that lines ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
are equidistant from the origin, if
f 4 − g 4 = c (bf 2 − ag 2 ). Also, find the product of
their distances from the origin.
44. Prove that if two of the lines represented by
ax 4 + bx 3 y + cx 2 y 2 + dxy 3 + ay 4 = 0
bisects the angle between the other two, then
c + 6a = 0 and b + d = 0.
Chap 03 Pair of Straight Lines
#L
231
Pair of Straight Lines Exercise 8 :
Question Asked in Previous 13 Years Exams
n
This section contains questions asked in IIT-JEE, AIEEE,
JEE Main & JEE Advanced from year 2005 to 2017.
46. If one of the lines of my 2 + (1 − m 2 )xy − mx 2 = 0 is a
bisector of the angle between the lines xy = 0, then m is
45. If the pair of lines ax + 2(a + b )xy + by = 0 lie long
2
2
[AIEEE 2007, 3M]
1
(a) −
2
(c) 1
diameters of a circle and divide the circle into four
sectors such that the area of one of the sector is thrice
the area of the another sector, then
[AIEEE 2005, 3M]
(a) 3a 2 + 2ab + 3b 2 = 0
(c) 3a 2 − 2ab + 3b 2 = 0
(b) −2
(d) 2
(b) 3a 2 + 10ab + 3b 2 = 0
(d) 3a 2 − 10ab + 3b 2 = 0
Answers
Exercise for Session 5
Exercise for Session 1
1. (c)
2. (a)
3. (b)
4. (b)
5. (c, d)
8
7.  ,
3
8
6. bx + ay = 0, ax − by = 0

3
9. (ab′ − a′ b)2 = 4 (ah′ − a′ h) (h′ b − hb′ )
2. (c, d)
1
6. tan −1  
 3
3. (a)
4. (c)
8. sec−1 (± p)
5. (b, c)
9.
1
3 sq units
6
Exercise for Session 3
1. (b)
2. (a, c)
3. (a, c)
4. (b)
9. ± 2
Exercise for Session 4
1. (b)
2. (d)
3. (a, d)
6. 1 sq unit 7. x − y = 0
10. x2 − 5xy + 4 y2 + 3x − 3 y = 0
4. (b)
9.
3. (d)
7. ± 1
4. (a)
5. (b)
10. x2 − xy − 2 y2 = 0
Chapter Exercises
Exercise for Session 2
1. (b)
1. (c)
2. (d)
6. 9x2 + 10xy − 9 y2 = 0
5. (a)
| g 2 − ac |
| a | h2 − ab
1. (b)
2. (c)
3. (d)
4. (c)
5. (b)
6. (b)
7. (c)
8. (a)
9. (c)
10. (c)
11. (d)
12. (c)
13. (a, c) 14. (a, b, d) 15. (c, d)
16. (a, d) 17. (a, b) 18. (a, d)
19. (b)
20. (c)
21. (d)
22. (b)
23. (a)
24. (b)
25. (a)
26. (b)
27. (c)
28. (7)
29. (1)
30. (2)
31. (7)
32. (2)
33. (A) → (p, r); (B) → (r); (C) → (p, q, s); (D) → (q, r)
34. (b)
35. (b)
36. (a)
37. (b)
 x(cosθ − cosθ′ ) + y (sin θ − sin θ′ ) − p + p′ = 0
41. 
and x(cosθ + cosθ′ ) + y(sin θ + sin θ′ ) − p − p′ − a = 0
| c|
43.
(a − b ) 2 + 4 h 2
45. (a)
46. (c)
Solutions
1. Qm1 + m2 = m1m2 Þ
\
2. For perpendicular lines
3a + (a 2 - 2 ) = 0
Þ
( x - 2y ) ( x - 2y + l) = 0
\ Lines are x - 2y = 0 and x - 2y + l = 0
Distance between lines = 3
| l - 0|
=3
Þ
(1 + 4 )
- 2l
4
=
-7
-7
8. \
l = -2
a=
ax 2 + 2hx( - y ) + b( - y ) 2 = 0
ax 2 - 2hxy + by 2 = 0
or
4. Q
15 x 2 + 56 xy + 48y 2 = 0
...(i)
Þ (3 x + 4y ) (5 x + 12y ) = 0
Equation of lines represented by Eq. (i) are
3 x + 4y = 0 and 5 x + 12y = 0
æ - 7l -2 ö
Let any point on 7 x + 4y + 2 = 0 is ç l,
÷.
è
4 ø
According to questions,
æ - 7l - 2 ö
æ - 7l - 2 ö
5 l + 12 ç
3l + 4ç
÷
÷
è
ø
è
ø
4
4
=
13
5
1
1
| - 4 l - 2 | = | - 16 l - 6 |
5
13
| 2l + 1 | | 8l + 3 |
Þ
=
5
13
Þ
13(2 l + 1 ) = ± 5 (8 l + 3 )
1 14
l=- ,\
7 33
5. Given lines are xy - 3x - 5y + 15 = 0
Þ
Y
(0,3) B
y=3
3x+
5y=
15
C (5,3)
x=5
X
A(5,0)
O
Þ
( x - 5 ) (y - 3 ) = 0
\
x = 5 and y = 3
Hence, orthocentre is (5, 3).
6. 3x 3 + 3x 2y - 3xy 2 + dy 3 = ( x 2 + pxy - y 2 ) (3x - dy )
2
2
On comparing coefficients of x y and xy , we get
and
\
3p - d = 3
- 3 - pd = - 3
p = 0, d = -3
(given)
| l | =3 5
\
- 3 ± (9 + 8 ) - 3 ± 17
=
6
6
3. For mirror image with respect to y = 0 replace y by - y, then
image of the pair of lines ax 2 + 2hxy + by 2 = 0 is
Þ
( x - 2y ) 2 + l( x - 2y ) = 0
7. Q
2
12 x + 7 xy - 12y 2 = 0
...(i)
Þ (3 x + 4y ) ( 4 x - 3y ) = 0
\Lines represented by Eq. (i) are
3 x + 4y = 0 and 4 x - 3y = 0
2
and 12 x + 7 xy - 12y 2 - x + 7y - 1 = 0
...(ii)
Þ
(3 x + 4y - 1 ) ( 4 x - 3y + 1 ) = 0
\ Lines represented by Eq. (ii) are
3 x + 4y - 1 = 0 and 4 x - 3y + 1 = 0
Distance between parallel lines 3 x + 4y = 0 and
1
3 x + 4y - 1 = 0 is .
5
And distance between parallel lines 4 x - 3y = 0 and
1
4 x - 3y + 1 = 0 is .
5
Hence, all sides along a square.
9. Given,
4 x 2 + 20 xy + 25y 2 + 2 x + 5y - 12 = 0
...(i)
(2 x + 5y ) 2 + (2 x + 5y ) - 12 = 0
Þ
Þ
(2 x + 5y + 4 ) (2 x + 5y - 3 ) = 0
\ Lines represented by Eq. (i) are
2 x + 5y + 4 = 0 and 2 x + 5y - 3 = 0
| 4 - (- 3) |
7
Hence, distance between parallel lines =
.
=
4 + 25
29
10. Let f ( x, y ) º 2x 2 - 5xy + 2y 2 + 3x + 3y + 1 = 0
¶f
¶f
= 4 x - 5y + 3 and
= - 5 x + 4y + 3
¶x
¶y
¶f
¶f
For point of intersection
= 0 and
=0
¶x
¶y
\
Þ 4 x - 5y + 3 = 0 and - 5 x + 4y + 3 = 0, we get
x = 3, y = 3
\ Point of intersection is (3, 3 ).
11. ax 2 + 4g xy + by 2 + 4p( x + y + 1) = 0
represents a pair of straight lines.
\D=0
Þ 4 abp + 16 p 2g - 4 p 2a - 4 p 2b - 16 g 2p = 0
Þ
(16 p ) g 2 - 16 p 2g + 4 p( pa + pb - ab) = 0
B 2 - 4 AC ³ 0
\
2 2
Þ
(16 p ) - 4 × (16 p ) × 4 p ( pa + pb - ab) ³ 0
Þ
p 2( p 2 - pa - pb + ab) ³ 0
Þ
\
( p - a) ( p - b) ³ 0
p £ a or p ³ b
[Q a < b]
Chap 03 Pair of Straight Lines
12. Equation of first line is y - 1 = tan q( x - 1)
18. The separate equations of the sides are x + y = 0, x - y = 0 and
x y
+ = 1.
3 2
Equation of second line is y - 1 = cot q( x - 1 )
So, their joint equation is
[(y - 1 ) - tan q ( x - 1 )] [(y - 1 ) - cot q ( x - 1 )] = 0
Þ(y - 1 ) 2 - ( x - 1 ) (y - 1 ) (tan q + cot q) + ( x - 1 ) 2 = 0
x=–2
Y
2x+
0
y=
x+
+ cot q - 2 ) ( x + y - 1 ) = 0
On comparing with the given equation, we get
tan q + cot q = a + 2
1
or
=a + 2
sin q cos q
2
or
sin 2 q =
(a + 2 )
13. For image w.r.t., Y -axis replace x by - x, then required image
of lines is
y = | - x - 1 | or y = | x + 1 |
and on squaring both sides, then
y 2 = x 2 + 2x + 1
x 2 - y 2 + 2x + 1 = 0
0
6=
B
3y–
x 2 - (tan q + cot q)xy + y 2 + (tan q
Þ
233
y
x–
A
X′
=0
y=1
X
O
Y′
Intersection of x = - 2 with y = - x and 2 x + 3y - 6 = 0 gives
the range of values of a.
10
2 <a <
\
3
and intersection of y = 1 with y = x and 2 x + 3y - 6 = 0 given
the range of values of b
[Q(b, 1 ) is exterior point]
-1 <b <1
14. Given equation is
ax 2 + by 2 + cx + cy = 0
...(i)
Here,
...(i)
Equation (i) represents a pair of perpendicular straight lines
\ D = 0 and coefficient of x 2 + coefficient of y 2 = 0
Þ
- a 2a - b 2 = 0 and 1 + a = 0
\
a = - 1 and b = ± a
16. Given pair of lines is
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
...(i)
Q Point of intersection of lines represented by (i) is
æ hf - bg gh - af ö æ æ f 2 - bc ö æ g 2 - ac ö ö
,
÷÷
÷, ç
ç
÷ or ç ç
è ab - h 2 ab - h 2 ø çè è h 2 - ab ø è h 2 - ab ø ÷ø
For Y -axis put x = 0
\
hf = bg, f 2 = bc and h 2 ¹ ab
Also, D = 0
\ abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
For
5 3
3
9
25
9
´ ´ - -1 ´ - l ´ -2 ´ = 0
2 2
2
4
4
4
45 25 9 l 9
2l +
- =0
Þ
4
4
4 2
l 1
Þ
- + =0
4 2
\
l =2
20. Let f ( x, y ) = x 2 - 3xy + 2y 2 + 3x - 5y + 2
[Q l = 2 ]
¶f
¶f
\
= 2 x - 3y + 3,
= - 3 x + 4y - 5
¶x
¶y
¶f
¶f
For point of intersection
= 0 and
= 0, we get
¶x
¶y
1 ´ l ´2 + 2 ´ -
15. Given, equation is
x 2 + ay 2 + 2by - a 2 = 0
a = 1, b = l , c = 2,
5
3
3
f = - , g = and h = - .
2
2
2
19. D = 0
c = 0 or a + b = 0
\
Sol. (Q. Nos. 19 to 21)
Given equation is x 2 - 3 xy + ly 2 + 3 x - 5y + 2 = 0
Eq. (i) represents a pair of straight lines.
\
D=0
2
c
c2
Þ
0 + 2× 0 - a ´
-b ´
=0
4
4
or
c 2(a + b ) = 0
f 2 = bc, 2 fgh = bg 2 + ch 2
17. Let y = mx be the common line, then m 2 + m - 12 = 0 and
bm 2 + 2hm + a = 0, then from first equation m = -4, 3
Substitute m = -4 in second equation, then a = 8(h - 2b ) and
substitute m = 3 in second equation, then a = - 3(2h + 3b ).
x = - 3, y = - 1
\
a = - 3 and b = - 1
\ The value of a 2 + b 2 is 10.
2 (h 2 - ab )
21. tan q =
=
(a + b )
ö
æ9
2 ç - 2÷
ø
è4
=
1+2
1
11 - tan 2 q
9 =4
\ cos 2 q =
=
1 + tan 2 q 1 + 1 5
9
1
3
234
Textbook of Coordinate Geometry
Solutions (Q. Nos. 22 to 24)
22. Q f1( x, y ) º ax 2 + 2hxy + by 2 = 0
Also, intersection on X -axis put y = 0.
...(i)
C
bisectors of f1( x, y ) is f 2( x, y )
\ Equation of bisectors of (i) is
x 2 - y 2 xy
=
a -b
h
2
Þ
hx - (a - b ) xy - hy = 0
\
f 2( x, y ) º hx 2 - (a - b ) xy - hy 2 = 0
M
...(ii)
Now, equation of bisectors of (ii) is
x2 - y 2
xy
=
h - ( - h ) - (a - b )
2
x2 - y 2
2 xy
=Þ
(a - b )
2h
2
2
\ f 3( x, y ) º (a - b ) x + 4hxy - (a - b )y = 0
23. f 2( x, y ) º bx 2 - 2hxy + ay 2= 0
2
2
and f 3( x, y ) º ax + 2hxy + by = 0
is same as f1( x, y ) = 0
24. For all n ³ 2
5
Now,
å
n=2
fn + 2( x, y )
fn ( x, y )
å1 = 1 + 1 + 1 + 1 = 4
n=2
Sol. (Q. Nos. 25 to 27)
For perpendicular straight lines, then
coefficient of x 2 + coefficient of y 2 = 0
Þ
2+b=0
\
b = -2
25. 2x 2 + 3xy - 2y 2 - 11x + 13y + c = 0
represents a pair of straight lines, then we have
13
11 3
121
9
169
2 ´ -2 ´ c + 2 ´
+2 ´
-c ´ = 0
´´ -2 ´
2
2 2
4
4
4
429 169 121 9c
Þ
- 4c +
=0
4
2
2
4
Þ
c = -2
26. | b + 2c | = | - 2 - 4 | = 6
2
2
27. Now, pair of lines is 2x + 3xy - 2y - 11x + 13y - 2 = 0
2
2
Let f ( x, y ) º 2 x + 3 xy - 2y - 11 x + 13y - 2 = 0
¶f
¶f
= 4 x + 3y - 11 and
= 3 x - 4y + 13
\
¶x
¶y
¶f
¶f
For point of intersection,
= 0 and
=0
dx
¶y
Þ
\
\
æ 11 ö
M º ç , 0÷
è4 ø
and orthocentre is
æ 1 19 ö
C ºç , ÷
è5 5 ø
11
2
2
19 ö
æ 11 1 ö
æ
\Required distance = ç - ÷ + ç 0 - ÷
è 4 5ø
è
5ø
4 x + 3y - 11 = 0 and 3 x - 4y + 13 = 0
æ 1 19 ö
C ºç , ÷
è5 5 ø
2
8377
= 20.94 = 4.57
400
[ l ] = [ 4.57 ] = 4
28. Given, 4y 3 - xy 2 - 10x 2y + lx 3 = 0
l=
\
Þ
5
=
2 x 2 - 11 x + 2 = 0 Þ x1 + x 2 =
3
fn + 2( x, y ) = fn ( x, y )
fn + 2( x, y )
=1
fn ( x, y )
\
\
æ x + x2 ö
\ Circumcentre of DABC Þ M ç 1
, 0÷
è 2
ø
(a - b ) x 2 + 4hxy - (a - b )y 2 = 0
Þ
A(x1,0)
(x2,0)B
2
2
æy ö
æy ö
æy ö
4 ç ÷ - ç ÷ - 10 ç ÷ + l = 0
èxø
èxø
èxø
Let y = mx, then
4m 3 - m 2 - 10m + l = 0
and let m1, m2, m3 be the slopes of three lines, then
1
...(i)
m1 + m2 + m3 =
4
10
...(ii)
m1m2 + m2m3 + m3m1 = 4
l
...(iii)
m1m2m3 = 4
Since, two lines are equally inclined with y = x, so let m2m3 = 1
l
From Eq. (iii),
m1 = 4
l+1
From Eq. (i),
...(iv)
m2 + m3 =
4
10
æ lö æ l + 1ö
and from Eq. (ii), 1 + ç - ÷ ç
÷=è 4ø è 4 ø
4
Þ
16 - l2 - l = - 40
Þ
l2 + l - 56 = 0
Þ
(l + 8) (l - 7) = 0
l = - 8, 7
Hence,
l =7
29. Qy 2 - 5xy + 6x 2 + 3x - y = 0
Þ
(3 x - y ) (2 x - y + 1 ) = 0
and y 2 - 5 xy + 6 x 2 + 2 x - y = 0
Þ
(3 x - y + 1 ) (2 x - y ) = 0
[Q l > 0]
235
Chap 03 Pair of Straight Lines
33. (A) ® (p, r); (B) ® (r); (C) ® (p, q, s); (D) ® (q, r)
2x – y+1=0
C
3x–y
+1 =
=0
3x–y
A
\Required Area =
æ 2 x + 2y ö
9 x 2 + 16y 2 = 144 ç
÷
è -l ø
\
l =1
30. Given pairs are
2
and
2
( l + 1 ) x - 8 xy + y = 0
...(i)
...(ii)
Q Equations of angle bisectors of (i) and (ii) are equal
xy
x 2 - y 2 xy
x2 - y 2
and
must be same
=
Þ
=
1 -2
l
(l + 1) - 1 - 4
l
4
Þ l2 = 4 Þ l = ± 2
=Þ
-1
l
\
\
and
or
...(i)
x 2 + y 2 - (6 x + 8y ) + 25 - l2 = 0
...(ii)
Homogenising Eq. (ii) with Eq. (i), we get
2
...(i)
...(ii)
Homogenising Eq. (ii) with Eq. (i), we get
æ y - nx ö
x2 + y 2 = ç
÷
è 2 ø
2
2
Þ
ö
æn
æ1
ö
ç - 1 ÷ x 2 - nxy + ç - 1 ÷y 2 = 0
è4
ø
ø
è 4
or
(n 2 - 4 ) x 2 - 4 nxy - 3y 2 = 0
Þ
16 (25 - l2 ) x 2 + 9(25 - l2 )y 2 + ( - 600 - 24 l2 ) xy = 0
The lines will be at right angles, when
16(25 - l2 ) + 9(25 - l2 ) = 0
Þ
625 - 25 l2 = 0
Þ
l2 = 25
| l| =5
(C) The lines pairs are (y - 1 ) (y - 3 ) = 0 i.e. y = 1 and y = 3
the other line pair is x 2 + 4 xy + 4y 2 - 5 x - 10y + 4 = 0
i.e.
or
32. Q x 2 - y 2 + 2y = 1
( x + 2y - 4 ) ( x + 2y - 1 ) = 0
x + 2y - 4 = 0 and x + 2y - 1 = 0
y=3
D
x = (y - 2y + 1 ) = (y - 1 )
2
C
0
y–4=
x+2
2
y
x+2
2
( 4 x + 3y )
æ 4 x + 3y ö
+ (25 - l2 ) ç
÷ =0
è 24 ø
24
\
The lines will be at right angles, when
(n 2 - 4 ) + ( - 3 ) = 0 Þ n 2 = 7
Þ
| l | = 10
4 x + 3y
=1
24
( x - 3 ) 2 + (y - 4 ) 2 = l2
x 2 + y 2 - (6 x + 8y )
y - nx
=1
2
x2 + y 2 = 1
and
l2 = 100
or
(B) Given,
| l| =2
31. Given,
144 l4 - 576 l2(25 ) = 0
Þ
x + 2 lxy + 2y = 0
2
Þ (9 l2 - 576 ) x 2 - 1152 xy + (16 l2 - 576 )y 2 = 0
The lines are coincident, then
( - 576 ) 2 = (9 l2 - 576 ) (16 l2 - 576 )
2 -1
2
...(ii)
Homogenising Eq. (ii) with Eq. (i), we get
(1 - 0 ) (1 - 0 )
= 1 sq unit
3 -1
2
...(i)
9 x 2 + 16y 2 = 144
and
B
2x–y=0
2 x + 2y
=1
-l
(A) Given
0
D
–1=
Þ
x = ± (y - 1 )
i.e.
x - y + 1 = 0 and x + y - 1 = 0
\ Bisectors are y = 1 and x = 0
0
A
Y
3
Required area =
y=1
B
(3 - 1 ) ( 4 - 1 )
= 6 sq units
1 2
0 1
1
o
(2,1)
X
So, area between x = 0, y = 1 and x + y = 3 is given by
1
´ 2 ´ 2 = 2 sq units
2
\
l =2
\
l =6
(D) Equation of the bisectors of 3 x 2 - 2 pxy - 3y 2 = 0 is
x2 - y 2
xy
=
Þ px 2 + 6 xy - py 2 = 0
3 - (- 3) - p
Which is same as 5 x 2 - 2qxy - 5y 2 = 0
p
6
-p
i.e.
=
=
Þ pq = - 15
5 - 2q - 5
\
l = | pq | = | - 15 | = 15
236
Textbook of Coordinate Geometry
34. Q Lines represented by 6x 2 + 5xy - 6y 2 = 0 are l1 : 2x + 3y = 0,
l 2 : 3 x - 2y = 0
and lines represented by
6 x 2 + 5 xy - 6y 2 - x + 5y - 1 = 0
are
l 3 : 2 x + 3y - 1 = 0
l 4 : 3 x - 2y + 1 = 0
Q l1 ^ l 2, l 2 ^ l 3, l 3 ^ l 4 and l 4 ^ l1.
Also, l1, l 2 intersect at ( 0, 0 ) and ( 0, 0 ) is equidistant from l 3 and
l 4 . These lines form sides of square.
\Statement I is true.
Statement II is true but Statement II is not a correct
explanation for Statement I.
35. dx 3 + cx 2y + bxy 2 + ay 3 = ( x 2 + pxy - y 2 ) (dx - ay )
On comparing, we get
c+a
d
-b -d
and
b = - ap - d Þ p =
a
From Eq. (i) and Eq. (ii),
c + a -b -d
=
d
a
Þ
d 2 + bd + ac + a 2 = 0
c = - a + pd Þ p =
38. From (y - mx ) 2 = a 2 (1 + m 2 )
Þ
y - mx = ± a (1 + m 2 )
\ Lines
y - mx = a 1 + m 2
…(i)
and
y - mx = - a 1 + m 2
…(ii)
and from
...(ii)
y - nx = ± a (1 + n 2 )
\ Lines
y - nx = a 1 + n 2
…(iii)
and
y - nx = - a 1 + n 2
…(iv)
Since, lines (i) and (ii) are parallel, the distance between
lines (i) and (ii) is
|a 1 + m 2 + a 1 + m 2 |
1 + m2
Hence, lines (i), (ii), (iii) and (iv) form a rhombus.
39. Dividing by x 3, the given equation can be written as
2
3
æ
æy ö
æy ö ö æy ö
m çç1 - 3ç ÷ ÷÷ + ç ÷ - 3ç ÷ = 0
èx ø
ø
è
ø
è
x
x
è
ø
writing
y
= tan q, we have
x
Y
2′
...(i)
3
π/3
X′
1′
ab = - 1
Hence, both statements are true and Statement II is correct
explanation for Statement I.
37. Put 2h = - (a + b ) in ax 2 + 2hxy + by 2 = 0
Þ
ax 2 - (a + b ) xy + by 2 = 0
Þ
( x - y ) (ax - by ) = 0
Þ One of the line bisects the angle between coordinate axes
in positive quadrant.
Also, put
b = - 2h - a in ax - by , we have
ax - by = ax - ( - 2h - a )y
= ax + (2h + a )y
Hence, ax + (2h + a )y is a factor of
ax 2 + 2hyx + by 2 = 0
\ Both statements are true but Statement II is not correct
explanation for Statement I.
= | 2a |
Similarly, (iii) and (iv) are parallel lines and the distance
between them = | 2a |. Therefore, distance between parallel
lines are same.
Comparing it with y 2 + 2bxy - x 2 = 0 , then
a 1
l
- = =
1 b -1
Þ
2
Þ
x2 - y 2
xy
=
1 - (- 1) - a
ax 2 + 2 xy - ay 2 = 0
2
(y - nx ) = a (1 + n )
...(i)
\ Statement I is true.
Hence, both statements are true but Statement II is not correct
explanation for Statement I.
36. Pair of bisectors between the lines x 2 - 2axy - y 2 = 0 is
Þ
2
2π/3
2π/3
π/3
O
2
1
φ/3
2π/3
Y′
X
3′
m(1 - 3 tan 2 q ) + tan 3 q - 3 tan q = 0
\
Let
then
\
Þ
m=
3 tan q - tan 3 q
1 - 3 tan 2 q
= tan 3q
m = tan f,
tan 3q = tan f
3q = f, f + p , f + 2p
f p f 2p
f
+ ,
+
q= ,
3 3 3
3
3,
Thus, the given equation represents three lines through the
f f p f 2p
origin and they are inclined at angles , + , +
with
3 3 3 3
3
the X-axis. Clearly, they are equally inclined to each other.
Chap 03 Pair of Straight Lines
40. Given pair of lines is
2
2
2
2
2
2
Given line is
…(ii)
Ax + By + C = 0
2
a = A 2 - 3B 2 , h = 4 AB, b = B 2 - 3A 2
If q be the acute angle between the lines represented by
Eq. (i), then
tan q =
2 (h 2 - ab ) 2 16A 2 B 2 - ( A 2 - 3B 2 )( B 2 - 3A 2 )
=
|a + b |
|( A 2 - 3B 2 + B 2 - 3A 2 )|
4
=
\
4
2
2
2 ( 3A + 3B + 6A B )
2
2
| -2( A + B )|
2
=
p=
2
(A + B )
= 3
q = 60°
Y
then remaining third angle is 180° - (60° + 60° ) = 60°.
\ The DAOB is equilateral.
In DOAB,
0+0+C
Length of ^ from O on AB = p = OM =
(A 2 + B2 )
2
3( A + B )
2
+
By
+
Ax
60°
\
C=
M
C
2
(A + B2 )
and in DOAM,
sin60° =
B
- 3A 3 + A 2 B - 3AB 2 + B 3
= tan -1 ( 3 ) = 60°
2
Comparing Eq. (i) with ax + 2hxy + by = 0
-3A 2 + 3 A 2 B - 3AB 2 + 3B 3
= tan -1
…(i)
( A - 3B )x + 8ABxy + ( B - 3A ) y = 0
p
OA
OA =
Þ OA =
2p
3
2| C |
0
3 (A 2 + B2 )
p
60°
(0,0)
O
\ Area of the equilateral triangle =
A
60°
=
2
æy ö
æy ö
( B - 3A )ç ÷ + 8AB ç ÷ + ( A 2 - 3B 2 ) = 0
èx ø
èx ø
2
4C
3
C2
´
=
4
3 ´ (A 2 + B 2 )
3(A 2 + B 2 )
41. Lines of parallelogram are
L = 0, L = a
L¢ = 0, L ¢ = a
Solving for y/x, we have
L=a
y -8AB ± 64 A 2 B 2 - 4( B 2 - 3A 2 )( A 2 - 3B 2 )
=
x
2( B 2 - 3A 2 )
=
y -4 AB ± 3( A 2 + B 2 )
=
x
( B 2 - 3A 2 )
Þ
ì -4 AB ± 3 ( A 2 + B 2 )
y =í
B 2 - 3A 2
î
L′=0
B
2
where,
ü
ýx
þ
3 ( A 2 + B 2 ) - 4 AB
L = x cos q + y sin q - p
L ¢ = x cos q ¢ + y sin q ¢ - p ¢
Equation of the line AC, through the point of intersection of
L = 0 and L ¢ = 0 is
L + lL ¢ = 0
It also passes through
B 2 - 3A 2
L = a and L ¢ = a,
and slope of the line Ax + By + C = 0 is – A/B.
Acute angle between these lines
2
2
3( A + B ) - 4 AB
= tan -1
L=0
A
Taking the positive sign, slope of one of the lines
=
L′=a
( B - 3A )
Þ
C
D
- 4 AB ± 16A 2 B 2 - ( A 2 B 2 - 3B 4 - 3A 4 + 9 A 2 B 2 )
2
3
3
(OA )2
(side)2 =
4
4
2
X
Eq. (i) can be written as
2
2
B - 3A
237
2
+
A
B
2
2
æ A ö é 3( A + B ) - 4 AB ù
1 - ç ÷ê
ú
2
è B øë
B - 3A 2
ûú
then,
a + la = 0
\
l = -1
Hence, diagonal of AC is L - L ¢ = 0
i.e.
x (cos q - cos q ¢ ) + y (sin q - sin q ¢ ) - p + p ¢ = 0
Equation of the line BD through point of intersection of
L = 0 and L ¢ = a is
L + l( L ¢ - a ) = 0
…(i)
238
Textbook of Coordinate Geometry
Since, the lines lx + my + n = 0 and l1x + m1y + n1 = 0 are
equidistant from the origin, then
|n |
| n1 |
=
2
2
2
l +m
l1 + m12
it also passes through L = a and L ¢ = 0.
\
a + l(0 - a ) = 0
\
l =1
Equation of diagonal BD is L + L ¢ - a = 0
Þ x (cos q + cos q¢ ) + y (sin q + sin q¢ ) - p - p ¢ - a = 0
42. Let the first equation represents lines
(lx + my + n ) (l1x + m1y + n1 ) = 0
then the second equation represents lines
(lx + my - n )(l1x + m1y - n1 ) = 0
On comparing coefficients, ll1 = a , mm1 = b, nn1 = c
mn1 + m1n = 2 f , nl1 + n1l = 2g , lm1 + l1 m = 2h .
Let the angle between two non-parallel lines be q, then
2 (h 2 - ab )
|a + b |
tan q =
sin q =
\
( a - b ) + 4h
p 1p 2 =
=
=
\
+l
2
m12
(nl1 - n1l )2 (nl1 + n1l )2 = (n1m - nm1 )2 (n1m + nm1 )2
Þ [(nl1 + n1l )2 - 4ll1nn1 ] [nl1 + n1l ]2
= [(n1m + nm1 )2 - 4mm1nn1 ] [n1m + nm1 ]2
( 4 g 2 - 4ac ) ( 4 g 2 ) = ( 4 f
2
Þ
2
f
is
|n |
2
2
.
(l + m )
=
4| nn1|
=
=
4 |c |
( a - b ) 2 + 4h 2
=
pp
Area = 1 2
sinq
4| c |
=
æ 2 h 2 - ab
( ( a - b ) 2 + 4h 2 ) ç
ç ( a - b ) 2 + 4h 2
è
=
- 4bc ) ( 4 f 2 )
=
ö
÷
÷
ø
2|c |
(h 2 - ab )
43. Given, equation
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
represents two lines
lx + my + n = 0 and l1x + m1y + n1 = 0
then, ax 2 + 2hxy + by 2 + 2gx + 2 fy + c
= (lx + my + n ) (l1x + m1y + n1 )
Comparing the coefficients of similar terms, we get
ll1 = a , mm1 = b, nn1 = c
ü
ý …(i)
lm1 + ml1 = 2h , n1l + nl1 = 2g ,nm1 + n1m = 2 f þ
2
4
4
- g = c (bf
2
- bc ) f
2
[from Eq. (i)]
2
- ag )
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
{l 2l12 + m 2m12 + (lm1 + l1m )2 - 4l l1mm1 }
a 2 + b 2 + 4h 2 - 2ab
2
Also, product of distances from origin to the lines
represented by
+ m 2l12 + m 2m12 )
4 |c |
2
( g - ac )g = ( f
Þ
| 4nn1|
(l 2l12
Þ (nl1 - n1l ) (nl1 + n1l ) = (n1m - nm1 ) (n1m + nm1 )
On squaring both sides, we get
[using the six relations]
pp
\ Area of parallelogram = 1 2 ,
sinq
where, p 1 and p 2 are distance between two parallel sides.
| n - ( - n )|
| n - ( -n1 )|
and p 2 = 1
p1 =
\
2
2
2
l +m
l1 + m12
\
n 2l12 - n12l 2 = n12m 2 - n 2m12
Þ
Þ
2 (h 2 - ab )
2
n 2l12 + n 2m12 = n12l 2 + n12m 2
Þ
=
| n1|
(l12
+ m12 )
| nn1|
(l 2l12
+l
2
m12
+l
2
+ m 2l12 + m 2m12 )
| nn1|
(l 2l12
m12
+ m 2l12 + m 2m12 )
| nn1|
(l 2l12
+m
2
2
2
m12
+ (lm1 + ml1 )2 - 2ll1mm1 )
|c |
(a + b + 4h 2 - 2ab )
[from Eq. (i)]
|c |
(a - b ) 2 + 4h 2
44. We know that bisectors are mutually perpendicular to each
other, then
ax 4 + bx 3y + cx 2y 2 + dxy 3 + ay 4 = 0
represents two pairs of mutually perpendicular lines.
Let ax 4 + bx 3y + cx 2y 2 + dxy 3 + ay 4
= (ax 2 + pxy - ay 2 ) ( x 2 + qxy - y 2 )
where, p and q are constants.
Comparing the coefficients of similar terms, we get
b = aq + p
c = pq - 2a
d = - p - aq
…(i)
…(ii)
…(iii)
Chap 03 Pair of Straight Lines
On adding Eqs. (i) and (iii), we have
b+d =0
From Eq. (ii),
…(iv)
pq = c + 2a
But given that bisectors of one pair are given by the other,
i.e.
x2 - y2
xy
=
a - ( -a ) ( p / 2 )
x2 - y2 a
=
p
4 xy
Þ
…(v)
A + 3 A + A + 3 A = pr 2
p
Þ
A = r2
8
1 2
p 2
1
[Q area of sector = r 2q]
r q= r
Þ
2
8
2
p
q=
\
4
\ Angle between lines represented by
p
ax 2 + 2(a + b ) xy + by 2 = 0 is .
4
\
is the same as the other pair.
2
Þ
2
x + qxy - y = 0
or
x2 - y2
q
=4 xy
4
a
q
=p
4
From Eqs. (v) and (vi),
\
Again from Eq. (iv),
\
…(vi)
pq = - 4a
-4a = c + 2a
c + 6a = 0
Þ
Þ
\
O
r
θ r
2
p 2 (a + b ) - ab
=
4
|a + b|
1=
2 (a + b ) 2 - ab
|a + b|
(a + b ) 2 = 4(a + b ) 2 - 4ab
3a 2 + 2ab + 3b 2 = 0
Putting y = ± x in
my 2 + (1 - m 2 ) xy - mx 2 = 0, we get
± (1 - m 2 ) x 2 = 0
Þ
r
tan
46. The equation of the bisectors of the lines xy = 0 are y = ± x.
45. Let A be the area of small sector, then area of major sector is 3A.
r
θ
239
\
m2 = 1
CHAPTER
04
Circle
Learning Part
Session 1
●
●
Definition
Locus of the Mid-point of the Chords of the Circle
that Subtends an Angle of 2q at its Centre
●
Equation of Circle in Different Forms
●
Equation of Circle Passing Through Three
Non-Collinear Points
●
Different Forms of the Equations of a Circle
Maximum and Minimum Distance of a Point from
the Circle
Session 2
●
Diametric Form of a Circle
Session 3
●
●
Intercepts Made on the Axes by a Circle
Position of a Point with Respect to Circle
●
Session 4
●
●
●
●
Intersection of a Line and a Circle
The Length of Intercept Cut-off from a Line by a Circle
Tangent to a Circle at a Given Point
Normal to a Circle at a Given Point
●
Product of the Algebraical Distances PA and PB is
Constant when from P, A Secant be Drawn to Cut the
Circle in the Points A and B
●
Length of the Tangent from a Point to a Circle
Chord of Contact
Pair of Tangents
Session 5
●
●
●
●
Tangents from a Point to the Circle
Power of a Point with Respect to a Circle
Chord Bisected at a Given Point
Director Circle
●
●
Session 6
●
●
●
Diameter of a Circle
Common Tangents to Two Circles
Family of Circles
●
●
Two Circles Touching Each Other
Common Chord of Two Circles
Session 7
●
●
●
Angle of Intersection of Two Circles
Radical Centre
Limiting Point
●
●
●
Radical Axis
Co-axial System of Circles
Image of the Circle by the Line Mirror
Practice Part
●
●
JEE Type Examples
Chapter Exercises
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Session 1
Definition, Equation of Circle in Different Forms,
Locus of the Mid-point of the Chords of the Circle
that Subtends an Angle of 2q at its Centre
Definition
A circle is the locus of a point which moves in a plane, so
that its distance from a fixed point in the plane is always
constant.
The fixed point is called the centre of the circle and the
constant distance is called its radius.
Equation of Circle
in Different Forms
1. Centre-radius form
Let a be the radius and C (h, k ) be the centre of any circle.
If P ( x , y ) be any point on the circumference.
CP = a Þ (CP ) 2 = a 2
Then,
us
radi
fixed
point
P
(moving point)
constant
distance
C (centre)
( x - h ) 2 + (y - k ) 2 = a 2
Þ
This equation is known as the central form of the
equation of a circle.
P (x, y)
a
i.e.
CP = constant distance = Radius
Equation of a Circle : The curve traced by the moving
point is called its circumference. i.e. the equation of any
circle is satisfied by co-ordinates of all points on its
circumference.
OR
The equation of the circle is meant the equation of the
circumference.
OR
It is the set of all points lying on the circumference of the
circle.
Chord and Diameter : The line joining any two points
on the circumference is called a chord. If any chord
passing through its centre is called its diameter.
C (h, k)
Remark
When, C ( h, k ) = C ( 0, 0 ), then equation of circle becomes
x 2 + y 2 = a2 which is known as standard form of the circle.
2. Parametric form
If the radius of a circle whose centre is at C (0, 0 ) makes an
angle q with the positive direction of X-axis, then q is
called the parameter.
Let
CP = a
\
CM = x , PM = y
Þ x = a cos q, y = a sinq
Y
Q
P(x,y)
C
a
B
X′
P
θ
C (0, 0) M
A
AB = Chord, PQ = Diameter
where, C is centre of the circle.
Y′
X
Chap 04 Circle
Hence, (a cos q, a sinq ) or ‘q’ are the parametric
coordinates of the circle x 2 + y 2 = a 2 and x = a cos q
and y = a sinq are called parametric equations of the
circle x 2 + y 2 = a 2 with parameters a and q.
(0 £ q < 2 p ).
Remarks
1. The parametric coordinates of any point on the circle
( x - h) 2 + ( y - k ) 2 = a2 are given by ( h + a cos q, k + a sin q)
( 0 £ q < 2p ) and parametric equations of the circle
( x - h) 2 + ( y - k ) 2 = a2 are x = h + a cos q, y = k + a sinq.
2. Equation of the chord of the circle x 2 + y 2 = a2 joining
( a cos a, a sin a) and ( a cos b , a sinb ) is
…(i)
which is of the form
x 2 + y 2 + 2 gx + 2 fy + c = 0
...(ii)
This is known as the general equation of a circle
comparing Eqs. (i) and (ii), we get
h = - g, k = - f
y Example 1. Find the centre and radius of the circle
2x 2 + 2y 2 = 3x - 5 y + 7
3
5
7
x2 + y2 - x + y - = 0
2
2
2
If centre is (a , b ), then
1 æ 3ö 3
a = - ç- ÷ =
2 è 2ø 4
or
General form The equation of the circle with centre
(h, k ) and radius a is ( x - h ) 2 + (y - k ) 2 = a 2
x 2 + y 2 - 2hx - 2ky + h 2 + k 2 - a 2 = 0
(ii) If g 2 + f 2 - c = 0, then the radius of circle will be real.
Hence in this case, circle is called a point circle.
(iii) If g 2 + f 2 - c < 0, then the radius of circle will be
imaginary number. Hence in this case, circle is called a
virtual circle or imaginary circle.
4. Concentric circle Two circles having the same centre
C ( h, k ) but different radii r1 and r2 respectively are called
concentric circles. Thus, the circles ( x - h) 2 + ( y - k ) 2 = r12
and ( x - h) 2 + ( y - k ) 2 = r22, r1 ¹ r2 are concentric circles.
Therefore, the equations of concentric circles differ only in
constant terms.
Sol. The given equation of circle is
2x 2 + 2y 2 = 3x - 5y + 7
a+ bö
æ a + b ö = a cos æ a - b ö .
x cos æç
÷ + y sin ç
÷
ç
÷
è 2 ø
è 2 ø
è 2 ø
or
and a = ( g 2 + f 2 - c )
5ö
æ3
\ Centre of circle is (a , b ) i.e. ç , - ÷
è4
4ø
and
1. Rule for finding the centre and radius of a circle
(i) Make the coefficients of x 2 and y 2 equal to 1 and right
hand side equal to zero.
(ii) Then, coordinates of centre will be ( a, b ),
1
1
where, a = - (coefficient of x) and b = - (coefficient of y)
2
2
(iii) Radius = a2 + b 2 - (constant term)
2. Conditions for a circle A general equation of second
degree
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0
in x, y represent a circle, if
(i) coefficient of x 2 = coefficient of y 2
i.e.
a= b
(ii) coefficient of xy is zero
i.e.
h=0
3. Nature of the circle Radius of the circle
( g 2 + f 2 - c)
Now. the following cases are possible :
(i) If g 2 + f 2 - c > 0, then the radius of circle will be real. Hence
in this case, real circle is possible.
radius of the circle
= a 2 + b 2 - (constant term)
9 25 7
9 + 25 + 56 3 10
+
+ =
=
16 16 2
16
4
=
Radius = ( g 2 + f 2 - c ) ( g 2 + f 2 ³ c )
Remarks
1 æ5ö
5
b = - ç ÷= 2 è2ø
4
and
\ Coordinates of the centre are ( - g, - f ) and
x 2 + y 2 + 2gx + 2fy + c = 0 is
243
y Example 2. Prove that the radii of the circles
x 2 + y 2 = 1, x 2 + y 2 - 2 x - 6 y = 6 and
x 2 + y 2 - 4x - 12 y = 9 are in AP.
x2 + y2 = 1
Sol. Given circles are
2
...(ii)
2
2
...(iii)
x + y - 2x - 6y - 6 = 0
and
...(i)
2
x + y - 4 x - 12y - 9 = 0
Let r1, r2 and r3 be the radii of the circles Eqs. (i), (ii) and (iii),
respectively.
Then,
r1 = 1
r 2 = ( - 1) 2 + ( - 3) 2 + 6 = 4
and
r 3 = ( - 2) 2 + ( - 6) 2 + 9 = 7
Clearly, r 2 - r1 = 4 - 1 = 3 = r 3 - r 2
Hence, r1, r 2 , r 3 are in AP.
y Example 3. Find the equation of the circle whose
centre is the point of intersection of the lines
2x - 3y + 4 = 0 and 3x + 4 y - 5 = 0 and passes
through the origin.
244
Textbook of Coordinate Geometry
Sol. The point of intersection of the lines 2x - 3y + 4 = 0 and
æ 1 22 ö
3x + 4y - 5 = 0 is ç - , ÷.
è 17 17 ø
æ 1 22 ö
Therefore, the centre of the circle is at ç - , ÷.
è 17 17 ø
y Example 5. A circle has radius 3 units and its centre
lies on the line y = x - 1. Find the equation of the circle
if it passes through (7, 3).
Since, the origin lies on the circle, its distance from the
centre of the circle is radius of the circle, therefore,
2
2
485
ö
æ 22
ö
æ 1
r = ç- 0÷ + ç - 0÷ =
ø
è 17
ø
è 17
289
Sol. Let the centre of the circle be (h, k). Since, the centre lies
on y = x - 1, we get
...(i)
k =h -1
Since, the circle passes through the point ( 7, 3), therefore
the distance of the centre from this point is the radius r of
the circle. We have,
\ The equation of the circle becomes
2
2
1ö
22 ö
485
æ
æ
ç x + ÷ + çy - ÷ =
è
è
17 ø
17 ø
289
or
4 + 49 + 16 - 42 + l = 0
\
l = -27
From Eq. (i), required circle is
x 2 + y 2 - 8x + 6y - 27 = 0
17 ( x 2 + y 2 ) + 2x - 44y = 0
r = ( h - 7 ) 2 + ( k - 3) 2
Aliter : Q Point of intersection of the lines 2x - 3y + 4 = 0
æ 1 22 ö
and 3x + 4y - 5 = 0 is ç - , ÷.
è 17 17 ø
æ 1 22 ö
Therefore, the centre of the circle is at ç - , ÷
è 17 17 ø
Let required circle is
x 2 + y 2 + 2gx + 2 fy + c = 0
…(i)
1
22
, -f = ,c = 0
17
17
[Q Circle passes through origin]
2x 44
2
2
From Eq. (i),
- y =0
x +y +
17 17
or
17( x 2 + y 2 ) + 2x - 44y = 0
Here, - g = -
y Example 4. Find the equation of the circle
concentric with the circle x 2 + y 2 - 8 x + 6 y - 5 = 0
and passing through the point ( -2, - 7 ) .
Sol. The given equation of circle is
x 2 + y 2 - 8x + 6y - 5 = 0
or
3 = ( h - 7 ) 2 + ( h - 1 - 3) 2
2
Þ
9 = (h - 7 ) + (h - 4 )
Þ
h 2 - 11h + 28 = 0
or
(h - 7 ) (h - 4 ) = 0
or
h = 7 and h = 4
For h = 7, we get k = 6 from Eq. (i)
and for h = 4, we get k = 3, from Eq. (i).
Hence, there are two circles which satisfy the given
conditions. They are
( x - 7 ) 2 + ( y - 6) 2 = 9
or
and
or
x 2 + y 2 - 14 x - 12y + 76 = 0
( x - 4 ) 2 + ( y - 3) 2 = 9
x 2 + y 2 - 8x - 6y + 16 = 0
y Example 6. Find the area of an equilateral triangle
inscribed in the circle
x 2 + y 2 + 2 gx + 2 fy + c = 0.
Sol. Given circle is
Therefore, the centre of the circle is at ( 4, - 3). Since, the
required circle is concentric with this circle, therefore, the
centre of the required circle is also at ( 4, - 3). Since, the
point ( -2, - 7 ) lies on the circle, the distance of the centre
from this point is the radius of the circle. Therefore, we get
x 2 + y 2 + 2gx + 2 fy + c = 0
A
60°
Hence, the equation of the circle becomes
O
( x - 4 )2 + (y + 3)2 = 52
60°
x 2 + y 2 - 8x + 6y - 27 = 0
Aliter : Equation of concentric circle is
x 2 + y 2 - 8x + 6y + l = 0
which pass through ( -2, - 7 ), then
...(i)
Let O be the centre and ABC be an equilateral triangle
inscribed in the circle Eq. (i).
r = ( 4 + 2)2 + ( -3 + 7 )2 = 52
or
[from Eq. (i)]
2
B
M
C
O º (-g, - f )
…(i)
and
OA = OB = OC = g 2 + f
2
-c
...(ii)
Chap 04 Circle
In DBOM, sin60° =
BM
OB
Þ
BM = OB sin60° = (OB )
\
BC = 2BM = 3 (OB )
\
3
( BC )2
4
3
3 (OB )2
=
4
3 3 2
=
(g + f
4
3
2
...(iii)
Area of D ABC =
[from Eq. (iii)]
2
Locus of the Mid-point of
the Chords of the Circle that
Subtends an Angle of 2q at
its Centre
Let mid-point M ( x 1 ,y 1 ) and centre, radius of circle are
(h,k ), r respectively, then
cos q =
- c ) sq units.
( x 1 - h ) 2 + (y 1 - k ) 2
CM
=
r
r
A
y Example 7. Find the parametric form of the equation
of the circle
x 2 + y 2 + px + py = 0.
M
2
(x
1,
y1 )
q
q
r
C (h , k )
Sol. Equation of the circle can be re-written in the form
2
B
2
pö
pö
p
æ
æ
ç x + ÷ + çy + ÷ =
è
è
2ø
2ø
2
Therefore, the parametric form of the equation of the given
circle is
p
p
p
x=- +
cos q = ( -1 + 2 cos q )
2
2
2
p
p
p
and
y=- +
sin q = ( -1 + 2 sin q )
2
2
2
where, 0 £ q < 2p.
y Example 8. If the parametric of form of a circle is
given by
(a) x = - 4 + 5 cosq and y = - 3 + 5 sin q
(b) x = a cosa + b sina and y = a sin a - b cos a
find its cartesian form.
Sol. (a) The given equations are
x = - 4 + 5 cosq
and
y = - 3 + 5 sinq
or
( x + 4 ) = 5 cos q
and
(y + 3) = 5 sin q
Squaring and adding Eqs. (i) and (ii), then
( x + 4 ) 2 + ( y + 3) 2 = 52
or
( x + 4 )2 + (y + 3)2 = 25
(b) The given equations are
...(iii)
x = a cos a + b sin a
...(iv)
y = a sin a - b cos a
Squaring and adding Eqs. (iii) and (iv), then
x 2 + y 2 = (a cos a + b sin a )2 + (a sin a - b cos a )2
Þ
x 2 + y 2 = a2 + b2
\ Required locus is
( x - h ) 2 + (y - k ) 2 - r 2
r
2
= -sin2 q
Remembering Method :
First make coefficient of x 2 = coefficient of y 2 = 1
LHS of circle
and RHS of circle is zero, then
= -sin2 q
(radius) 2
y Example 9. Find the locus of mid-points of the
chords of the circle 4 x 2 + 4 y 2 - 12x + 4 y + 1 = 0 that
2p
at its centre.
subtend an angle of
3
2p
p
Þ q=
3
3
Equation of circle can be written as
1
x 2 + y 2 - 3x + y + = 0
4
\Required locus is
1
x 2 + y 2 - 3x + y +
4 = - sin 2 æç p ö÷ = - 3
2
è3ø
4
æ 9 1 1ö
+ - ÷
ç
è 4 4 4ø
Sol. Here, 2q =
…(i)
...(ii)
245
Þ
or
1
27
=4
16
2
2
16( x + y ) - 48x + 16y + 31 = 0
x 2 + y 2 - 3x + y +
246
Textbook of Coordinate Geometry
Exercise for Session 1
1.
If x 2 + y 2 - 2x + 2ay + a + 3 = 0 represents a real circle with non-zero radius, then most appropriate is
(a) a Î (- ¥, - 1)
(c) a Î (2, ¥)
2.
If the equation ax 2 + (2 - b )xy + 3y 2 - 6bx + 30y + 6b = 0 represents a circle, then a 2 + b 2 is
(a) 5
(c) 25
3.
4.
(b) 13
(d) 41
The equation of the circle passing through (4, 5) having the centre at (2, 2) is
(a) x 2 + y 2 + 4x + 4y - 5 = 0
(b) x 2 + y 2 - 4x - 4y - 5 = 0
(c) x 2 + y 2 - 4x - 13 = 0
(d) x 2 + y 2 - 4x - 4y + 5 = 0
Equation of the diameter of the circle is given by x 2 + y 2 - 12x + 4y + 6 = 0 is given by
(a) x + y = 0
(c) x = y
5.
3
2
5
(c)
2
(b) 3
(d) 5
Area of a circle in which a chord of length 2 makes an angle
p
4
(c) p
8.
9.
p
2
(d) 2p
The lines 2x - 3y - 5 = 0 and 3x - 4y = 7 are diameters of a circle of area 154 sq units, then the equation of the
circle is :
(a) x 2 + y 2 + 2x - 2y - 62 = 0
(b) x 2 + y 2 + 2x - 2y - 47 = 0
(c) x 2 + y 2 - 2x + 2y - 62 = 0
(d) x 2 + y 2 - 2x + 2y - 47 = 0
If the lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 lie along diameters of a circle of circumference 10p, then the
equation of the circle is
(a) x 2 + y 2 - 2x + 2y - 23 = 0
(b) x 2 + y 2 - 2x - 2y - 23 = 0
(c) x 2 + y 2 + 2x + 2y - 23 = 0
(d) x 2 + y 2 + 2x - 2y - 23 = 0
The triangle PQR is inscribed in the circle x 2 + y 2 = 25. If Q and R have coordinates (3, 4) and ( -4,3)
respectively, then ÐQPR is equal to
p
2
p
(c)
4
p
3
p
(d)
6
(a)
10.
p
at the centre is
2
(b)
(a)
7.
(b) x + 3y = 0
(d) 3x + 2y = 0
If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to a circle, then the diameter of the circle is
(a)
6.
(b) a Î (-1, 2)
(d) a Î (- ¥, - 1) È (2, ¥)
(b)
If a circle is concentric with the circle x 2 + y 2 - 4x - 6y + 9 = 0 and passes through the point ( -4, - 5), then its
equation is
(a) x 2 + y 2 + 4x + 6y - 87 = 0
2
2
(c) x + y - 4x - 6y - 87 = 0
(b) x 2 + y 2 - 4x + 6y + 87 = 0
(d) x 2 + y 2 + 4x + 6y + 87 = 0
Chap 04 Circle
11.
Let AB be a chord of the circle x 2 + y 2 = r 2 subtending a right angle at the centre. Then, the locus of the
centroid of the DPAB as P moves on the circle is
(a) a parabola
(c) an ellipse
12.
247
(b) a circle
(d) a pair of straight lines
Let PQ and RS be tangents extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a
point X on the circumference of the circle, then 2r equals
(a) PQ × RS
(b)
2PQ.RS
PQ + RS
(d)
(c)
PQ + RS
2
(PQ )2 + (RS )2
2
13.
Find the centre and radius of the circle 5x 2 + 5y 2 + 4x - 8y = 16.
14.
Prove that the centres of the circles x 2 + y 2 = 1, x 2 + y 2 + 6x - 2y - 1 = 0 and x 2 + y 2 - 12 x + 4y = 1 are
collinear.
15.
Find the equation of the circle whose centre is (1, 2) and which passes through the point of intersection of
3x + y = 14 and 2x + 5y = 18.
16.
Find the equation of the circle passing through the centre of the circle x 2 + y 2 - 4x - 6y = 8 and being
concentric with the circle x 2 + y 2 - 2x - 8y = 5 .
1 2
( x + y 2 ) + x cos q + y sin q - 4 = 0 is x 2 + y 2 = 1.
2
17.
Prove that the locus of the centre of the circle
18.
Find the equation of the following curves in cartesian form. If the curve is a circle, then find its centre and radius
x = - 1 + 2 cos a, y = 3 + 2 sin a
(0 £ a < 2p )
Session 2
Diametric Form of a Circle, Equation of Circle
Passing Through Three Non-Collinear Points
Y
, y)
P (x
y 2)
x 2,
(
B
90°
,k
C(h
)
)A
, y1
X´
(x 1
X
O
Y′
Now,
Slope of AP =
and
Slope of BP =
y - y1
x - x1
y - y2
x - x2
Since, Ð APB = 90°
\ Slope of AP ´ Slope of BP = -1
(y - y 1 ) (y - y 2 )
´
= -1
Þ
(x - x 1 ) (x - x 2 )
Þ
( x - x 1 ) ( x - x 2 ) + (y - y 1 ) (y - y 2 ) = 0
Remark
The diameteric form of a circle can also be written as
x 2 + y 2 - x ( x1 + x 2 ) - y ( y1 + y2 ) + x1x 2 + y1 y2 = 0
or
x 2 + y 2 - x (sum of abscissae) - y (sum of ordinates)
+ product of abscissae + product of ordinates = 0
y Example 10. Find the equation of the circle the end
points of whose diameter are the centres of the circles
x 2 + y 2 + 6 x - 14 y = 1 and x 2 + y 2 - 4 x + 10y = 2.
and x 2 + y 2 - 4 x + 10y - 2 = 0 are ( - 3, 7 ) and (2, - 5),
respectively.
According to the question, the points ( -3, 7 ) and (2, - 5) are
the extremities of the diameter of required circle.
Hence, equation of circle is
( x + 3) ( x - 2) + ( y - 7 ) ( y + 5) = 0
Þ
x 2 + y 2 + x - 2y - 41 = 0
y Example 11. The sides of a square are
x = 2, x = 3, y = 1 and y = 2 . Find the equation of the
circle drawn on the diagonals of the square as its
diameter.
Sol. Let ABCD be a square and equation of its sides
AB, BC , CD and DA are y = 1, x = 3, y = 2, and x = 2,
respectively.
(2, 2)
D
y=2
A
(2, 1)
(3, 2)
C
x=3
Theorem : The equation of the circle on the line segment
joining ( x 1 , y 1 ) and ( x 2 , y 2 ) as diameter is
( x - x 1 ) ( x - x 2 ) + (y - y 1 ) (y - y 2 ) = 0.
Proof : Let A ( x 1 , y 1 ) and B ( x 2 , y 2 ) be the end points of
a diameter and let P ( x , y ) be any point on the circle
Sol. The centres of the given circles
x 2 + y 2 + 6x - 14y - 1 = 0
x=2
Diametric Form of a Circle
y=1
B
(3, 1)
Then, A º (2, 1), B º (3, 1), C º (3, 2) and D º (2, 2)
Since, diagonals of squares are the diameters of the circle,
then equation of circle is
( x - 2) ( x - 3) + (y - 1)(y - 2) = 0
Þ
x 2 + y 2 - 5x - 3y + 8 = 0 (If AC as diameter).
y Example 12. The abscissae of two points A and B
are the roots of the equation x 2 + 2ax - b 2 = 0 and
their ordinates are the roots of the equation
x 2 + 2px - q 2 = 0. Find the equation and the radius
of the circle with AB as diameter.
Sol. Given equations are
x 2 + 2ax - b 2 = 0
...(i)
Chap 04 Circle
x 2 + 2px - q 2 = 0
and
Let the roots of the Eq. (i) be a and b and those of Eq. (ii) be
g and d. Then,
g + d = - 2p ü
a + b = - 2a ü
ý
ý and
gd = - q 2 þ
a b = - b2þ
Let A º (a , g ) and B º (b, d ) .
Now, equation of circle whose diameter is AB will be
( x - a ) ( x - b ) + (y - g ) (y - d ) = 0
Þ
x 2 + y 2 - (a + b ) x - ( g + d ) y + ab + gd = 0
Þ
x 2 + y 2 + 2ax + 2py - b 2 - q 2 = 0
and
radius = (a 2 + p 2 + b 2 + q 2 )
C
y Example 13. Find the equation of the circle which
passes through the points (4, 1),(6, 5) and has its centre
on the line 4 x + y = 16 .
...(i)
...(ii)
x 22 + y 22 + 2 gx 2 + 2 fy 2 + c = 0,
...(iii)
x 23 + y 23 +2 gx 3 + 2 fy 3 + c = 0,
...(iv)
g, f , c are obtained from Eqs. (ii), (iii) and (iv). Then, to
find the circle Eq. (i).
Aliter : Eliminate g, f , c from Eqs. (i), (ii), (iii) and (iv)
with the help of determinant
1
1
=0
1
1
which is the required equation of circle
Applying R 1 ® R 1 - R 4 , R 2 ® R 2 - R 4 and R 3 ® R 3 - R 4
then, we get
x 2 + y 2 - x 23 - y 23
x 12 + y 12 - x 23 - y 23
x 22 + y 22 - x 23 - y 23
x 23 + y 23
x -x3
x1 - x 3
x2 - x 3
x3
y -y3
y1 - y 3
y2 - y 3
y3
0
0
=0
0
1
O´
D
x 12 + y 12 + 2 gx 1 + 2 fy 1 + c = 0,
y
y1
y2
y3
Remarks
O
If three points ( x 1 , y 1 ), ( x 2 , y 2 ), ( x 3 , y 3 ) lie on the circle
Eq. (i), their coordinates must satisfy its equation. Hence,
solving equations
x
x1
x2
x3
y -y3
y1 - y 3 = 0
y2 - y 3
A
x 2 + y 2 + 2 gx + 2 fy + c = 0
+y2
+ y 12
+ y 22
+ y 23
x -x3
x1 - x 3
x2 - x 3
B
Let the equation of circle be
x2
x 12
x 22
x 23
i.e.
1. Cyclic quadrilateral If all four vertices of a quadrilateral lie
on a circle, then the quadrilateral is called a cyclic
quadrilateral. The four vertices are said to be concyclic.
2. Concyclic points If A, B, C, D are concyclic, then
OA× OD = OB × OC
where, O¢ be the centre of the circle.
Equation of Circle Passing
Through Three Non-Collinear
Points
and
x 2 + y 2 - x 23 - y 23
x 12 + y 12 - x 23 - y 23
x 22 + y 22 - x 23 - y 23
...(ii)
249
Sol. Let the equation of circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
...(i)
Since, the centre of Eq. (i) i.e. ( - g , - f ) lies on 4 x + y = 16
then,
- 4 g - f = 16
or
...(ii)
4 g + f + 16 = 0
Since, the points ( 4, 1) and (6, 5) lie on circle
x 2 + y 2 + 2gx + 2 fy + c = 0, we get the equations
16 + 1 + 8g + 2 f + c = 0
or
17 + 8g + 2 f + c = 0
and
36 + 25 + 12g + 10 f + c = 0
or
61 + 12g + 10 f + c = 0
Subtracting Eq. (iii) from Eq. (iv), then
44 + 4 g + 8 f = 0
Solving Eqs. (ii) and (v), we get
f = - 4 and g = - 3
Now, from Eq. (iii), 17 - 24 - 8 + c = 0
Þ
c = 15
Hence, the equation of circle becomes
x 2 + y 2 - 6x - 8y + 15 = 0
...(iii)
...(iv)
...(v)
y Example 14. Find the equation of the circle passing
through the three non-collinear points (1, 1),(2, - 1) and
( 3, 2 ) .
Sol. Let the equation of circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
Since, the three given points lie on circle Eq. (i), we get
1 + 1 + 2g + 2 f + c = 0
...(i)
250
Textbook of Coordinate Geometry
or
...(ii)
2g + 2 f + c + 2 = 0
Þ
4 + 1 + 4g - 2f + c = 0
or
...(iii)
4g - 2f + c + 5 = 0
Þ
9 + 4 + 6g + 4 f + c = 0
or
...(iv)
6g + 4 f + c + 13 = 0
Subtracting Eq. (ii) from Eq. (iii) and subtracting Eq. (iii)
from Eq. (iv), then
...(v)
2g - 4 f + 3 = 0
and
...(vi)
2g + 6 f + 8 = 0
Solving Eq. (v) and Eq. (vi), we get
1
5
and g = f =2
2
Now, from Eq. (ii), -5 - 1 + c + 2 = 0
\
c =4
Hence, from Eq. (i), equation of circle is
x 2 + y 2 - 5x - y + 4 = 0
Aliter I Equation of circle passing through three points
(1, 1), (2, - 1) and (3, 2) is
x2 + y2
12 + 12
2
2 + ( - 1) 2
32 + 22
Þ
x2 + y2
2
5
13
x
1
2
3
x
1
2
3
y
1
-1
2
y
1
-1
2
1
1
=0
1
1
1
1 =0
1
1
2
x -1
...
1
y -1
...
1
Þ
3
11
1
2
-2
1
0
.
.
.
... 1
.
. =0
.
0
.
.
.
0
Expand with respect to fourth column, then
x2 + y2 - 2 x - 1 y - 1
3
1
-2 =0
11
2
1
Expand with respect to first now, then
( x 2 + y 2 - 2)(5) - ( x - 1)(25) + (y - 1)( -5) = 0
or
C(3,2)
x 2 + y 2 - 5x - y + 4 = 0
Aliter II The centre of the circumcircle is the point of
intersection of the right bisectors of the sides of the triangle
and the radius is the distance of the circumcentre from any
of the vertices of the triangle.
E
(1, 1)
A
O
X´
D
X
O
Y´
B(2,–1)
Let D and E are the mid-points of BC and CA, then
æ5 1ö
æ 3ö
D º ç , ÷ and E º ç2, ÷
è2 2ø
è 2ø
Slope of BC =
2 - ( - 1)
=3
3-2
1
3
1
1æ
5ö
\ Equation of OD, y - = - ç x - ÷
è
2
3
2ø
\
Slope of OD = -
Þ
\
or
6y - 3 = - 2x + 5
2x + 6y - 8 = 0
x + 3y - 4 = 0
1-2 1
Slope of CA =
=
1-3 2
and
Applying R1 ® R1 - R 2 , R 3 ® R 3 - R 2 and R 4 ® R 4 - R 2 ,
then
x2 +y2 -2
Y
…(i)
Slope of OE = - 2
\
\ Equation of OE,
3
y - = - 2 ( x - 2)
2
Þ
2y - 3 = - 4 x + 8
Þ
4 x + 2y - 11 = 0
...(ii)
5
1
Solving Eq. (i) and Eq. (ii), we get x = and y =
2
2
æ5 1ö
\ Circumcentre is ç , ÷ and radius
è2 2ø
2
2
1ö
5
5ö
æ
æ
OC = ç3 - ÷ + ç2 - ÷ =
è
è
2ø
2
2ø
\ Equation of circle is
( x - 5 / 2) 2 + ( y - 1 / 2) 2 = 5 / 2
Þ
x 2 + y 2 - 5x - y + 4 = 0
y Example 15. Show that the four points
(1, 0), (2, - 7 ), (8, 1) and (9, - 6 ) are concyclic.
Sol. Since, the given four points are concyclic, we are to show
that they lie on a circle. Let the general equation of circle
is
…(i)
x 2 + y 2 + 2gx + 2 fy + c = 0
Chap 04 Circle
has three parameters, it is sufficient to obtain the equation
of the circle passing through any three of these points. For
concyclic, the fourth point should lie on this circle.
Let three points A (1, 0), B (2, - 7 ) and D (8, 1) lie on Eq. (i),
then
...(ii)
1 + 0 + 2g + 0 + c = 0 or 1 + 2g + c = 0
Y
D (8, 1)
X´
O
X
A (1, 0)
C (9, –6)
Y´
Now, subtracting Eq. (ii) from Eq. (iii), we get
52 + 2g - 14 f = 0
or
...(v)
26 + g - 7 f = 0
and subtracting Eq. (iii) from Eq. (iv), we get
12 + 12g + 16 f = 0
...(vi)
Þ
3 + 3g + 4 f = 0
Solving Eq. (v) and Eq. (vi), we get
g = - 5 and f = 3
From Eq. (ii), 1 - 10 + c = 0
\
c =9
Therefore, equation of circle passing through these points is
x 2 + y 2 - 10x + 6y + 9 = 0
B (2, –7)
…(iii)
Substituting the fourth point in the equation of this circle,
we get
(9 )2 + ( -6)2 - 10 (9 ) + 6 ( -6) + 9 = 0
…(iv)
Hence, the point C (9, - 6) lies on the circle, that is, the four
points are concyclic.
( 2) 2 + ( - 7 ) 2 + 2g ( 2) + 2 f ( - 7 ) + c = 0
or
and
53 + 4 g - 14 f + c = 0
( 8) + ( 1) + 2g ( 8) + 2 f ( 1) + c = 0
2
2
Þ
65 + 16g + 2 f + c = 0
251
Exercise for Session 2
1.
If the line x + 2ly + 7 = 0 is a diameter of the circle x 2 + y 2 - 6x + 2y = 0, then the value of l is
(a) 1
(c) 5
2.
If one end of a diameter of the circle 2x 2 + 2y 2 - 4x - 8y + 2 = 0 is (-1, 2), then the other end of the diameter is
(a) (2, 1)
(c) (4, 3)
3.
4.
(b) (3, 2)
(d) (5, 4)
If a circle passes through the points (0,0), (a, 0) and (0, b ), then centre of the circle is
(a) (a,b )
a b
(b) æç , ö÷
è 2 2ø
a b
(c) æç , ö÷
è 2 4ø
a b
(d) æç , ö÷
è 4 2ø
A circle passes through the points ( -1, 3) and (5, 11) and its radius is 5. Then, its centre is
(a) (-5, 0)
(c) (2, 7)
5.
(b) 3
(d) 7
(b) (-5, 7)
(d) (5, 0)
The radius of the circle, having centre at (2, 1) whose one of the chord is a diameter of the circle
x 2 + y 2 - 2x - 6y + 6 = 0 is
(a) 3
(c) 1
6.
(b) 2
(d) 3
The centre of the circle inscribed in the square formed by the lines x 2 - 8x + 12 = 0 and y 2 - 14y + 45 = 0 is
(a) (4, 7)
(c) (9, 4)
(b) (7, 4)
(d) (4, 9)
252
Textbook of Coordinate Geometry
7.
8.
ABCD is a square whose side is a. The equation of the circle circumscribing the square, taking AB and AD as
the axes of reference is
(a) x 2 + y 2 + ax + ay = 0
(b) x 2 + y 2 - ax + ay = 0
(c) x 2 + y 2 - ax - ay = 0
(d) x 2 + y 2 + ax - ay = 0
The locus of the centre of the circle for which one end of the diameter is (3, 3) while the other end lies on the
line x + y = 4 is
(a) x + y = 3
(c) x + y = 7
9.
(b) x + y = 5
(d) x + y = 9
The equation of the circle which passes through (1, 0) and (0, 1) and has its radius as small as possible is
(a) x 2 + y 2 + x + y = 0
2
2
(c) x + y + x - y = 0
10.
(b) -1
14
3
(a) 1
14
3
12.
(d) x 2 + y 2 - x - y = 0
If the points (2, 0), (0, 1), (4,5) and (0, c ) are concyclic, then the value of c is
(d) -
(c)
11.
(b) x 2 + y 2 - x + y = 0
The point on a circle nearest to the point P(2,1) is at a distance of 4 units and farthest point is (6, 5), then the
centre of the circle is
(a) (3 +
2, 2 +
2)
(b) (2 +
2, 3 +
2)
(c) (4 +
2, 3 +
2)
(d) (3 +
2, 4 +
2)
The intercept on the line y = x by the circle x 2 + y 2 - 2x = 0 is AB. Equation of the circle on AB as a diameter is
(a) x 2 + y 2 - x - y = 0
(b) x 2 + y 2 - x + y = 0
(c) x 2 + y 2 + x + y = 0
(d) x 2 + y 2 + x - y = 0
13.
Find the equation of the circle, the end points of whose diameter are (2, - 3) and ( -2, 4). Find the centre and
radius.
14.
If (4, 1) be an extremity of a diameter of the circle x 2 + y 2 - 2x + 6y - 15 = 0, find the coordinates of the other
extremity of the diameter.
15.
Find the equation of the circle drawn on the diagonal of the rectangle as its diameter whose sides are
x = 4, x = - 2, y = 5 and y = - 2.
16.
Find the equation of the circle which passes through the points (1, 1), (2, 2) and whose radius is 1.
17.
Find the equation of the circle which passes through the points (3, 4), (3, - 6) and (1, 2).
Session 3
Intercepts Made on the Axes by a Circle, Different
Forms of the Equations of a Circle, Position of a Point
with Respect to Circle, Maximum and Minimum
Distance of a Point from the Circle
Intercepts Made on the Axes
by a Circle
Let the circle x 2 + y 2 + 2 gx + 2 fy + c = 0
Remarks
...(i)
Length of intercepts on X-axis and Y-axis are
| AB | = | x 2 - x 1 | and | CD | = | y 2 - y 1 | respectively.
Y
Y´
A
(x1, 0)
B
(x2, 0)
X
The circle intersects the X-axis, when y = 0
then
x 2 + 2 gx + c = 0
Since, the circle intersects the X-axis at A ( x 1 , 0 ) and
B ( x 2 , 0)
then,
x 1 + x 2 = - 2 g, x 1 x 2 = c
\
| AB | = | x 2 - x 1 | = ( x 2 + x 1 ) 2 - 4 x 1 x 2
= 2 (g 2 - c )
and the circle intersects the Y-axis, when x = 0, then
y 2 + 2 fy + c = 0
Since, the circle intersects the Y-axis at C (0, y 1 ) and
D (0, y 2 )
then,
y 1 + y 2 = - 2 f , y 1y 2 = c
\
and if circle touches Y-axis, then| CD| = 0
\
c = f2
3. If circle touches both axes, then| AB| = 0 =| CD|
\
c = g2 = f 2
y Example 16. Find the equation of the circle whose
diameter is the line joining the points ( -4 , 3) and
(12, - 1) . Find also the intercept made by it on Y-axis.
(0, y2)D
(0, y1)C
X´
O
1. Intercepts are always positive.
2. If circle touches X-axis, then| AB| = 0
\
c = g2
| CD | = | y 2 - y 1 | = (y 2 + y 1 ) 2 - 4y 2 y 1
= 2 (f 2 - c)
Sol. Equation of circle having ( -4, 3) and (12, - 1) as the ends
of a diameter is
( x + 4 ) ( x - 12) + (y - 3) (y + 1) = 0
...(i)
Þ
x 2 + y 2 - 8x - 2y - 51 = 0
Comparing Eq. (i) with standard equation of circle
x 2 + y 2 + 2gx + 2 fy + c = 0
then,
g = - 4, f = - 1, c = - 51
\ Intercept on Y-axis = 2 ( f
2
- c ) = 2 (1 + 51) = 4 13.
Different Forms of the
Equations of a Circle
(i) When the circle passes through the origin
(0, 0) and has intercepts 2a and 2b on the
X-axis and Y -axis, respectively
Here,
OA = 2a, OB = 2b
then,
OM = a and ON = b
Centre of the circle is C(a, b) and radius
OC = (a 2 + b2 )
254
Textbook of Coordinate Geometry
Y
Remark
If the circle x 2 + y 2 + 2gx + 2fy + c = 0 touches the Y-axis, then
B
| - g| = g 2 + f 2 - c
c = f2
or
C (α, β)
N
β
(iv) When the circle touches both axes
α M
O
Here,
| OM | = | ON |
Since, length of tangents are equal from any point on
circle.
X
A
then, equation of circle is
( x - a ) 2 + (y - b) 2 = a 2 + b2
Y
x 2 + y 2 - 2ax - 2by = 0
or
Remark
α
N
If a circle is passing through origin, then constant term is absent
i.e.
x 2 + y 2 + 2gx + 2fy = 0
α
(ii) When the circle touches X-axis
( x - a ) 2 + (y - b) 2 = b2
\ Let centre is (a, a ) also radius = a
\ Equation of circle is ( x - a ) 2 + (y - a ) 2 = a 2
x 2 + y 2 - 2ax - 2ay + a 2 = 0
Þ
Y
X
M
O
Let (a, b) be the centre of the circle, then radius =| b |
\ Equation of circle is
C (α, α)
Remarks
1. If the circle x 2 + y 2 + 2gx + 2fy + c = 0 touches both the axes,
C (α, β)
then
β
X
x 2 + y 2 - 2ax - 2by + a 2 = 0
Þ
Remark
If the circle x 2 + y 2 + 2gx + 2fy + c = 0 touches the X-axis, then
| - f| = g2 + f 2 - c
c = g2
or
c = g2 = f 2
\
M
O
|- g| =| - f| = g 2 + f 2 - c
\
g=f =± c
\ Equation of circle is
x 2 + y2 ± 2 c x ± 2 c y + c = 0
Þ
(x ±
c )2 + ( y ±
c )2 = c 2
2. If a > 0, then centres for I, II, III and IV quadrants are
( a, a),( - a, a),( - a, - a) and ( a, - a), respectively.
Then, equation of circles in these quadrants are
( x - a) 2 + ( y - a) 2 = a2, ( x + a) 2 + ( y - a) 2 = a2,
(iii)When the circle touches Y-axis
Let (a, b) be the centre of the circle, then
radius = | a |
Y
( x + a) 2 + ( y + a) 2 = a2 and ( x - a) 2 + ( y + a) 2 = a2,
respectively.
(v) When the circle touches X-axis and cut-off
intercepts on Y-axis of length 2l
Let centre be (a, b)
N
α
C (α, β)
Y
P
X
O
M
l
\ Equation of circle is
2
2
( x - a ) + (y - b) = a
2
l
Þ x + y 2 - 2ax - 2by + b2 = 0
2
Q
O
α
β
α
C (α, β)
β
N
X
Chap 04 Circle
Let centre be (a, b)
radius = b
CQ = CN = b
\
255
Y
In DCMQ, b2 = a 2 + l 2 , a = (b2 - l 2 ) (for I quadrant)
k
\ Equation of circle is
α
N
[ x - (b2 - l 2 ) ]2 + (y - b) 2 = b2
Remark
P
Q Length of intercepts on Y-axis of the circle
O Q
2
C (α, β)
k
β
l
l
M
X
2
2
x + y + 2gx + 2fy + c = 0 is 2l = 2 ( f - c )
\
l2 = f 2 - c
i.e.
2
2
(CP ) = (CQ ) = l
and also circle touches X-axis
then,
c = g2
(say)
2
2
a + k 2 = b2 + l 2 = l2
l 2 = f 2 - g 2 or l 2 = ( - f ) 2 - ( - g ) 2
\
radius = CP = CQ = l
\ Locus of centre is y 2 - x 2 = l 2 (rectangular hyperbola)
\
(vi) When the circle touches Y-axis and cut-off
intercept on X-axis of length 2k
Let centre be (a, b)
a = l2 - k 2 and b = l2 - l 2
\ Equation of circle is
2
(for I quadrant)
2 2
2
( x - l - k ) + (y - l - l 2 ) 2 = l2
(viii) When the circle passes through the origin
and centre lies on X-axis
Y
Let centre of circle be C (a, 0 )
Y
α
N
C (α, β)
α
β
k M k
O Q
X
C
(a, 0)
O
radius = a
\
X
CN = CQ = a
a 2 = b2 + k 2
In DCMQ,
b = (a 2 - k 2 )
(for I quadrant)
\ Equation of circle is
2
2
2 2
( x - a ) + (y - a - k ) = a
2
Remarks
( x - a ) 2 + (y - 0 ) 2 = a 2 or
(ix) When the circle passes through origin and
centre lies on Y-axis
Y
2
2
x + y + 2gx + 2fy + c = 0 is 2k = 2 ( g - c )
i.e.
2
x 2 + y 2 - 2ax = 0
Let centre of circle be C (0, a )
Q Length of intercept on X-axis of the circle
2
\ radius = a
\ Equation of circle is
2
k =g -c
and also circle touches Y-axis
then,
c = f2
\
k 2 = g 2 - f 2 = ( - g )2 - ( - f )2
2
2
C (0, a)
2
\ Locus of centre is x - y = k (rectangular hyperbola)
(vii)When the circle cut-off intercepts on X-axis
and Y-axis of lengths 2l and 2k and not
passing through origin
O
X
256
Textbook of Coordinate Geometry
From Eq. (i), the equation of circle can be written as
x 2 + y 2 ± 2 (c + 9 ) x ± 2 cy + c = 0
\ radius = a
\ Equation of circle is
The circle touches the Y-axis
\
x =0
\
y 2 ± 2 cy + c = 0
( x - 0 ) 2 + (y - a ) 2 = a 2
x 2 + y 2 - 2ay = 0
or
y Example 17. Find the equation of the circle which
touches the axis of y at a distance of 4 units from the
origin and cuts the intercept of 6 units from the axis
of x.
CM = NO = 4
Sol. Q
( PC )2 = (3)2 + ( 4 )2
In DPCM,
\
PC = 5
radius of circle = 5
\
NC = 5
Centre of circle is (5, 4 ).
5
4
X´
4
5
O P
Y´
3 M
3
Q
2
2
2
2
( x - 5) + (y + 4 ) = 25
Hence, there are 4 circles which satisfy the given
conditions. They are
( x ± 5)2 + (y ± 4 )2 = 25
2
x + y ± 10x ± 8y + 16 = 0
Aliter : Let the equation of the circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
Since, the circle touches the Y-axis
\
c =f2
f =± c
Also given the circle makes an intercept of 6 units along
X-axis. Therefore,
2 g2 - c = 6
or
or
Since, the circle touches the Y-axis at a distance of 4 units
from the origin, we have
y=m c =4
or
therefore,
c = 16
f =± c =±4
and
g = ± c + 9 = ± 16 + 9 = ± 5
g2 - c = 9
g = ± (c + 9 )
...(i)
then, 2 ( g 2 - c ) = a
( x + 5) + (y + 4 ) = 25
or
y=m c
Since, the circle passes through the origin, we get c = 0 and
given the intercepts on X and Y axes are a and b
If centre in II, III and IV quadrant, then equations are
( x + 5)2 + (y - 4 )2 = 25,
or
y ± c =0
Sol. Let the equation of the circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
X
( x - 5)2 + (y - 4 )2 = 25
2
\
y Example 18. Find the equation of the circle which
passes through the origin and makes intercepts of
length a and b on the X and Y axes, respectively.
C (5, 4)
\ Equation of circle, if centre in I quadrant
and
(y ± c )2 = 0
Hence, there are 4 circles which satisfy the given
conditions. They are
x 2 + y 2 ± 10x ± 8y + 16 = 0
Y
N
or
or
2 ( g 2 - 0) = a
\
g = ± a /2
and
2 (f
2
- c) = b
or
2 (f
2
- 0) = b
\
f = ± b /2
Hence, the equation of circle from Eq. (i) becomes
x 2 + y 2 ± ax ± by = 0
...(i)
y Example 19. Find the equation of the circle which
touches the axes and whose centre lies on the line
x - 2y = 3 .
Sol. Since, the circle touches both the axes, let the radius of
the circle by a, then
Case I If centre (a, a ) but given centre lies on
x - 2y = 3
\
a - 2a = 3
\
a= -3
Centre = ( -3, - 3)
\
Chap 04 Circle
and
radius = | - 3 | = 3
Y
(– a, a)
(a , a )
a
X´
a
a
a
a
O
a
(– a, –a)
X
Y´
\Equation of circle is
( x + 3) 2 + ( y + 3) 2 = 32
and
x 2 + y 2 + 6x + 6y + 9 = 0
Case II If centre ( -a, a ) but centre lies on x - 2y = 3
\
-a - 2a = 3
\
a= -1
then, centre = (1, - 1) and radius = | - 1 | = 1
\ Equation of circle is ( x - 1)2 + (y + 1)2 = 1
or
x 2 + y 2 - 2x + 2y + 1 = 0
Case III If the centre = ( -a , - a )
but centre lies on x - 2y = 3
\
- a + 2a = 3
\
a=3
then centre ( -3, - 3) and radius = | 3 | = 3
\ Equation of circle is
( x + 3) 2 + ( y + 3) 2 = 32
or
x 2 + y 2 + 6x + 6y + 9 = 0
Case IV If centre = (a , - a ) but centre lies on x - 2y = 3
or
a + 2a = 3
\
a=1
then centre = (1, - 1) and radius = 1
\ Equation of circle is
( x - 1) 2 + ( y + 1) 2 = 1
or
x 2 + y 2 - 2x + 2y + 1 = 0
Aliter I : Since, the circle touches both the axes, therefore
its centre will be (a , ± a ) and radius will be | a |, where a is
positive or negative number.
Case I If centre = (a , a )
Since, centre lies on x - 2y = 3
\
a - 2a = 3
a= -3
\ Centre of circle is ( -3, - 3) and radius = | - 3 | = 3.
Hence, equation of circle will be
( x + 3) 2 + ( y + 3) 2 = 32
or
Case II If centre = (a , - a )
Since, centre lies on x - 2y = 3
\
a + 2a = 3
\
a=1
\ Centre of circle is (1, - 1) and radius = | 1 | = 1
Hence, equation of circle will be
( x - 1) 2 + ( y + 1) 2 = 1
a
a
( a, –a)
x 2 + y 2 + 6x + 6y + 9 = 0
257
or
x 2 + y 2 - 2x + 2y + 1 = 0
Aliter II : Let the equation of circle is
x 2 + y 2 + 2gx + 2 fy + c = 0
centre = ( - g , - f )
Since, centre ( - g , - f ) lies on x - 2y = 3
or
-g + 2f = 3
Since, circle touches both axes
\
g2 = f
2
...(i)
...(ii)
= c or g = ± f
if g = f , then from Eq. (ii), - f + 2 f = 3
\
f = 3 and g = 3
but
c =f
2
= g2 = 9
\ Equation of circle from Eq. (i) is
x 2 + y 2 + 6x + 6y + 9 = 0
and if g = - f , then from Eq. (ii)
f + 2f = 3
\
f = 1 and g = - 1
but
c = g2 = f
2
=1
\ Equation of circle from Eq. (i) is
x 2 + y 2 - 2x + 2y + 1 = 0
Aliter III : Since, centre of circle lies on x - 2y = 3, also
since circle touches the axes, therefore, its centre will lie on
the line y = x or y = - x
Case I When the centre lies on the line y = x
but
x - 2y = 3
or
x - 2x = 3
\
x = -3=y
Hence, the centre = ( -3, - 3) and radius = | - 3 | = 3
Therefore, the equation of circle in this case will be
( x + 3) 2 + ( y + 3) 2 = 32
or
x 2 + y 2 + 6x + 6y + 9 = 0
Case II When the centre lies on the line y = - x
but
x - 2y = 3
or
x + 2x = 3
\
x = 1 then y = - 1
\Centre of circle (1, - 1) and radius is | 1 | = 1
Hence, equation of circle will be
( x - 1) 2 + ( y + 1) 2 = 1
or
x 2 + y 2 - 2x + 2y + 1 = 0
258
Textbook of Coordinate Geometry
y Example 20. A circle of radius 2 lies in the first
quadrant and touches both the axes of coordinates.
Find the equation of the circle with centre at (6, 5) and
touching the above circle externally.
Sol. Given, AC = 2 units
and
Position of a Point with
Respect to Circle
Theorem : A point ( x 1 , y 1 ) lies outside, on or inside a
circle
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
A º (2, 2), B º (6, 5)
AB = (2 - 6)2 + (2 - 5)2
then
where, S 1 =
Y
B (6, 5)
X´
x 12
CP = ( x 1 + g ) 2 + (y 1 + f ) 2
\
C
If r be the radius of the circle, then
A
(2, 2)
X
O
r = (g 2 + f 2 - c )
Y´
Since
AC + CB = AB
\
2 + CB = 5
\
CB = 3
Hence, equation of required circle with centre at (6, 5) and
radius 3 is
( x - 6) 2 + ( y - 5) 2 = 32
x 2 + y 2 - 12x - 10y + 52 = 0
or
y Example 21. A circle of radius 5 units touches the
coordinate axes in first quadrant. If the circle makes
one complete roll on X-axis along the positive direction
of X-axis, find its equation in the new position.
Sol. Let C be the centre of the circle in its initial position and
D be its centre in the new position.
5
5
L
M
O
5
X
10 π
Y´
Since, the circle touches the coordinates axes in first
quadrant and the radius of circle be 5 units.
\Centre of circle is (5, 5)
Moving length of circle = circumference of the circle
= 2pr = 2p (5) = 10p
Now, centre of circle in new position is (5 + 10p, 5) and
radius is 5 units, therefore, its equation will be
( x - 5 - 10p )2 + (y - 5)2 = 52
or
Þ
(CP ) 2 >, =, or
Þ
( x 1 + g ) 2 + (y 1 + f ) 2 >, =,
Þ
x 12 + y 12 + 2 gx 1 + 2 fy 1 + c >, =, or
Þ
S 1 >, =,
or
x 12
+ y 12
where, S 1 =
<r2
or
< g2 + f 2 - c
<0
<0
+ 2 gx 1 + 2 fy 1 + c .
y Example 22. Discuss the position of the points (1, 2)
and (6, 0) with respect to the circle
x 2 + y 2 - 4x + 2 y - 11 = 0.
S1 = 12 + 22 - 4 × 1 + 2 × 2 - 11 = - 6
D (5+10 π, 5)
C (5, 5)
5
X´
The point P lies outside, on or inside the circle according
as
CP >, = , or < r
Sol. Let S º x 2 + y 2 - 4 x + 2y - 11 = 0 for the point (1, 2)
Y
N
+ y 12 + 2 gx 1 + 2 fy 1 + c .
Proof : Let P ( x 1 , y 1 ) be the given point and let C be the
centre of the circle
Then,
C º ( -g, - f )
r
2
S 1 > , =, or < 0
according as
= 16 + 9 = 5
x 2 + y 2 - 10 (1 + 2p ) x - 10y + 100p 2 + 100p + 25 = 0
\
S1 < 0
and for the point (6, 0)
S 2 = 62 + 0 - 4 × 6 + 2 × 0 - 11
= 36 - 24 - 11
= 36 - 35 = 1
\
S2 > 0
Hence, the point (1, 2) lies inside the circle and the point
(6, 0) lies outside the circle.
y Example 23. The circle x 2 + y 2 - 6 x - 10y + l = 0
does not touch or intersect the coordinate axes and
the point (1, 4) is inside the circle. Find the range of
values of l.
Chap 04 Circle
Sol. Let S º x 2 + y 2 - 6x - 10y + l = 0
and the maximum distance of P from circle = PB
= CB + CP = r + CP
Case II If P outside the circle
In this case S 1 > 0 the minimum distance of P from
circle
= PA = CP - CA =CP - r
and the maximum distance of P from the circle
= PB = CP + CB = r + CP
Q Point (1, 4) is inside the circle, then S1 < 0
Y
C (3, 5)
r
r
L
X′
O
X
M
P
Y′
1 + 16 - 6 - 40 + l < 0
…(i)
Þ
l < 29
Centre and radius of the circle are (3, 5) and (34 - l ),
respectively.
Q Circle does not touch or intersect the coordinate axes.
\
5 > r and 3 > r
or
5 > (34 - l ) and 3 > (34 - l )
Þ
25 > 34 - l and 9 > 34 - l
Þ
l > 9 and l > 25
\
l > 25
Also,
34 - l > 0
\
l < 34
From Eqs. (i), (ii) and (iii), we get 25 < l < 29
…(ii)
…(iii)
A
C
B
Case III If P on the circle
In this case S 1 = 0
the minimum distance of P from the circle = 0
and the maximum distance of P from the circle
= PA = 2r
P
Maximum and Minimum
Distance of a Point from
the Circle
C
A
Remark
Let any point P ( x 1 , y 1 ) and circle
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
...(i)
The centre and radius of the circle are
C ( -g, - f ) and
( g 2 + f 2 - c ) respectively
Case I If P inside the circle
In this case S 1 < 0
Q
259
If point P inside or outside or on the circle and centre of circle at
C and radius r, then minimum distance of P from the circle
=|CP - r| and maximum distance of P from the circle = CP + r
y Example 24. Find the shortest and largest
distance from the point (2, - 7 ) to the circle
x 2 + y 2 - 14 x - 10y - 151 = 0
Sol. Let
r = ( g 2 + f 2 - c ) = CA = CB
\
S º x 2 + y 2 - 14 x - 10y - 151 = 0
S1 = (2)2 + ( -7 )2 - 14 (2) - 10 ( -7 ) - 151 = - 56 < 0
Y
B
A
P
C
C (7, 5)
B
X′
The minimum distance of P from circle = PA = CA - CP
= r - CP
X
O
P
A
Y′
260
Textbook of Coordinate Geometry
\ P (2, - 7 ) inside the circle
2
2
radius of the circle, r = ( -7 ) + ( -5) + 151 = 15
Q Centre of circle C º (7, 5)
CP = (7 - 2)2 + (5 + 7 )2 = 13
\
\ Shortest distance = PA = r - CP = 15 - 13 = 2
and Largest distance = PB = r + CP = 15 + 13 = 28
y Example 25. Find the points on the circle
x 2 + y 2 - 2x + 4 y - 20 = 0 which are farthest and
nearest to the point ( -5,6 ).
Sol. The given circle is S º x 2 + y 2 - 2x + 4y - 20 = 0
Let
P º ( -5,6)
P
A
r
C
r
B
For the point P
S1 = 25 + 36 + 10 + 24 - 20
= 75 > 0
lies
outside the circle.
\Point P( -5,6)
The centre and radius of the circle are (1, - 2) and 5,
respectively.
Q
CP = (1 + 5)2 + ( -2 - 6)2 = 10
Now, point A divides CP in the ratio
AP CP - r 10 - 5
=
=
=1
AC
r
5
\A is mid-point of CP.
æ 1 - 5 -2 + 6 ö
,
Aºç
\
÷
è 2
2 ø
or
A º ( -2,2)
and C is the mid-point of AB.
\
B º (2 ´ 1 - ( -2),2 ´ -2 - 2)
or
B º ( 4, - 6)
Hence, point A( -2,2) is nearest to P and B ( 4, - 6) is farthest
from P.
Exercise for Session 3
1.
The length of intercept, the circle x 2 + y 2 + 10x - 6y + 9 = 0 makes on the X-axis is
(a) 2
2.
(b) 4
2
(b) 3
(c) y 2 + 2cy = b 2 + c 2
(d) x 2 + cy = b 2 + c 2
(c) 8
(d) 10
If a circle of constant radius 3k passes through the origin and meets the axes at A and B, the locus of the
centroid of DOAB is
(b) x 2 + y 2 = 2k 2
(c) x 2 + y 2 = 3k 2
(d) x 2 + y 2 = 4k 2
The centre of the circle touching Y-axis at (0, 3) and making an intercept of 2 units on positive X-axis is
(a) (10, 3 )
7.
(b) x 2 + cx = b 2 + c 2
(b) 5
(a) x 2 + y 2 = k 2
6.
(d) 7
If a straight line through C( - 8, 8 ) making an angle of 135° with the X-axis cuts the circle x = 5cos q, y = 5sin q
at points A and B, then the length of AB is
(a) 3
5.
(c) 5
The locus of the centre of a circle which passes through the origin and cuts-off a length 2b from the line x = c is
(a) y 2 + 2cx = b 2 + c 2
4.
(d) 8
The circle x + y + 4x - 7y + 12 = 0 cuts an intercept on Y-axis is of length
(a) 1
3.
(c) 6
2
(b) ( 3, 10)
(c) ( 10, 3)
(d) (3, 10 )
A circle passes through the points A(1, 0) and B(5,0) and touches the Y-axis at C(0, l ). If ÐACB is maximum,
then
(a) |l |=
5
(b) |l |= 2 5
(c) |l |= 3 5
(d) |l |= 4 5
Chap 04 Circle
8.
9.
10.
11.
The equation of a circle whose centre is (3, - 1) and which intercept chord of 6 units length on straight line
2x - 5y + 18 = 0 is
(a) x 2 + y 2 - 6x + 2y - 28 = 0
(b) x 2 + y 2 + 6x - 2y - 28 = 0
(c) x 2 + y 2 + 4x - 2y + 24 = 0
(d) x 2 + y 2 + 2x - 2y - 12 = 0
The locus of the centre of a circle which touches externally the circle x 2 + y 2 - 6x - 6y + 14 = 0 and also
touches the Y-axis, is given by the equation
(a) x 2 - 6x - 10y + 14 = 0
(b) x 2 - 10x - 6y + 14 = 0
(c) y 2 - 6x - 10y + 14 = 0
(d) y 2 - 10x - 6y + 14 = 0
The locus of the centre of a circle of radius 2 which rolls on the outside of circle x 2 + y 2 + 3x - 6y - 9 = 0 is
(a) x 2 + y 2 + 3x - 6y + 5 = 0
(b) x 2 + y 2 + 3x - 6y - 31 = 0
(c) x 2 + y 2 + 3x - 6y + 11 = 0
(d) x 2 + y 2 + 3x - 6y - 36 = 0
The point ([ l + 1],[ l ]) is lying inside the circle x 2 + y 2 - 2x - 15 = 0. Then, the set of all values of l is (where [.]
represents the greatest integer function)
(a) [-2, 3]
12.
261
(b) (-2, 3)
(c) [-2, 0) È (0, 3)
(d) [0, 3)
The greatest distance of the point (10, 7) from the circle x 2 + y 2 - 4x - 2y - 20 = 0 is
(a) 5
(b) 10
(c) 15
(d) 20
13.
Find equations to the circles touching Y-axis at (0, 3) and making intercept of 8 units on the X-axis.
14.
Show that the circle x 2 + y 2 - 2ax - 2ay + a 2 = 0 touches both the coordinate axes.
15.
If the point ( l, - l) lies inside the circle x 2 + y 2 - 4x + 2y - 8 = 0, then find range of l.
16.
Find the equation of the circle which passes through the origin and cuts-off chords of lengths 4 and 6 on the
positive side of the X-axis and Y-axis, respectively.
Session 4
Intersection of a Line and a Circle, Product of the
Algebraical Distances PA and PB is Constant when
from P, A Secant be Drawn to Cut the Circle in the
Points A and B, The Length of Intercept Cut-off from
a Line by a Circle, Tangent to a Circle at a Given Point,
Normal to a Circle at a Given Point
X′
Intersection of a
Line and a Circle
Let the equation of the circle be
x 2 + y 2 = a2
...(i)
and the equation of the line be
y = mx + c
From Eq. (i) and Eq. (ii)
...(ii)
from (0, 0 ) to y = mx + c
Þ a > length of perpendicular from (0, 0 ) to y = mx + c
Thus, a line intersects a given circle at two distinct points
if radius of circle is greater than the length of
perpendicular from centre of the circle to the line.
Case II When the points of intersection are coincident,
then Eq. (iii) has two equal roots
x 2 + (mx + c ) 2 = a 2
or
(1 + m 2 ) x 2 + 2mcx + c 2 - a 2 = 0
M
...(iii)
y = mx + c
a
Case I When points of intersection are real and distinct,
then Eq. (iii) has two distinct roots.
O
(0, 0)
M
y=
O
(0, 0)
+c
\
Þ
B 2 - 4 AC = 0
4m 2 c 2 - 4 (1 + m 2 ) (c 2 - a 2 ) = 0
\
a2 =
or
a=
2
\
B - 4 AC > 0
2 2
4m c - 4 (1 + m ) (c 2 - a 2 ) > 0
or
2
a2 >
or
or
mx
a>
|c |
(1 + m 2 )
c
2
1 +m2
= length of perpendicular
c2
(1 + m 2 )
|c |
(1 + m 2 )
a = length of the perpendicular from the point (0, 0 ) to
y = mx + c
Thus, a line touches the circle if radius of circle is equal to
the length of perpendicular from centre of the circle to the
line.
Chap 04 Circle
Case III When the points of intersection are imaginary.
In this case (iii) has imaginary roots
M
y = mx + c
O
(0, 0)
Product of the Algebraical
Distances PA and PB is
Constant when from P, A
Secant be Drawn to Cut the
Circle in the Points A and B
If a straight line through P (a, b) makes an angle q with
the positive direction of X-axis, then its equation is
x -a y -b
=
=r
cos q sin q
B 2 - 4 AC < 0
\
B
4m 2 c 2 - 4 (1 + m 2 ) (c 2 - a 2 ) = 0
a2 <
\
c2
1 +m2
A
|c |
= length of perpendicular from (0, 0 ) to
1 +m2
y = mx + c
or a < length of perpendicular from (0, 0 ) to y = mx + c
Thus, a line does not intersect a circle if the radius of
circle is less than the length of perpendicular from centre
of the circle to the line.
or a <
y Example 26. Find the points of intersection of the
line 2x + 3y = 18 and the circle x 2 + y 2 = 25.
Sol. We have,
and
From Eq. (i),
2
P
2
...(ii)
x + y = 25
18 - 2x
y=
3
2
Þ
9 x 2 + 4 (9 - x )2 = 225
Þ
9 x 2 + 4 (81 - 18x + x 2 ) = 225
Þ
13x 2 - 72x + 324 - 225 = 0
Þ
13x 2 - 72x + 99 = 0
( x - 3) (13x - 33) = 0
33
Þ
x = 3 or x =
13
56
From Eq. (i),
y = 4 or y =
13
Hence, the points of intersection of the given line and the
æ 33 56 ö
given circle are (3, 4 ) and ç , ÷.
è 13 13 ø
θ
(α, β)
where, r is the algebraical distance of the point ( x , y ) from
the point P (a, b) .
\ ( x , y ) = (a + r cos q, b + r sinq )
If this point lies on the circle x 2 + y 2 + 2 gx + 2 fy + c = 0
or
(a + r cos q ) 2 + (b + r sinq ) 2 + 2g (a + r cos q )
Þ
+ 2 f (b + r sinq ) + c = 0
r + 2r (a cos q + b sinq + g cos q + f sinq )
2
+ (a 2 + b2 + 2 ga + 2 f b + c ) = 0
....(i)
2x + 3y = 18
æ 18 - 2x ö
Substituting in Eq. (ii), then x 2 + ç
÷ = 25
è 3 ø
Þ
263
This is quadratic equation in r, then PA and PB are the
roots of this equation.
\
PA × PB = a 2 + b2 + 2 ga + 2 f b + c = constant
Since, RHS is independent of q .
Remark
Secants are drawn from a given point Ato cut a given circle at the
pairs of points P1, Q1 ; P2, Q2;...; Pn, Qn , then
AP1 × AQ1 = AP2 × AQ2 = K = APn × AQn
The Length of Intercept
Cut-off from a Line by a Circle
Theorem : The length of the intercept cut-off from the
line y = mx + c by the circle x 2 + y 2 = a 2 is
ìa 2 (1 + m 2 ) - c 2 ü
2 í
ý
(1 + m 2 )
þ
î
264
Textbook of Coordinate Geometry
Proof : Draw OM perpendicular to PQ
Now, OM = length of perpendicular from O (0, 0 ) to
|c |
(y = mx + c ) =
(1 + m 2 )
and
then
Hence, line 4 x - 3y - 10 = 0 passes through the centre of the
circle.
Hence, intercepted length = diameter of the circle
= 2 ´ 5 = 10
OP = radius of the circle = a
P
4 x - 3y - 10 = 0
| 4 ´ 1 - 3 ´ ( -2) - 10 |
OM =
=0
4 2 + ( - 3) 2
y=
m
M x+c
y Example 28. Find the coordinates of the middle point
of the chord which the circle x 2 + y 2 + 4 x - 2y - 3 = 0
cuts-off the line x - y + 2 = 0.
Q
O
(0, 0)
Sol. Centre and radius of the circle x 2 + y 2 + 4 x - 2y - 3 = 0
are ( -2, 1) and 4 + 1 + 3 = 2 2 respectively.
x–y+2=0
In DOPM,
2
PM = (OP ) - (OM )
2
Q
2
2
2
O (–2, 1)
2ü
ìa (1 + m ) - c
= í
ý
(1 + m )
1 +m2
þ
î
c
= a2 -
2
M
ìa 2 (1 + m 2 ) - c 2 ü
PQ = 2 PM = 2 í
ý
1 +m2
þ
î
\
Remarks
1. If the line y = mx + c touches the circle x 2 + y 2 = a2, then
intercepted length is zero
2
2
2
ïì a ( 1 + m ) - c ïü
Þ2 í
ý=0
2
ïî
ïþ
1+ m
i.e.
PQ = 0
\
c 2 = a2 ( 1 + m2 )
which is the required condition for tangency.
2. If a line touches the circle, then length of perpendicular
from the centre upon the line is equal to the radius of the
circle.
y Example 27. Find the length of the intercept on
the straight line 4 x - 3y - 10 = 0 by the circle
2
2
x + y - 2x + 4 y - 20 = 0.
P
Draw perpendicular from O upon x - y + 2 = 0 is OM.
Equation of OM which is perpendicular to x - y + 2 = 0 is
x + y = l, it passes through ( -2, 1)
Then,
-2 + 1 = l
\
l = -1
then equation of OM is x + y + 1 = 0
Since, M is the mid-point of PQ which is point of
intersection of x - y + 2 = 0 and x + y + 1 = 0, coordinates of
æ 3 1ö
M is ç - , ÷.
è 2 2ø
Aliter : Let M º (a , b ), then
( - 2 - 1 + 2)
a +2 b -1
==
1
1+1
-1
Þ
2
2
Sol. Centre and radius of the circle x + y - 2x + 4y - 20 = 0
are (1, - 2) and 1 + 4 + 20 = 5 respectively.
or
\
4x
–3
y–
10
=0
O (1, – 2)
Let OM be the perpendicular from O on the line
(Here, M is foot of perpendicular)
a +2 b -1 1
=
=
1
2
-1
3
1
a = - and b =
2
2
æ 3 1ö
M º ç- , ÷
è 2 2ø
y Example 29. For what value of l will the line
y = 2x + l be a tangent to the circle x 2 + y 2 = 5 ?
Sol. Comparing the given line with y = mx + c , we get
m = 2 , c = l and given circle with x 2 + y 2 = a 2
then
a2 = 5
265
Chap 04 Circle
Q Condition for tangency is
c 2 = a 2 (1 + m 2 )
Different forms of the equations of tangents
1. Point form :
l 2 = 5 (1 + 4 )
Þ
l 2 = 25
\
l = ±5
Aliter : Since, line y = 2x + l
or
2x - y + l = 0
is the tangent to the circle x 2 + y 2 = 5 then length of
perpendicular from centre upon the line is equal to the
radius of the circle
Theorem : The equation of tangent at the point P ( x 1 , y 1 )
to a circle
x 2 + y 2 + 2 gx + 2 fy + c = 0 is
xx 1 + yy 1 + g ( x + x 1 ) + f (y + y 1 ) + c = 0
Proof : Since, P ( x 1 , y 1 ) be a point on the circle
x 2 + y 2 + 2 gx + 2 fy + c = 0
...(i)
Let Q ( x 2 , y 2 ) be any other point on the circle Eq. (i). Since,
points P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) lie on the circle, therefore
P (x1 , y )
1
5
C (0, 0)
T
M
C
(–g, –f )
y = 2x + λ
\
Q (x2 , y
2)
| CM | = 5
|0 - 0 + l |
or
4 +1
x 12
= 5
and
|l|
Þ
or
...(ii)
x 22 + y 22 + 2 gx 2 + 2 fy 2 + c = 0
...(iii)
( x 22 - x 12 ) + (y 22 - y 12 ) + 2 g
( x 2 - x 1 ) + 2 f (y 2 - y 1 ) = 0
Þ
Tangent to a Circle at a
Given Point
Let PQ be a chord and AB be a secant passing through P.
Let P be the fixed point and move along the circle towards
P, then the secant PQ turns about P. In the limit, when Q
coincides with P, then the secant AB becomes a tangent to
the circle at the point P.
A
P
A
+ 2 gx 1 + 2 fy 1 + c = 0
On subtracting Eq. (ii) from Eq. (iii), we have
= 5
5
|l|=5
l = ±5
Þ
+ y 12
B
Q3
Q2
Q1
O
Q
B
( x 2 - x 1 ) ( x 2 + x 1 + 2 g ) + (y 2 - y 1 )
(y 2 + y 1 + 2 f ) = 0
Þ
æ y2 - y1 ö
æ x 1 + x 2 + 2g ö
ç
÷ =-ç
÷
è x2 - x1 ø
è y1 + y2 + 2f ø
...(iv)
Now, the equation of the chord PQ is
æ y - y1 ö
y - y1 = ç 2
÷ (x - x 1 )
è x2 - x1 ø
...(v)
æ y - y1 ö
Putting the value of ç 2
÷ from Eq. (iv) in Eq. (v),
è x2 - x1 ø
then equation PQ becomes
æ x + x 2 + 2g ö
y - y1 = - ç 1
÷ (x - x 1 )
è y1 + y2 + 2f ø
...(vi)
Now, when Q ® P (along the circle), line PQ becomes
tangent at P, we have x 2 ® x 1 , y 2 ® y 1 . So, the equation
of tangent at P ( x 1 , y 1 ) is :
266
Textbook of Coordinate Geometry
Remarks
æ x + x 1 + 2g ö
y - y1 = - ç 1
÷ (x - x 1 )
è y1 + y1 + 2f ø
1. For equation of tangent of circle at ( x1, y1 ), substitute xx1 for
x + x1
y + y1
xy + x1 y
for y and 1
for x,
for xy and
x 2, yy1 for y 2,
2
2
2
keep the constant as such.
Þ
æx + gö
y - y1 = - ç 1
÷ (x - x 1 )
èy1 + f ø
Þ
(y - y 1 ) (y 1 + f ) + ( x - x 1 ) ( x 1 + g ) = 0
Þ
2. This method of tangent at ( x1, y1 ) is applied only for any
conics of second degree. i.e. equation of tangent of
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 at ( x1, y1 )
xx 1 + yy 1 + gx + fy = x 12 + y 12 + gx 1 + fy 1
is axx1 + h( xy1 + x1 y ) + byy1 + g ( x + x1 ) + f ( y + y1 ) + c = 0
On adding gx 1 + fy 1 + c to both sides, we get
xx 1 + yy 1 + g ( x + x 1 ) + f (y + y 1 ) + c
= x 12 + y 12 + 2 gx 1 + 2 fy 1 + c = 0
[from Eq. (ii)]
Þ
xx 1 + yy 1 + g ( x + x 1 ) + f (y + y 1 ) + c = 0
This is the required equation of the tangent PT to the
circle at the point ( x 1 , y 1 ).
Aliter : Since, circle is
Equation of tangent at ( x 1 , y 1 )
æ x + x1 ö
æy + y1 ö
2
Þ
aç
÷ xx 1 + b ç
÷ yy 1 = c
è 2 ø
è 2 ø
x 2 + y 2 + 2 gx + 2 fy + c = 0
x 12 +y 12 + 2 gx 1 + 2 fy 1 + c = 0
a ( x ) ( x 2 ) + b (y ) (y 2 ) = c
or
P ( x 1 , y 1 ) lie on the circle
\
Wrong process : Mostly students use wrong process
Suppose any curve
ax 3 + by 3 = c
...(i)
P (x1 , y )
1
T
which is a second degree conic not the equation of
tangent.
Reason : This method is applicable only for second degree
conic, its a third degree conic. (find its tangent only by
calculus)
y Example 30. Prove that the tangents to the circle
x 2 + y 2 = 25 at ( 3, 4 ) and (4, - 3) are perpendicular
C
(–g, –f )
to each other.
Its centre is C ( -g, - f )
y - (- f ) y 1 + f
Slope of CP = 1
\
=
x 1 - ( -g ) x 1 + g
Sol. The equations of tangents to x 2 + y 2 = 25 at (3, 4 ) and
( 4, - 3) are
...(i)
3x + 4y = 25
and
...(ii)
4 x - 3y = 25
respectively.
3
Now,
slope of Eq. (i) = - = m1
(say)
4
4
and
slope of Eq. (ii) = = m 2
(say)
3
Clearly,
m1m 2 = - 1
Hence, Eq. (i) and Eq. (ii) are perpendicular to each other.
Since, tangent PT is perpendicular to CP.
\
æx + gö
Slope of tangent = - ç 1
÷
èy1 + f ø
\ Equation of tangent at P ( x 1 , y 1 ) is
æx + gö
y - y1 = - ç 1
÷ (x - x 1 )
èy1 + f ø
Þ
Þ
(y - y 1 ) (y 1 + f ) + ( x 1 + g ) ( x - x 1 ) = 0
xx 1 + yy 1 + gx + fy =
x 12
+ y 12
+ gx 1 + fy 1
On adding gx 1 + fy 1 + c to both sides, we get
xx 1 + yy 1 + g( x + x 1 ) + f (y + y 1 ) + c
= x 12 + y 12 + 2 gx 1 + 2 fy 1 + c =0
[from Eq. (i)]
or
xx 1 + yy 1 + g( x + x 1 ) + f (y + y 1 ) + c = 0
This is the required equation of the tangent PT to the
circle at the point P ( x 1 , y 1 ).
y Example 31. Find the equation of tangent to the
circle x 2 + y 2 - 2ax = 0 at the point
[a (1 + cos a ), a sin a ].
Sol. The equation of tangent of x 2 + y 2 - 2ax = 0 at
[a (1 + cos a ), a sin a ] is
x × a (1 + cos a ) + y . a sin a - a [ x + a (1 + cos a )] = 0
Þ
ax cos a + ay sin a - a 2 (1 + cos a ) = 0
or
x cos a + y sin a = a (1 + cos a )
Chap 04 Circle
y Example 32. Show that the circles
x 2 + y 2 - 4 x + 6 y + 8 = 0 and x 2 + y 2 - 10x
- 6 y + 14 = 0 touch at ( 3, - 1) .
3x + ( - 1) y - 2 ( x + 3) + 3 ( y - 1) + 8 = 0
or
x + 2y - 1 = 0
and equation of tangent at (3, - 1) of the circle
x 2 + y 2 - 10x - 6y + 14 = 0 is
3 × x + ( -1) × y - 5 ( x + 3) - 3 (y - 1) + 14 = 0
or
- 2 x - 4y + 2 = 0
or
x + 2y - 1 = 0
which is the same as Eq (i).
Hence, the given circles touch at (3, - 1).
æ
æ q + fö ö
æ q + fö
ç a cos ç
÷÷
÷ a sin ç
è 2 ø÷
è 2 ø
ç
,
ç
æ q - fö ÷
æ q - fö
cos
÷ cos ç
ç
÷
ç
è 2 ø
è 2 ø ÷ø
è
We get
Sol. Equation of tangent at (3, - 1) of the circle
x 2 + y 2 - 4 x + 6y + 8 = 0 is
...(i)
Corollary 3 : The angle between a pair of tangents from a
point P to the circle x 2 + y 2 = a 2 is a. Then, the locus of
the point P is
x 2 +y 2 =
...(ii)
a2
æa ö
sin2 ç ÷
è2ø
Proof
Y
R
a
2. Parametric form :
Theorem : The equation of tangent to the circle
x 2 + y 2 = a 2 at the point (a cos q, a sinq ) is
x cos q + y sinq = a
(f)Q
P(q)
f
X¢
Proof : The equation of tangent of x 2 + y 2 = a 2 at ( x 1 , y 1 )
q
O
X
is xx 1 + yy 1 = a 2 (using point form of the tangent)
Putting
x 1 = a cosq, y 1 = a sinq
then, we get
x cos q + y sinq = a
Corollary 1 : Equation of chord joining (a cos q, a sinq )
and (a cos f, a sin f) is
æ q + fö
æ q + fö
æ q - fö
x cos ç
÷ + y sin ç
÷ = a cos ç
÷
è 2 ø
è 2 ø
è 2 ø
Corollary 2 : Point of intersection of tangents at
(a cos q, a sinq ) and (a cos f, a sin f) is
æ
æ q + fö
æ q + fö ö
ç a cos ç
÷ a sin ç
÷÷
è
ø
è 2 ø÷
2
ç
,
ç
æ q - fö
æ q - fö ÷
÷ cos ç
÷
ç cos çè
è 2 ø ÷ø
2 ø
è
Remembering method :
æ q + fö
æ q + fö
æ q - fö
x cos ç
Q
÷ +y sin ç
÷ =a cos ç
÷
è 2 ø
è 2 ø
è 2 ø
or
ì
æ q + fö ü
æ q + fö ü ì
ïa cos çè 2 ÷ø ï ïa sin çè 2 ÷ø ï
ï 2
ï
ï ï
xí
ý =a
ý +y í
q
f
q
f
æ
öï
ö ï ï
ï cos æç
cos ç
÷
÷
è 2 ø ïþ
è 2 ø ïþ ïî
ïî
267
Y¢
Q
\
or
f - q + a = 180°
q-f
aö
æ
= -ç 90° - ÷
è
2
2ø
æ q - fö
æa ö
cos ç
÷ = sin ç ÷
è2ø
è 2 ø
Now, point of intersection is
æ
æ q + fö
æ q + fö ö
ç a cos ç
÷ a sin ç
÷÷
è
ø
è 2 ø÷
2
ç
,
ç
æa ö
æa ö ÷
sin ç ÷ ÷
ç sin çè ÷ø
è2ø ø
2
è
Let
\
x=
x 2 +y 2 =
æ q + fö
a cos ç
÷
è 2 ø
æa ö
sin ç ÷
è2ø
a2
æa ö
sin2 ç ÷
è2ø
and y =
æ q + fö
a sin ç
÷
è 2 ø
æa ö
sin ç ÷
è2ø
268
Textbook of Coordinate Geometry
Remarks
1. The angle between a pair of tangents from a point P to the
circle x 2 + y 2 + 2gx + 2fy + c = 0 is 2q, then the locus of P is
x 2 + y 2 + 2gx + 2fy + c = ( g 2 + f 2 - c )cot 2 q
2. If angle between a pair of tangents from a point P to the
p
circle x 2 + y 2 = a2 is , then the locus of P is
2
æHere, a = p ö
x 2 + y 2 = 2a2
ç
÷
è
2ø
2
2
2
which is director circle of x + y = a .
(Qlocus of point of intersection of perpendicular tangents is
director circle)
3. The equation of the tangent to the circle
( x - a) 2 + ( y - b) 2 = r 2 at the point( a + r cos q, b + r sin q) is
( x - a) cos q + ( y - b) sin q = r .
y Example 33. The angle between a pair of tangents
p
from a point P to the circle x 2 + y 2 = 25 is . Find
3
the equation of the locus of the point P.
p
3
\ Required locus is x 2 + y 2 =
Sol. Here, a =
or q =
2q =
3
\Required locus is
6
x 2 + y 2 - 6x - 8y + 9 = (9 + 16 - 9 )cot 2
or
or
p
6
x 2 + y 2 - 6x - 8y + 9 = 16 ´ 3
x 2 + y 2 - 6x - 8y - 39 = 0
3. Slope form :
Theorem : The equation of a tangent of slope m to the
circle x 2 + y 2 = a 2 is y = mx ± a (1 + m 2 ) and the
coordinates of the point of contact are
æ
ö
am
a
ç±
÷
,m
ç
2
2 ÷
(1 + m )
(1 + m ) ø
è
\ Length of perpendicular from centre of circle (0, 0 ) on
(y = mx + c ) = radius of circle
\
(1 + m 2 )
=a
...(i)
which are the required equations of tangents.
Also, let ( x 1 , y 1 ) be the point of contact, then equation of
tangent at ( x 1 , y 1 ) to the circle x 2 + y 2 = a 2 is
xx 1 + yy 1 = a 2
...(ii)
On comparing Eq. (i) and Eq. (ii), we get
x1 y1
a2
=
=
m -1 ± a (1 + m 2 )
Þ
x1
y
a
=- 1 =±
1
m
(1 + m 2 )
Þ
x1 = ±
am
1 +m2
and y 1 = m
a
(1 + m 2 )
Corollary : It also follows that y = mx + c is a tangent to
x 2 + y 2 = a 2 , if c 2 = a 2 (1 + m 2 ) which is condition of
tangency.
Remarks
1. The reason why there are two equations y = mx ± a 1 + m2 ,
there are two tangents, both are parallel and at the ends of
diameter.
2. The line ax + by + c = 0 is tangent to the circle x 2 + y 2 = r 2 if
and only if c 2 = r 2 ( a2 + b2 ).
3. If the line y = mx + c is the tangent to the circle x 2 + y 2 = r 2,
æ mr 2 r 2 ö
then point of contact is given by çç , ÷
c c ÷ø
è
4. If the line ax + by + c = 0 is the tangent to the circle
æ ar 2 br 2 ö
÷.
x 2 + y 2 = r 2, then point of contact is given by çç ,c ÷ø
è c
5. The condition that the line lx + my + n = 0 touches the circle
x 2 + y 2 + 2gx + 2fy + c = 0 is
( lg + mf - n) 2 = ( l 2 + m2 ) ( g 2 + f 2 - c ).
6. Equation of tangent of the circle x 2 + y 2 + 2gx + 2fy + c = 0 in
terms of slope is
Proof : Let y = mx + c is the tangent of the circle
x 2 + y 2 = a2.
|c |
y = mx ± a (1 + m 2 )
æ
ö
am
a
÷
Hence, ( x 1 , y 1 ) = ç ±
,m
ç
2
2 ÷
(1 + m )
(1 + m ) ø
è
25
= 100
æpö
sin 2 ç ÷
è6ø
y Example 34. The angle between a pair of tangents
from a point P to the circle x 2 + y 2 - 6 x - 8 y + 9 = 0
p
is . Find the equation of the locus of the point P.
3
p
p
Sol. Here,
On substituting this value of c in y = mx + c , we get
2
Þ c = ± a (1 + m )
y + f = m ( x + g ) ± ( g 2 + f 2 - c ) ( 1 + m2 )
7. The equation of tangents of slope m to the circle
( x - a) 2 + ( y - b) 2 = r 2 are given by
( y - b) = m ( x - a) ± r ( 1 + m2 )
and the coordinates of the points of contact are
ö
æ
mr
r
÷
ça ±
,bm
ç
( 1 + m2 )
( 1 + m2 ) ÷ø
è
Chap 04 Circle
y Example 35. Find the equations of the tangents to
the circle x 2 + y 2 = 9, which
(i) are parallel to the line 3 x + 4y - 5 = 0
(ii) are perpendicular to the line 2 x + 3 y + 7 = 0
(iii) make an angle of 60° with the X-axis
Sol. (i) Slope of 3x + 4y - 5 = 0 is -
then, perpendicular distance from (0, 0) on Eq. (ii) = radius
|l|
=3
2
3 + ( - 2) 2
or
| l | = 3 13
or
l = ± 3 13
From Eq. (ii), equations of tangents are
3x - 2y ± 3 13 = 0
3
4
3
4
and equation of circle is x 2 + y 2 = 9
(iii) Let equation of tangent which makes an angle of 60°
with the X-axis is
…(iii)
y = 3x + c
Let m = -
\ Equations of tangents
or
æ
3
y = - x ± 3 çç1 +
4
è
269
2ö
æ 3ö ÷
ç- ÷ ÷
è 4ø ø
3x - y + c = 0
and circle
x2 + y2 = 9
Þ
then, perpendicular distance from (0, 0) to Eq. (iii) = radius
|c |
=3
2
( 3 ) + ( - 1) 2
3
\ Slope of perpendicular to 2x + 3y + 7 = 0 is = m (say)
2
and given circle is x 2 + y 2 = 9
or
|c | =6
or
c =± 6
From Eq. (iii), equations of tangents are
3x - y ± 6 = 0
4y = - 3x ± 15 or 3x + 4y ± 15 = 0
2
(ii) Slope of 2x + 3y + 7 = 0 is 3
\ Equations of tangents perpendicular to 2x + 3y + 7 = 0 is
3
y = x ±3 1+
2
æ3ö
ç ÷
è2ø
Þ
2y = 3x ± 3 13
or
3x - 2y ± 3 13 = 0
2
Sol. If the line lx + my + n = 0 touches the circle
( x - a )2 + (y - b )2 = r 2 , then length of the perpendicular
from the centre = radius
| la + mb + n |
=r
(l 2 + m 2 )
(iii) Since, tangent make an angle 60° with the X-axis
\
m = tan60° = 3
and given circle
x2 + y2 = 9
\ Equation of tangents y = 3x ± 3 1 + ( 3 )2
or
Þ
3 x -y ±6=0
Aliter :
(i) Let tangent parallel to 3x + 4y - 5 = 0 is
3x + 4y + l = 0
and circle
x2 + y2 = 9
…(i)
then perpendicular distance from (0, 0) on Eq. (i) = radius
|l|
=3
2
(3 + 4 2 )
or
| l | =15
\
l = ± 15
From Eq. (i), equations of tangents are
3x + 4y ± 15 = 0
(ii) Let tangent perpendicular to 2x + 3y + 7 = 0 is
3x - 2y + l = 0
and circle
x2 + y2 = 9
y Example 36. Prove that the line lx + my + n = 0
touches the circle ( x - a ) 2 + ( y - b ) 2 = r 2 if
(al + bm + n ) 2 = r 2 (l 2 + m 2 ) .
…(ii)
(la + mb + n )2 = r 2 (l 2 + m 2 )
Aliter :
Here, line is lx + my + n = 0 and circle is
( x - a )2 + (y - b )2 = r 2 . Here, centre of circle (a, b ) shift at
(0, 0), then replacing x by x + a and y by y + b in the
equation of straight line lx + my + n = 0 and circle
( x - a )2 + (y - b )2 = r 2 , the new form of straight line and
circle are
l ( x + a ) + m (y + b ) + n = 0
or
…(i)
lx + my + (al + mb + n ) = 0
…(ii)
and
x2 + y2 = r 2
respectively.
On comparing Eq. (i) with y = Mx + C
l
then
M=m
(al + bm + n )
and
C =m
270
Textbook of Coordinate Geometry
Since, Eq. (i) is the tangent of Eq. (ii), then
C 2 = r 2 (1 + M 2 )
or
or
(al + bm + n )2
m2
æ
l2 ö
= r 2 ç1 + 2 ÷
m ø
è
(al + bm + n )2 = r 2 (l 2 + m 2 )
y Example 37. Show that the line 3x - 4 y = 1 touches
the circle x 2 + y 2 - 2x + 4 y + 1 = 0. Find the
coordinates of the point of contact.
Sol. The centre and radius of the circle
x 2 + y 2 - 2x + 4y + 1 = 0 are (1, - 2)
y Example 38. If lx + my = 1 touches the circle
x 2 + y 2 = a 2 , prove that the point (l , m ) lies on the
circle x 2 + y 2 = a -2 .
Sol. Since, lx + my = 1 touches the circle x 2 + y 2 = a 2 .
Then, length of perpendicular from (0, 0) on lx + my = 1 is
equal to radius
| -1|
then,
= a or l 2 + m 2 = a -2
l 2 + m2
Hence, locus of (l , m ) is x 2 + y 2 = a -2
Aliter : Let the point of contact of line lx + my = 1 and
circle x 2 + y 2 = a 2 is ( x 1, y1 ), then tangent of circle at
and ( -1) + (2)2 - 1 = 2 respectively.
( x 1, y1 ) is xx 1 + yy1 = a 2
Since, length of perpendicular from centre (1, - 2) on
3x - 4y = 1 is
| 3 ´ 1 - 4 ´ ( -2) - 1 | 10
=
5
( 3) 2 + ( - 4 ) 2
Since,
= 2 = radius of the circle
Hence, 3x - 4y = 1 touches the circle
x 2 + y 2 - 2x + 4y + 1 = 0
Second part : Let point of contact is ( x 1, y1 ), then tangent
at ( x 1, y1 ) on x 2 + y 2 - 2x + 4y + 1 = 0 is
xx 1 + yy1 - ( x + x 1 ) + 2 (y + y1 ) + 1 = 0
…(i)
Þ
x ( x 1 - 1) + y (y1 + 2) - x 1 + 2y1 + 1 = 0
and given line
…(ii)
3x - 4y - 1 = 0
Since, Eq. (i) and Eq. (ii) are identical, then comparing Eq. (i)
and Eq. (ii), we get
x 1 - 1 y1 + 2 - x 1 + 2y1 + 1
=
=
-1
3
-4
1
2
or
and y1 = x1 = 5
5
æ 1 2ö
\ Point of contact is ç - , - ÷.
è 5 5ø
Aliter for second part : Since, perpendicular line to
tangent always passes through the centre of the circle,
perpendicular line to
...(i)
3x - 4y = 1
is
...(ii)
4 x + 3y = l
which passes through (1, - 2), then
4 -6= l
\
l = -2
From Eq. (ii),
...(iii)
4 x + 3y = - 2
Solving Eq. (i) and Eq. (iii), we get the point of contact i.e.
1
2
x = - and y = 5
5
æ 1 2ö
Hence, point of contact is ç - , - ÷.
è 5 5ø
xx 1 + yy1 = a 2
are identical, then
\
and lx + my = 1
x 1 y1 a 2
=
=
1
l
m
x 1 = la 2 , y1 = ma 2
but ( x 1, y1 ) lie on x 2 + y 2 = a 2
then,
\
l 2a 4 + m 2a 4 = a 2
l 2 + m 2 = a -2
\Locus of (l , m ) is x 2 + y 2 = a -2
y Example 39. Show that the line
( x - 2) cos q + ( y - 2) sin q = 1 touches a circle for all
values of q. Find the circle.
Sol. Given line is ( x - 2) cos q + (y - 2) sin q
1 = cos 2 q + sin 2 q
On comparing
x - 2 = cosq
and
y - 2 = sinq
Squaring and adding Eq. (i) and Eq. (ii), then
( x - 2)2 + (y - 2)2 = cos 2 q + sin 2 q
Þ
or
...(i)
...(ii)
( x - 2) 2 + ( y - 2) 2 = 1
x 2 + y 2 - 4 x - 4y + 7 = 0
Aliter : Since, tangent at (cos q , sin q ) of
x2 + y2 = 1
...(i)
is
...(ii)
x cos q + y sin q = 1
replacing x by x - 2 and y by y - 2 in Eqs. (i) and (ii), then
...(iii)
( x - 2) 2 + ( y - 2) 2 = 1
( x - 2) cos q + (y - 2) sin q = 1
Hence, Eq. (iv) touches the circle Eq. (iii).
\ Equation of circle is
( x - 2) 2 + ( y - 2) 2 = 1
or
x 2 + y 2 - 4 x - 4y + 7 = 0
…(iv)
Chap 04 Circle
write first two rows as ax1 + hy1 + g and hx1 + by1 + f
Normal to a Circle at a
Given Point
Then, normal at ( x1, y1 ) of conic (i)
x - x1
y - y1
=
ax1 + hy1 + g hx1 + by1 + f
The normal of a circle at any point is a straight line which
is perpendicular to the tangent at the point and always
passes through the centre of the circle.
Different form of the Equation of Normals
Corollary 1 : Equation of normal of x 2 + y 2 = a 2 at
( x 1 , y 1 ) is
x - x1
y - y1
=
1× x 1 + 0 + 0 0 + 1× y 1 + 0
(Here, g, f = 0 and a = b = 1)
x - x1 y - y1
=
x1
y1
1. Point form :
Theorem : The equation of normal at the point P ( x 1 , y 1 )
to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 is
x - x1 y - y1
=
x1 + g y1 + f
Þ
y
x
=
x1 y1
or
Corollary 2 : Equation of normal of
x 2 + y 2 + 2 gx + 2 fy + c = 0 at ( x 1 , y 1 ) is
Proof :
Equation of the given circle is
x 2 + y 2 + 2gx + 2 fy + c = 0
x - x1 y - y1
=
x1 + g y1 + f
...(i)
Its centre C is ( -g, - f )
1. Normal always passes through the centre of the circle.
Just write the equation of the line joining ( x1, y1 ) and the
centre of the circle.
2. The equations of the normals show that they pass through
the centre i.e. the normals are the radii which we know from
Euclidean geometry.
T
C
(–g, –f )
Let P ( x 1 , y 1 ) be the given point.
Q Normal of the circle at P ( x 1 , y 1 ) passes through centre
C ( -g, - f ) of the circle.
Then, equation of normal CP passes through the points
C ( -g, - f ) and P ( x 1 , y 1 ) is
(y + f )
y - y1 = 1
(x - x 1 )
(x 1 + g )
x - x1 y - y1
=
x1 + g y1 + f
Remark
Easy method to find normal at ( x1, y1 ) of second degree conics
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0
y Example 40. Find the equation of the normal to
the circle x 2 + y 2 = 2x , which is parallel to the line
x + 2y = 3.
Sol. Given circle is
x 2 + y 2 - 2x = 0
Centre of given circle is (1, 0)
Since, normal is parallel to x + 2y = 3
let the equation of normal is x + 2y = l
Since, normal passes through the centre of the circle i.e.
(1, 0)
then
This is the required equation of normal at P ( x 1 , y 1 ) of the
given circle.
a h g
then, according to determinant h b f
g f c
(Here, a = b = 1 and h = 0)
Remarks
P (x1, y1)
or
271
\
1+0= l
l =1
then, equation of normal is x + 2y = 1
or
x + 2y - 1 = 0
Aliter Equation of normal at ( x 1, y1 ) of x 2 + y 2 - 2x = 0 is
...(i)
x - x 1 y - y1
=
x 1 - 1 y1 - 0
or
Slope =
y1
= m1
x1 - 1
(say)
272
Textbook of Coordinate Geometry
Since normal is parallel to x + 2y = 3
1
Slope = - = m 2
\
2
but given m1 = m 2
y1
1
= - or x 1 + 2y1 - 1 = 0
x1 - 1
2
Aliter II : Equation of tangent at (5, 6) is
5
5 × x + 6 × y - ( x + 5) + (y + 6) - 48 = 0
2
Þ
10x + 12y - 5x - 25 + 2y + 12 - 96 = 0
Þ
5x + 14y - 109 = 0
5
Slope of tangent = 14
14
Slope of normal =
\
5
14
is
\Equation of normal at (5, 6) with slope
5
14
( x - 5)
y - 6=
5
Þ
5y - 30 = 14 x - 70
or
14 x - 5y - 40 = 0
(say)
\ Locus of ( x 1, y1 ) is x + 2y - 1 = 0
y Example 41. Find the equation of the normal to the
circle x 2 + y 2 - 5x + 2y - 48 = 0 at the point ( 5, 6 ) .
Sol. Equation of the normal at (5, 6) is
x -5 y -6
x -5 y -6
2x - 10 y - 6
Þ
=
Þ
=
=
5 6+1
5
7
5
7
52
2
Þ
14 x - 70 = 5y - 30
\
14 x - 5y - 40 = 0
Aliter I : Since, centre of the circle
æ5
ö
x 2 + y 2 - 5x + 2y - 48 = 0 is ç , - 1÷, normal at (5,6) is the
è2
ø
ö
æ5
equation of a line, which passes through ç , - 1÷ and (5, 6) is
ø
è2
6+1 æ
çx 5 è
52
7
y + 1 = ( 2x - 5)
5
5y + 5 = 14 x - 35
y +1=
Þ
Þ
5ö
14 æ
5ö
÷ Þ y +1=
çx - ÷
2ø
5 è
2ø
2. Parametric form
Since, parametric coordinates of circle x 2 + y 2 = a 2 is
(a cos q, a sinq ).
\ Equation of normal at (a cos q, a sinq ) is
y
x
=
a cos q a sinq
y
x
or y = x tanq
=
cos q sinq
or
or
or 14 x - 5y - 40 = 0
y = mx , where m = tanq
which is slope form of normal.
Exercise for Session 4
1.
The length of the chord cut-off by y = 2x + 1from the circle x 2 + y 2 = 2 is
(a)
2.
5
6
6
5
(d)
5
6
(b) 9 < l < 20
(c) -35 < l < 15
(d) -16 < l < 30
(b) -20
(c) 20
(d) 22
Tangent which is parallel to the line x - 3y - 2 = 0 of the circle x 2 + y 2 - 4x + 2y - 5 = 0, has point/points of
contact
(a) (1, - 2)
5.
(c)
If the line 3x - 4y + l = 0, ( l > 0) touches the circle x 2 + y 2 - 4x - 8y - 5 = 0 at (a,b ), then l + a + b is equal to
(a) -22
4.
6
5
Circle x 2 + y 2 - 4x - 8y - 5 = 0 will intersect the line 3x - 4y = l in two distinct points, if
(a) -10 < l < 5
3.
(b)
(b) (-1, 2)
(c) (3, 4)
(d) (3, - 4)
If a circle, whose centre is ( -11
, ) touches the straight line x + 2y = 12, then the co-ordinates of the point of
contact are
7
(a) æç - , - 4ö÷
è 2
ø
18 21
(b) æç - , - ö÷
è 5
5ø
(c) (2, - 7)
(d) (-2, - 5)
Chap 04 Circle
6.
The area of the triangle formed by the tangent at the point (a, b ) to the circle x 2 + y 2 = r 2 and the coordinate
axes is
(a)
7.
r4
2ab
(b)
r4
2 | ab |
(c)
r4
ab
(d)
r4
| ab |
The equation of the tangent to the circle x 2 + y 2 + 4x - 4y + 4 = 0 which make equal intercepts on the positive
coordinate axes is
(b) x + y = 2 2
(a) x + y = 2
8.
273
(c) x + y = 4
(d) x + y = 8
If a > 2b > 0, then the positive value of m for which y = mx - b (1+ m 2 ) is a common tangent to x 2 + y 2 = b 2
and ( x - a )2 + y 2 = b 2 is
2b
(a)
2
2
(b)
(a - 4b )
9.
(a 2 - 4b 2 )
2b
(c)
2b
a - 2b
(d)
The angle between a pair of tangents from a point P to the circle x 2 + y 2 = 16 is
b
a - 2b
p
and locus of P is
3
x 2 + y 2 = r 2, then value of r is
(a) 5
10.
11.
(c) 7
(d) 8
The normal at the point (3, 4) on a circle cuts the circle at the point ( -1, - 2). Then, the equation of the circle is
(a) x 2 + y 2 + 2x - 2y - 13 = 0
(b) x 2 + y 2 - 2x - 2y - 11 = 0
(c) x 2 + y 2 - 2x + 2y + 12 = 0
(d) x 2 + y 2 - 2x - 2y + 14 = 0
The line ax + by + c = 0 is a normal to the circle x 2 + y 2 = r 2. The portion of the line ax + by + c = 0 intercepted
by this circle is of length
(a) r
12.
(b) 6
(b) r
(c) r 2
If the line ax + by + c = 0 touches the circle x 2 + y 2 - 2x =
(d) 2r
3
and is normal to the circle x 2 + y 2 + 2x - 4y + 1 = 0,
5
then (a,b ) are
(a) (1, 3)
(b) (3, 1)
(c) (1, 2)
(d) (2, 1)
13.
Show that for all values of q, x sin q - y cos q = a touches the circle x + y = a .
14.
Find the equation of the tangents to the circle x 2 + y 2 - 2x - 4y - 4 = 0
2
2
2
which are (i) parallel (ii) perpendicular to the line 3x - 4y - 1 = 0.
15.
Find the equation of the family of circle which touch the pair of straight lines x 2 - y 2 + 2y - 1 = 0.
16.
Find the value of l so that the line 3x - 4y = l may touch the circle x 2 + y 2 - 4x - 8y - 5 = 0.
17.
Show that the area of the triangle formed by the positive X-axis, the normal and tangent to the circle
x 2 + y 2 = 4 at (1, 3 ) is 2 3.
Session 5
Tangents from a Point to the Circle, Length of the
Tangent from a Point to a Circle, Power of a Point
with Respect to a Circle, Chord of Contact, Chord
Bisected at a Given Point, Pair of Tangents,
Director Circle
Tangent from a Point
to the Circle
If P outside the circle, then substituting these values of m
in Eq. (ii), we get the equation of tangents.
Theorem : From a given point two tangents can be drawn
to a circle which are real, coincident or imaginary
according as the given point lies outside,on or inside the
circle .
2
2
...(i)
y = mx + a (1 + m 2 )
…(ii)
Proof : If circle is
2
x +y =a
any tangent to the circle Eq. (i) is
If outside point is ( x 1 , y 1 )
2
2
or
Þ
2
m
+m
( x 12
2
x 12
2
2
- 2mx 1 y 1 = a + a m
2
- a ) - 2mx 1 y 1 + y 12 - a 2 = 0
(mx 1 - y 1 ) 2 = a 2 (1 + m 2 )
m 2 x 12 - 2m x 1 y 1 + y 12 = a 2 + a 2m 2
m 2 ( x 12 - a 2 ) - 2mx 1 y 1 + y 12 - a 2 = 0
which is quadratic in m which gives two values of m.
...(iii)
which is quadratic in m which gives two values of m.
(real coincident or imaginary) corresponding to any value
of x 1 and y 1 .
The tangents are real, coincident or imaginary according
as the values of m obtained from Eq. (iii) are real,
coincident or imaginary.
or Discriminant >, =, or <0
y Example 42. Find the equations of the tangents to
the circle x 2 + y 2 = 16 drawn from the point (1, 4 ) .
Sol. Given circle is
x 2 + y 2 = 16
...(i)
Any tangent of Eq. (i) in terms of slope is
y = mx + 4 (1 + m 2 )
which passes through (1, 4 )
Þ
4 x 12 y 12 - 4 ( x 12 - a 2 ) (y 12 - a 2 ) >, =, or < 0
then,
Þ
( x 12 + y 12 - a 2 ) >, = or < 0
Þ
( 4 - m )2 = 16 (1 + m 2 )
Þ
15m 2 + 8m = 0
i.e. P ( x 1 , y 1 ) lies outside, on or inside the circle
x 2 + y 2 = a2.
...(i)
2
length of perpendicular from centre (0, 0 ) to Eq. (i) = radius
of the circle
| mx 1 - y 1 |
then,
=a
(1 + m 2 )
Þ
2
2
which is tangent of the circle x + y = a , then
or
2
(y 1 - mx 1 ) = a (1 + m )
y 12
2
or
then, y 1 = mx 1 + a 1 + m 2
or
Aliter :
First write equation of line through ( x 1 , y 1 ) say
y - y 1 = m (x - x 1 )
\
4 = m + 4 (1 + m 2 )
m = 0, -
8
15
...(ii)
Chap 04 Circle
From Eq. (ii), equations of tangents drawn from (1, 4) are
y=4
and
y=-
64 ö
æ
ç1 +
÷
è
225 ø
8
x+4
15
or
8x + 15y = 68 respectively.
Aliter : Equation of line through (1, 4 ) is y - 4 = m ( x - 1)
...(i)
Þ
mx - y + 4 - m = 0
Then, perpendicular length from centre (0, 0) to
mx - y + 4 - m = 0 is equal to radius
|4 -m|
then,
=4
m 2 +`1
or
Þ
2
2
Length of the Tangent from a
Point to a Circle
Theorem : The length of tangent from the point
P ( x 1 , y 1 ) to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0 is
( x 12 + y 12 + 2 gx 1 + 2 fy 1 + c ) = S 1
Proof : Let PT and PT ¢ be two tangents from the given
point P ( x 1 , y 1 ) to the circle x 2 + y 2 + 2 gx + 2 fy + c = 0.
Then, the centre and radius are C ( -g, - f ) and
( g 2 + f 2 - c ) ( = CT = CT ¢ ) respectively.
( 4 - m ) = 16 (m + 1)
2
15m + 8m = 0
8
15
From Eq. (i), equation of tangents from (1, 4 ) are y = 4 and
8x + 15y = 68, respectively.
\
m = 0, -
y Example 43. The angle between a pair of tangents
from a point P to the circle x 2 + y 2 + 4 x - 6 y + 9 sin 2 a
+ 13 cos 2 a = 0 is 2a. Find the equation of the locus of
the point P.
Sol. Let coordinates of P be ( x 1, y1 ) and given circle is
x 2 + y 2 + 4 x - 6y + 9 sin 2 a + 13 cos 2 a = 0
T
C (– g, –
f)
T´
)
P (x 1, y 1
In D PCT,
PT = ( PC ) 2 - (CT ) 2
or
( x + 2)2 + (y - 3)2 - 4 - 9 + 9 sin 2 a + 13 cos 2 a = 0
= ( x 1 + g ) 2 + (y 1 + f ) 2 - g 2 - f 2 + c
Þ
( x + 2)2 + (y - 3)2 + 9 sin 2 a - 13 (1 - cos 2 a ) = 0
= ( x 12 + y 12 + 2 gx 1 + 2 fy 1 + c ) = S 1 = PT ¢
Þ
275
2
2
2
( x + 2) + (y - 3) = 4 sin a
\ Centre and radius are ( -2, 3) and 2 sina, respectively.
T
2 sin α
C (– 2, 3)
2 sin α
α
α
R
P (x1, y1)
where, S 1 = x 12 + y 12 + 2 gx 1 + 2 fy 1 + c
Remarks
1. To find length of tangent
let
S = x 2 + y 2 + 2gx + 2fy + c
then,
S1 = x12 + y12 + 2gx1 + 2fy1 + c
where, P ( x1, y1 )
\ length of tangent = S1
2. For S1 first write the equation of circle in general form i.e.
coefficient of x 2 = coefficient of y 2 = 1 and making RHS of
circle is zero, then let LHS by S.
Distance between P ( x 1, y1 ) and centre of circle C ( -2, 3) is
CP = ( x 1 + 2)2 + (y1 - 3)2
In D PCT , sin a =
2 sin a
CT
=
CP
( x 1 + 2) 2 + ( y 1 - 3) 2
or
( x 1 + 2) 2 + ( y 1 - 3) 2 = 2
or
( x 1 + 2) 2 + ( y 1 - 3) 2 = 4
The required locus of P ( x 1, y1 ) is
( x + 2) 2 + ( y - 3) 2 = 4
y Example 44. Find the length of tangents drawn
from the point ( 3, - 4 ) to the circle
2x 2 + 2y 2 - 7 x - 9 y - 13 = 0.
Sol. The equation of the given circle is
2x 2 + 2y 2 - 7 x - 9y - 13 = 0
Re-writing the given equation of the circle
7
9
13
i.e.
x2 + y2 - x - y =0
2
2
2
276
Textbook of Coordinate Geometry
7
9
13
x- y2
2
2
13
9
7
2
2
\
S1 = (3) + ( -4 ) - ´ 3 - ´ ( -4 ) 2
2
2
13
21
= 43 - 17 = 26
= 25 + 18 2
2
\ Length of tangent = S1 = 26
Let
S = x2 + y2 -
\
\
y Example 45. If the length of tangent from ( f , g ) to
the circle x 2 + y 2 = 6 be twice the length of the
tangent from ( f , g ) to circle x 2 + y 2 + 3x + 3y = 0, then
find the value of f 2 + g 2 + 4 f + 4 g .
Sol. According to the question
(g 2 + f
On squaring g 2 + f
or
3f
or
2
- 6) = 2 ( f
2
- 6 = 4f
2
+ g 2 + 3 f + 3g )
Sol. Let ( x 1, y1 ) be any point on
+ 4 g 2 + 12 f + 12g
x 2 + y 2 + 2gx + 2 fy + c = 0
+ 3g + 12 f + 12g + 6 = 0
2
then
+ g 2 + 4 f + 4g + 2 = 0
f
2
(let)
\
2
2
PC = ( 4 - 0) + (3 - 0) = 5 units
4
Q
α
3
M
4
(0, 0) C
3
P (4, 3)
α
R
...(i)
x 2 + y 2 + 2gx + 2 fy + c 1 = 0 is
Sol. Since, PQ = PR = 4 2 + 32 - 9 = 4 units
ÐCPQ = ÐCPR = a
x 12 + y12 + 2gx 1 + 2 fy1 + c = 0
\ Length of tangent from ( x 1, y1 ) to the circle
+ g 2 + 4 f + 4g = - 2
y Example 46. Show that the area of the triangle
formed by tangents from the point (4, 3) to the circle
x 2 + y 2 = 9 and the line segment joining their points of
17
contact is 7
square units in length.
25
\
y Example 47. Show that the length of the tangent
from any point on the circle x 2 + y 2 + 2gx + 2 fy + c = 0
to the circle x 2 + y 2 + 2gx + 2 fy + c 1 = 0 is (c 1 - c ) .
2
f
or
2
2
12
5
1
Area of D PQR = × QR × PM
2
1
= (2QM ) × PM = (QM ) ( PM )
2
æ 12 ö æ 16 ö 192
=ç ÷ç ÷=
è 5 ø è 5 ø 25
17
sq units
=7
25
QM =
x 12 + y12 + 2gx 1 + 2 fy1 + c 1
[From Eq. (i)]
= ( -c + c 1 ) = ( c 1 - c )
Power of a Point With
Respect to a Circle
Theorem : The power of a point P ( x 1 , y 1 ) with respect to
the circle
x 2 + y 2 + 2 gx + 2 fy + c = 0 is S 1
where, S 1 = x 12 + y 12 + 2 gx 1 + 2 fy 1 + c
Proof : Let P ( x 1 , y 1 ) be a point outside the circle and PAB
and PCD drawn two secants. The power of P ( x 1 , y 1 ) with
respect to
B
\In DPQC,
\
and
In DPMQ,
\
and
3
tana = ,
4
3
sina =
5
4
cosa =
5
PM 4
cosa =
=
4
5
16
PM =
5
QM 3
sina =
=
4
5
T
A
P (x 1, y 1)
C
D
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
is equal to PA × PB which is
x 12 + y 12 +2 gx 1 + 2 fy 1 + c = S 1
\ Power remains constant for the circle
Chap 04 Circle
277
i.e. independent of A and B.
\ PA × PB = PC × PD = ( PT ) 2 = S 1 = ( S 1 ) 2
(x′, y′ )
T
2
\ PA × PB = ( S 1 ) = square of the length of tangent.
Chord of
contact
Remark
T ′(x ′′, y ′′)
If P outside, inside or on the circle, then PA× PB is + ve, - ve
or zero, respectively.
y Example 48. Find the power of point (2, 4 ) with
2
P (x1, y1)
Then, equations of tangents PT and PT ¢ are
2
xx ¢ + yy ¢ = a 2 and xx ¢ ¢ + yy ¢ ¢ = a 2 respectively.
respect to the circle x + y - 6 x + 4 y - 8 = 0
Sol. The power of the point (2, 4 ) with respect to the circle
2
2
Since, both tangents pass through P ( x 1 , y 1 ), then
x 1 x ¢ + y 1y ¢ = a 2
2
x + y - 6 x + 4y - 8 = 0 is ( S 1 ) or S 1
where, S = x 2 + y 2 - 6x + 4y - 8
\
S1 = (2)2 + ( 4 )2 - 6 ´ 2 + 4 ´ 4 - 8
Q Points T ( x ¢ , y ¢ ) and T ¢ ( x ¢¢,y ¢¢ ) lie on
= 4 + 16 - 12 + 16 - 8 = 16
[Q(2, 4 ) is outside from the circle x 2 + y 2 - 6x + 4y - 8 = 0]
y Example 49. Show that the locus of the point, the
powers of which with respect to two given circles
are equal, is a straight line.
Sol. Let the given circles be
x 2 + y 2 + 2gx + 2 fy + c = 0
and
...(i)
x 2 + y 2 + 2g 1x + 2 f 1y + c 1 = 0
...(ii)
Let P ( x 1, y1 ) be a point, the powers of which with respect
to the circles Eqs. (i) and (ii) are equal. Then,
\
[ ( x 12 + y12 + 2gx 1 + 2 fy1 + c ) ]2
= [ ( x 12 + y12 + 2g 1x 1 + 2 f 1y1 + c 1 ) ]2
or
x 1 x ¢¢ + y 1 y ¢¢ = a 2
and
x 12 + y12 + 2gx 1 + 2 fy1 + c
xx 1 + yy 1 = a 2
\ Equation of chord of contact TT ¢ is xx 1 + yy 1 = a 2
Remark
Equation of chord of contact like as equation of tangent at
that point but point different.
Now, for chord of contact at ( x 1 , y 1 ), replacing x 2 by
x + x1
y + y1
xx 1 , y 2 by yy 1 , x by
, y by
2
2
xy 1 + x 1 y
.
and xy by
2
Corollary 1 : If R is the radius of the circle and L is the
length of the tangent from P ( x 1 ,y 1 ) on S = 0.
Here, L = S 1 , then
= x 12 + y12 + 2g 1x 1 + 2 f 1y1 + c 1
Þ
2 ( g - g 1 ) x 1 + 2 ( f - f 1 ) y1 + c - c 1 = 0
then, locus of P ( x 1, y1 ) is
2 ( g - g1 ) x + 2 ( f - f 1 ) y + c - c1 = 0
which is a straight line.
Chord of Contact
(a) Length of chord of contact TT ¢ =
2 LR
( R 2 + L2 )
(b) Area of triangle formed by the pair of tangents and its
RL3
chord of contact =
R 2 + L2
(c) Angle between the pair of tangents from P ( x 1 ,y 1 )
æ 2RL ö
= tan -1 ç
÷
è L2 - R 2 ø
From any external point, two tangents can be drawn to a
given circle. The chord joining the points of contact of the
two tangents is called the chord of contact of tangents.
Corollary 2 : Equation of the circle circumscribing the
Theorem : The equation of the chord of contact of
tangents drawn from a point ( x 1 , y 1 ) to the circle
x 2 + y 2 = a 2 is xx 1 + yy 1 = a 2 .
triangle PTT ¢ is
( x - x 1 )( x + g ) + (y - y 1 )(y + f ) = 0,
where, O ( -g, - f ) is the centre of the circle
Sol. Let T ( x ¢ , y ¢ ) and T ¢ ( x ¢¢, y ¢¢ ) be the points of contact of
2
2
2
tangents drawn from P ( x 1, y1 ) to x + y = a .
x 2 + y 2 + 2 gx + 2 fx + c = 0
278
Textbook of Coordinate Geometry
Sol. Equation of chord of contact AB is hx + ky = a 2
T
For equation of pair of tangents of OA and OB, make
homogeneous x 2 + y 2 = a 2 with the help of hx + ky = a 2 or
hx + ky
=1
a2
O(–g, –f)
P
(x1, y1)
T¢
A
i.e. Required circle always passes through the centre of
the given circle (Here, OP is the diameter of the required
circle).
y Example 50. If the pair of tangents are drawn from
the point (4, 5) to the circle x 2 + y 2 - 4 x - 2y - 11 = 0,
then,
(i) Find the length of chord of contact.
(ii) Find the area of the triangle formed by a pair of
tangents and their chord of contact.
(iii) Find the angle between the pair of tangents.
Sol. Here, P º ( 4,5),
R = ((2)2 + (1)2 + 11) = 4
or
(i) Length of chord of contact
2LR
2´2´ 4
8
unit
=
=
=
2
2
2
2
5
(R + L )
( 4 ) + ( 2)
RL3
4 ´8 8
=
= sq units
R + L2 16 + 4 5
(iii) Angle between the pair of tangents
æ2 ´ 4 ´ 2ö
= p + tan -1 ç 2
÷
è 2 - 42 ø
æ hx + ky ö
x 2 + y 2 = a2 ç
÷
è a2 ø
a 2 ( x 2 + y 2 ) = (hx + ky )2
x 2 (a 2 - h 2 ) - 2hkxy + y 2 (a 2 - k 2 ) = 0
p
but
Ð AOB =
2
\ Coefficient of x 2 + Coefficient of y 2 = 0
a 2 - h 2 + a 2 - k 2 = 0 or h 2 + k 2 = 2a 2
y Example 53. The chord of contact of tangents drawn
from a point on the circle x 2 + y 2 = a 2 to the circle
x 2 + y 2 = b 2 touches the circle x 2 + y 2 = c 2 . Show
that a, b , c are in GP.
Then, equation of chord of contact of tangents drawn from
P (a cos q , a sin q ) to the circle x 2 + y 2 = b 2 is
R
x2 + y2 = c 2
x2 + y2 = b2
x2 + y 2= a2
Sol. Here, P º ( -8,2) and O º (0,0)
\Equation of the required circle is
( x - ( -8))( x - 0) + (y - 2)(y - 0) = 0
or
2
x + y + 8 x - 2y = 0
(QOP is the diameter)
y Example 52. Find the condition that chord of
contact of any external point (h, k ) to the circle
x 2 + y 2 = a 2 should subtend right angle at the centre
of the circle .
T
P
(Q L < R)
y Example 51. Tangents PQ, PR are drawn to the circle
x 2 + y 2 = 36 from the point P( -8,2) touching the circle
at Q, R respectively. Find the equation of the
circumcircle of DPQR.
2
2
Sol. Let P (a cos q , a sin q ) be a point on the circle
x 2 + y 2 = a2.
2
æ4ö
= p - tan -1 ç ÷
è3ø
B
or
Þ
and L = S1 = (( 4 )2 + (5)2 - 4 ´ 4 - 2 ´ 5 - 11) = 2
(ii) Area of triangle =
P (h, k)
then,
O (0, 0)
π
2
ax cos q + ay sin q = b 2
This touches the circle x 2 + y 2 = c 2
...(i)
...(ii)
\ Length of perpendicular from (0, 0) to Eq. (i) = radius of
Eq. (ii)
| 0 + 0 - b2|
=c
\
(a 2 cos 2 q + a 2 sin 2 q )
or
b 2 = ac
Þ a, b, c are in GP.
Chap 04 Circle
279
Chord Bisected at a
Given Point
y Example 55. Find the middle point of the chord
intercepted on line lx + my + n = 0 by the circle
x 2 + y 2 = a2.
Theorem : The equation of the chord of the circle
x 2 + y 2 = a 2 bisected at the point ( x 1 , y 1 ) is given by
Sol. Let ( x 1, y1 ) be the middle point of the chord intercepted
by the circle x 2 + y 2 = a 2 on the line lx + my + n = 0.
Then, equation of the chord of the circle x 2 + y 2 = a 2 ,
whose middle points is ( x 1, y1 ), is
xx 1 + yy1 - a 2 = x 12 + y12 - a 2
xx 1 + yy 1 - a 2 = x 12 + y 12 - a 2
or
T = S1
Proof : Let any chord AB of the circle x 2 + y 2 = a 2 be
bisected at D ( x 1 , y 1 ).
If centre of circle is represented by C
A
C (0,0 )
D
,
(x 1
y 1)
or
\
0 - y1 y1
slope of DC =
=
0 - x1 x1
\ Slope of the chord AB is -
then, equation of AB is y - y 1 = or
or
or
x 12 + y12 = - nl
l 2 l2 + m 2 l2 = - nl
l=-
So, from Eq. (ii),
x1
y1
...(i)
Clearly, lx + my + n = 0 and Eq. (i) represented the same
line,
x 1 y1 x 12 + y12
(say)
=l
=
=
-n
l
m
x1 = l l ü
...(ii)
\
y1 = mlýþ
and
B
then,
xx 1 + yy1 = x 12 + y12
or
[from (ii)]
n
l 2 + m2
x1 = -
nl
2
l +m
2
, y1 = -
mn
2
l + m2
æ
nl
nm ö
Hence, the required point is ç - 2
,- 2
÷.
2
l + m2 ø
è l +m
x1
(x - x 1 )
y1
yy 1 - y 12 = - xx 1 + x 12
xx 1 + yy 1 = x 12 + y 12
xx 1 + yy 1 - a 2 = x 12 + y 12 - a 2 or T = S 1
y Example 56. Through a fixed point (h, k ) , secants are
drawn to the circle x 2 + y 2 = r 2 . Show that the locus
of mid-point of the portions of secants intercepted by
the circle is x 2 + y 2 = hx + ky .
Sol. Let P ( x 1, y1 ) be the middle point of any chord AB, which
passes through the point C (h , k ).
Remarks
1. The equation of chord of the circle x 2 + y 2 + 2gx + 2fy + c = 0,
which is bisected at ( x1, y1 ) ; is T = S1
where, T = xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c
and
S1 = x12 + y12 + 2gx1 + 2fy1 + c
O
C (h, k)
2. The chord bisected at point ( x1, y1 ) is the farthest from the
centre among all the chords passing through the point ( x1, y1 ).
Also, for such chord, the length of the chord is minimum.
y Example 54. Find the equation of the chord of
x 2 + y 2 - 6 x + 10y - 9 = 0 which is bisected at ( -2, 4 ).
Sol. The equation of the required chord is
- 2x + 4y - 3 ( x - 2) + 5 ( y + 4 ) - 9
= 4 + 16 + 12 + 40 - 9
Þ
-5x + 9y - 46 = 0
or
5x - 9y + 46 = 0
y 1)
x 1,
P(
B
A
Equation of chord AB is T = S1
\
xx 1 + yy1 - r 2 = x 12 + y12 - r 2
or
x 12 + y12 = xx 1 + yy1
But since AB passes through C (h , k ), then
x 12 + y12 = hx 1 + ky1
\ Locus of P ( x 1, y1 ) is x 2 + y 2 = hx + ky
280
Textbook of Coordinate Geometry
y Example 57. Find the locus of middle points of
chords of the circle x 2 + y 2 = a 2 , which subtend right
angle at the point (c , 0) .
Sol. Let N (h , k ) be the middle point of any chord AB, which
subtend a right angle at P (c , 0).
Since,
ÐAPB = 90°
\
NA = NB = NP
Theorem : The combined equation of the pair of tangents
drawn from a point P ( x 1 , y 1 ) to the circle x 2 + y 2 = a 2 is
( x 2 + y 2 - a 2 ) ( x 12 + y 12 - a 2 ) = ( xx 1 + yy 1 - a 2 ) 2
or
SS 1 = T 2
S = x 2 + y 2 - a 2 , S 1 = x 12 + y 12 - a 2
where,
Y
A
and
T = xx 1 + yy 1 - a 2
Proof : The given circle is x 2 + y 2 = a 2
(h, k)
N
X′
P (c, 0)
B
Pair of Tangents
X
O
Its centre and radius are C (0, 0 ) and a respectively . Given
external point be P ( x 1 , y 1 ).
From point P ( x 1 ,y 1 ) two tangents PT and PR be drawn to
the circle, touching circle at T and R respectively.
Y′
T
or
But also
\
Ð BNO = 90°
(OB )2 = (ON )2 + ( NB )2
Þ
- ( NB )2 = (ON )2 - (OB )2
Þ - [(h - c )2 + (k - 0)2 ] = (h 2 + k 2 ) - a 2
or
a
(since distance of the vertices from middle
point of the hypotenuse are equal)
2
2
...(i)
( NA ) = ( NB ) = (h - c )2 + (k - 0)2
2 (h 2 + k 2 ) - 2ch + c 2 - a 2 = 0
\ Locus of N (h , k ) is
2 ( x 2 + y 2 ) - 2cx + c 2 - a 2 = 0
y Example 58. Find the equation of the chord of the
circle x 2 + y 2 = r 2 passing through the point (2, 3)
farthest from the centre.
Sol. Let P º (2,3) be the given point and M be the middle point
of chord of circle x 2 + y 2 = r 2 through P.
C (0, 0)
a
(α, β)
Q
R
P (x1, y1)
Let Q (a, b) on PT, then equation of PQ is
b - y1
y - y1 =
(x - x 1 )
a - x1
or
y (a - x 1 ) - x (b - y 1 ) - ay 1 + bx 1 = 0
Length of perpendicular from C (0, 0 ) on PT = a (radius)
Þ
| bx 1 - ay 1 |
2
(a - x 1 ) + (b - y 1 )
or
2
=a
(bx 1 - ay 1 ) 2 = a 2 {(a - x 1 ) 2 + (b - y 1 ) 2 }
\ Locus of Q (a, b) is
(yx 1 - xy 1 ) 2 = a 2 {( x - x 1 ) 2 + (y - y 1 ) 2 }
O
P
Then,
M
(OM )2 = (OP )2 - ( PM )2
If OM maximum, then PM is minimum. i.e. P coincides with
M, which is middle point of the chord.
Hence, the equation of the chord is
T = S1
i.e.
2x + 3y - r 2 = 22 + 32 - r 2
or
2x + 3y = 13
Þ
y 2 x 12 + x 2 y 12 - 2 xy x 1 y 1 = a 2
{ x 2 + x 12 - 2 xx 1 + y 2 + y 12 - 2yy 1 }
Þ y 2 ( x 12 - a 2 ) + x 2 (y 12 - a 2 ) - a 2 ( x 12 + y 12 )
= 2 xyx 1 y 1 - 2a 2 xx 1 - 2a 2 yy 1
On adding both sides, ( x 2 x 12 + y 2 y 12 + a 4 ), then
y 2 ( x 12 + y 12 - a 2 ) + x 2 ( x 12 + y 12 - a 2 ) -a 2 ( x 12 + y 12 - a 2 )
= ( xx 1 + yy 1 - a 2 ) 2
Chap 04 Circle
Þ ( x 2 + y 2 - a 2 ) ( x 12 + y 12 - a 2 ) = ( xx 1 + yy 1 - a 2 ) 2
This is the required equation of pair of tangents drawn
from ( x 1 , y 1 ) to circle x 2 + y 2 = a 2 .
2
2
2
Aliter : Let circle be x + y = a with centre C (0, 0 ) and
radius a. Length of tangents from P ( x 1 , y 1 ) and Q (a, b)
are
and
2
QT = a + b - a
Remarks
1. Equation of pair of tangents in notation form is SS1 = T 2
where,
S1 º x12 + y12 - a2, T º xx1 + yy1 - a2
2. When circle is x 2 + y 2 + 2gx + 2fy + c = 0 and tangents are
drawn from ( x1, y1 ), then pair of tangents is
( x 2 + y 2 + 2gx + 2fy + c ) ( x12 + y12 + 2gx1 + 2fy1 + c)
where,
2
and
Q(
α,
β)
Students are advised that, if they do not want to use the
formula (SS 1 = T 2 ), then use the following method :
C (0,0)
a
R
Now, equation of TR (chord of contact is)
xx 1 + yy 1 - a 2 = 0
\
PN =
| x 12 + y 12 - a 2 |
( x 12 + y 12 )
and QM =
|ax 1 + by 1 - a 2 |
( x 12 + y 12 )
But from similar Ds PNT and QMT,
( PN ) 2 ( PT ) 2
PN PT
=
=
Þ
QM QT
(QM ) 2 (QT ) 2
( x 12 + y 12 - a 2 ) 2
Þ
( x 12 + y 12 )
(ax 1 + by 1 - a 2 ) 2
=
x 12 + y 12 - a 2
a 2 + b2 - a 2
Let y - y 1 = m( x - x 1 ) be any line through P ( x 1 ,y 1 ).
Then, use condition of tangency p = r i.e.
Length of perpendicular from the centre of circle or this
line = radius of the circle.
Gives the values of m. In such away, we can find the
equations of tangents from P.
y Example 59. Find the equations of the tangents
from the point A ( 3, 2) to the circle
x 2 + y 2 + 4 x + 6 y + 8 = 0.
Sol. Combined equation of the pair of tangents drawn from
A (3, 2) to the given circle x 2 + y 2 + 4 x + 6y + 8 = 0 can be
written in the usual notation.
T 2 = SS1 namely
Þ [3x + 2y + 2 ( x + 3) + 3 (y + 2) + 8]2
( x 12 + y 12 )
Þ
T º xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c
Advised for Students
a
M
N
(x1, y1)
S º x 2 + y 2 + 2gx + 2fy + c,
S1 º x12 + y12 + 2gx1 + 2fy1 + c,
T
P
S º x 2 + y 2 - a2
= [ xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c ] 2
PT = x 12 + y 12 - a 2
2
281
( x 12 + y 12 - a 2 ) (a 2 + b2 - a 2 ) = (ax 1 + by 1 - a 2 ) 2
\ Locus of Q (a, b) is
( x 2 + y 2 - a 2 ) ( x 12 + y 12 - a 2 )
= ( xx 1 + yy 1 - a 2 ) 2
This is the required equation of pair of tangents drawn
from ( x 1 , y 1 ) to circle x 2 + y 2 = a 2 .
= [ x 2 + y 2 + 4 x + 6y + 8] [9 + 4 + 12 + 12 + 8]
Þ
(5x + 5y + 20)2 = 45 ( x 2 + y 2 + 4 x + 6y + 8)
Þ
5 ( x + y + 4 )2 = 9 ( x 2 + y 2 + 4 x + 6y + 8)
Þ
5 ( x 2 + y 2 + 2xy + 8x + 8y + 16)
= 9 ( x 2 + y 2 + 4 x + 6y + 8)
Þ
or
4 x 2 + 4y 2 - 10xy - 4 x + 14y - 8 = 0
2x 2 + 2y 2 - 5xy - 2x + 7y - 4 = 0
Corollary : The angle between the two tangents from
æ a ö
÷ , where
( x 1 ,y 1 ) to the circle x 2 + y 2 = a 2 is 2 tan -1 ç
ç S ÷
è 1ø
or
(2x - y - 4 ) ( x - 2y + 1) = 0
Hence, the required tangents to the circle from A (3, 2) are
2x - y - 4 = 0 and x - 2y + 1 = 0
Aliter : Let S º x 2 + y 2 + 4 x + 6y + 8 = 0
S 1 = x 12 + y 12 - a 2 .
Centre C ( -2, - 3) and radius = 5
282
Textbook of Coordinate Geometry
Let the slope of a tangent from A to ‘ S =0’ be m, then
equation of tangent is
y - 2 = m ( x - 3)
or
mx - y + 2 - 3m = 0
Length of perpendicular from C ( -2, - 3) on Eq. (i)
= radius of circle.
m 2 (h 2 - a 2 ) - 2mkh + k 2 - a 2 = 0
or
…(i)
This is quadratic equation in m, let two roots are m 1 and
m2 .
But tangents are perpendiculars, then
P (h, k)
A (3, 2)
90°
T1
5
5
C (– 2, –3 )
T2
m 1m 2 = - 1
2
S=0
Þ
| - 2m + 3 + 2 - 3m |
2
Þ
25m 2 - 50m + 25 = 5m 2 + 5
Þ
20m 2 - 50m + 20 = 0
or
2m 2 - 5m + 2 = 0
SS 1 = T 2
S 1 = h2 + k 2 - a 2
x 2 + y 2 = 2a 2
...(i)
Let P (h, k ) be the point of intersection of tangents, then
P (h, k ) lies on Eq. (i)
k = mh + a (1 + m 2 )
(k - mh ) 2 = a 2 (1 + m 2 ) q
T = hx + ky - a 2
\
( x 2 + y 2 - a 2 ) (h 2 + k 2 - a 2 ) = (hx + ky - a 2 ) 2
Þ
h 2 + k 2 - a 2 - h 2 + h 2 + k 2 -a 2 - k 2 = 0
Þ
h 2 + k 2 - 2a 2 = 0 or h 2 + k 2 = 2a 2
Hence, the locus of (h, k ) is x 2 + y 2 = 2a 2
Remarks
1. The equation of the director circle of the circle
( x - h) 2 + ( y - k ) 2 = a2 is ( x - h) 2 + ( y - k ) 2 = 2a2
2. The equation of the director circle of the circle
x 2 + y 2 + 2gx + 2fy + c = 0 is x 2 + y 2 + 2gx + 2fy + c = g 2 + f 2 - c
Proof : The equation of any tangent to the circle
x 2 + y 2 = a2
y = mx + a (1 + m )
and
This equation will represent a pair of perpendicular lines
if, coefficient of x 2 + coefficient of y 2 = 0
Director circle : The locus of the point of intersection of
two perpendicular tangents to a given circle is known as
its director circle.
Theorem : The equation of the director circle of the circle
x 2 + y 2 = a 2 is
2
S = x 2 + y 2 - a2
where,
Director Circle
or
h 2 + k 2 = 2a 2
Aliter : The combined equation of the pair of tangents
drawn from (h, k ) to x 2 + y 2 = a 2 is
(2m - 1) (m - 2) = 0
1
\
m = or m = 2
2
Substituting these values of m in Eq. (i), we get the
equations of two tangents are x - 2y + 1 = 0 and
2x - y - 4 = 0.
\
= - 1 or k 2 - a 2 = - h 2 + a 2
Hence, locus of P (h, k ) is x 2 + y 2 = 2a 2
(5m - 5)2 = 5 (m 2 + 1)
or
is
h2 - a 2
or
= 5
( m + 1)
or
k - a2
Þ
3. If two tangents are drawn from a point on the director circle to
the circle, then angle between tangents is 90°.
y Example 60. If two tangents are drawn from a point
on the circle x 2 + y 2 = 50 to the circle x 2 + y 2 = 25,
then find the angle between the tangents.
Sol. Q x 2 + y 2 = 50 is the director circle of x 2 + y 2 = 25
Hence, angle between tangents = 90°
Chap 04 Circle
283
Exercise for Session 5
1.
If the tangent at the point P on the circle x 2 + y 2 + 6x + 6y = 2 meets the straight line 5x - 2y + 6 = 0 at a point Q
on the Y-axis, then the length PQ is
(a) 4
2.
(b) 2 5
2
l1 + l 2
2
(c) an ellipse
(d) a hyperbola
2
(b) x 2 + y 2 = 1
(d) x 2 + y 2 = 2
(c) x + y = 2
(b) 51
(c) 61
(d) 71
(b) g 2 + f 2 = 2c
(c) g 2 + f 2 = 5c
(d) g 2 + f 2 = 4c
1 1
(b) æç - , - ö÷
è 2 4ø
1 1
(c) æç , ö÷
è 2 4ø
(d) (-2, - 4)
5
(b) æç ö÷
è 2ø
(c)
5
2
(d) 2
The perpendicular tangents to the circle x 2 + y 2 = a 2 meet at P. Then, the locus of P has the equation
(b) x 2 + y 2 = 3a 2
(c) x 2 + y 2 = 4a 2
(d) x 2 + y 2 = 5a 2
The tangents to x 2 + y 2 = a 2 having inclinations a and b intersect at P. If cot a + cot b = 0, then the locus of P is
(a) x + y = 0
12.
(b) a parabola
The length of tangent from (0, 0) to the circle 2( x 2 + y 2 ) + x - y + 5 = 0 is
(a) x 2 + y 2 = 2a 2
11.
(d) (l 12 - l 22 )
The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to the circle
x 2 + y 2 = 1 pass through a fixed point
(a) 5
10.
(c) (l 12 + l 22 )
p
If the angle between the tangents drawn to x 2 + y 2 + 2gx + 2fy + c = 0 from (0,0) is , then
2
(a) (2, 4)
9.
l1 - l 2
2
The length of tangents from P(1, - 1) and Q(3,3) to a circle are 2 and 6 respectively, then the length of tangent
from R( -2, - 7) to the same circle is
(a) g 2 + f 2 = 3c
8.
(d) 288
The locus of the mid-points of a chord of the circle x + y = 4, which subtends a right angle at the origin is
(a) 41
7.
(b)
2
(a) x + y = 1
6.
(c) 144
If the chord of contact of tangents from a point ( x1, y1) to the circle x 2 + y 2 = a 2 touches the circle
( x - a )2 + y 2 = a 2, then the locus of ( x1, y1) is
(a) a circle
5.
(b) 72
The chord of contact of tangents from a point P to a circle passes through Q. If l1 and l 2 are the lengths of
tangents from P and Q to the circle, then PQ is equal to
(a)
4.
(d) 3 5
If the circle x + y + 2gx + 2fy + c = 0 is touched by y = x at P such that OP = 6 2, where O is origin, then the
value of c is
(a) 36
3.
(c) 5
2
(b) x - y = 0
(c) xy = 0
(d) xy = 1
The exhaustive range of values of a such that the angle between the pair of tangents drawn from (a,a ) to the
æp ö
circle x 2 + y 2 - 2x - 2y - 6 = 0 lies in the range ç , p ÷ is
è3 ø
(a) (-1, 3)
(b) (-5, - 3) È (3, 5)
(c) (-3, 5)
(d) (-3, - 1) È (3, 5)
13.
Distances from the origin to the centres of the three circles x + y - 2lx = c , where c is a constant and l is
available, are in GP. Prove that the lengths of tangents drawn from any point on the circle x 2 + y 2 = c 2 to the
three circles are also in GP.
14.
Find the area of the quadrilateral formed by a pair of tangents from the point (4, 5) to the circle
x 2 + y 2 - 4x - 2y - 11 = 0 and a pair of its radii.
15.
If the length of the tangent from a point (f , g ) to the circle x 2 + y 2 = 4 be four times the length of the tangent
2
2
2
from it to the circle x 2 + y 2 = 4x , show that 15f 2 + 15g 2 - 64f + 4 = 0 .
16.
17.
Find the equation of that chord of the circle x 2 + y 2 = 15 which is bisected at (3, 2).
The chords of contact of the pair of tangents to the circle x 2 + y 2 = 1 drawn from any point on the line
2x + y = 4 pass through the point ( a, b ), then find a 2 + b 2.
Session 6
Diameter of a Circle, Two Circles Touching
Each Other, Common Tangents to Two Circles,
Common Chord of Two Circles, Family of Circles
Diameter of a Circle
Aliter : Let (h, k ) be the middle point of the chord
y = mx + c of the circle x 2 + y 2 = a 2
The locus of the middle points of a system of parallel
chords of a circle is called a diameter of the circle.
then, T = S 1
Let the circle be x 2 + y 2 = a 2 and equation of parallel
chord is
y = mx + c
h
= m Þ h + mk = 0
k
Hence, locus of mid-point is x +my = 0.
slope = -
Remark
A (x1, y1)
Di
a
m
et
er
, y 2)
B (x 2
y=
m
x+
c
P
O
Let P (h, k ) be the middle point of the chord y = mx + c .
Since, P is the mid-point of A ( x 1 y 1 ) and B( x 2 , y 2 ), then
x1 + x2
y1 + y2
= h and
=k
2
2
…(i)
or
x 1 + x 2 = 2h and y 1 + y 2 = 2k
Q
P (h, k ) lie on y = mx + c
then,
k = mh + c
or
…(ii)
k - mh = c
Substituting y = mx + c in x 2 + y 2 = a 2
then, x 2 + (mx + c ) 2 = a 2
or
(1 + m 2 ) x 2 + 2mcx + c 2 - a 2 = 0
…(iii)
Let x 1 , x 2 are roots of Eq. (iii), then
2mc
x1 + x2 = 1 +m2
2m
Þ
(k - mh ) [from Eq. (i) and Eq. (ii)]
2h = (1 + m 2 )
Þ
h + m 2 h = -mk + m 2 h Þ h + mk = 0
Hence, locus of (h, k ) is x + my = 0
Þ xh + ky = h 2 + k 2
The diameter of circle always passes through the centre of the
circle and perpendicular to the parallel chords.
Let circle is x 2 + y 2 = a2 and parallel chord be y = mx + c, then
equation of line ^ to y = mx + c is
…(i)
my + x + l = 0
which passes through origin (centre)
then,
0 + 0 + l =0 \ l =0
Then, equation of diameter from Eq. (i) is x + my = 0 .
y Example 61. Find the equation of the diameter of
the circle x 2 + y 2 + 2gx + 2 fy + c = 0 which
corresponds to the chord ax + by + d = 0.
Sol. The diameter of circle passes through the centre of the
circle and perpendicular to the chord ax + by + d = 0 is
…(i)
bx - ay + l = 0
which passes through centre of circle i.e. ( - g , - f )
Then, -bg + af + l = 0
\
l = bg - af
From Eq. (i), the equation of the diameter is
bx - ay + bg - af = 0
Two Circles Touching Each Other
1. When two circles touch each other
externally
Then, distance between their centres = sum of their radii
i.e.
|C 1 C 2 | = r1 + r2
Chap 04 Circle
In such cases, the point of contact P divides the line
joining C 1 and C 2 internally in the ratio r1 : r2
\
C 1 º (1, 2),r1 = (1 + 4 ) or r1 = 5
and
C 2 º (0, 4 ),r 2 = 0 + 16 + 4 or r 2 = 2 5
285
Now, C 1C 2 = (1 - 0)2 + (2 - 4 )2
r1
C1
C 1C 2 = 5 = r 2 - r1
r2
P
Hence, the two circles touch each other internally.
C2
y Example 63. Prove that the circles
Þ
If
x 2 + y 2 + 2ax + c 2 = 0 and x 2 + y 2 + 2by + c 2 = 0
C 1 P r1
=
C 2 P r2
touch each other, if
and C 2 º ( x 2 , y 2 )
C1 º (x 1 , y 1 )
1
a
2
+
1
b
2
1
=
c2
.
Sol. Given circles are
x 2 + y 2 + 2ax + c 2 = 0
ær x +r x r y +r y ö
then, coordinate of P is ç 1 2 2 1 , 1 2 2 1 ÷
r1 + r2 ø
è r1 + r2
2
and
2
…(i)
2
…(ii)
x + y + 2by + c = 0
Let C 1 and C 2 be the centres of circles Eqs. (i) and (ii),
respectively and r1 and r 2 be their radii, then
C 1 = ( -a, 0), C 2 = (0, - b ),
2. When two Circles Touch each other
Internally
Then, distance between their centres =Difference of their
radii
i.e.
| C 1 C 2 | = | r1 - r2 |
In such cases, the point of contact P divides the line
joining C 1 and C 2 externally in the ratio r1 : r2
r1 = (a 2 - c 2 ) , r 2 = (b 2 - c 2 )
Here, we do not find the two circles touch each other
internally or externally.
For touch, | C 1 C 2 | = | r1 ± r 2 |
or
(a 2 + b 2 ) = | (a 2 - c 2 ) ± (b 2 - c 2 )|
On squaring
a 2 + b 2 = a 2 - c 2 + b 2 - c 2 ± 2 (a 2 - c 2 ) (b 2 - c 2 )
r2
C2
C1
or
P
c 2 = ± a 2b 2 - c 2 (a 2 + b 2 ) + c 4
Again, squaring,
c 4 = a 2b 2 - c 2 (a 2 + b 2 ) + c 4
r1
or
Þ
If
C 1 P r1
=
C 2 P r2
C 1 º ( x 1 , y 1 ) and C 2 º ( x 2 , y 2 )
ær x -r x r y -r y ö
then, coordinates of P is ç 1 2 2 1 , 1 2 2 1 ÷
r1 - r2 ø
è r1 - r2
y Example 62. Examine if the two circles
x 2 + y 2 - 2x - 4 y = 0 and x 2 + y 2 - 8 y - 4 = 0 touch
and
x 2 + y 2 - 8y - 4 = 0
Let centres and radii of circles Eqs. (i) and (ii) are
represented by C 1, r1 and C 2 , r 2 , respectively.
1
a
2
+
1
b
2
=
1
c2
Common Tangents to Two
Circles
Different Cases of Intersection of Two
Circles :
Let the two circles be
each other externally or internally.
Sol. Given circles are
x 2 + y 2 - 2x - 4y = 0
c 2 (a 2 + b 2 ) = a 2b 2 or
( x - x 1 ) 2 + (y - y 1 ) 2 = r12
2
2
= r22
…(i)
…(i)
and
…(ii)
with centres C 1 ( x 1 , y 1 ) and C 2 ( x 2 , y 2 ) and radii r1 and r2
respectively. Then following cases may arise :
( x - x 2 ) + (y - y 2 )
…(ii)
286
Textbook of Coordinate Geometry
Case I : When |C 1 C 2 | > r1 + r2 i.e. the distance between the
centres is greater than the sum of their radii.
How to find transverse common tangent
Direct common
tangents
Direct common
tangents
D
D
Q
r2
P´
P
r1
r1
C2
C2
r2
T
T Q
1
Transverse common
tangent
C1
C1
Transverse common
tangents
Q Equation of circles are
C 1 º ( x 1 , y 1 ) , C 2 º ( x 2 ,y 2 )
Q
\
ær x -r x r y -r y ö
D º ç 1 2 2 1 , 1 2 2 1 ÷ º (a, b) (say)
r1 - r2 ø
è r1 - r2
and
ær x +r x r y +r y ö
T ºç 1 2 2 1, 1 2 2 1÷
r1 + r2 ø
è r1 + r2
S 1 º ( x - x 1 ) 2 + (y - y 1 ) 2 - r12 = 0
and
S 2 º ( x - x 2 ) 2 + (y - y 2 ) 2 - r22 = 0
then, equation of common tangent is
S1 - S2 = 0
which is same as equation of common chord.
º ( g, d) (say)
How to find direct common tangents Let equation
of common tangent through D (a, b) is
y - b = m( x - a )
Remark
In this case circles touch at one point i.e. Number of solutions of
two circles is one.
…(i)
Now, length of ^ from C 1 or C 2 on Eq. (i) = r1 or r2
Then, we get two values of m.
Substituting the values of m in Eq. (i), we get two direct
common tangents.
Case III : When |r1 - r2 | < |C 1 C 2 | < r1 + r2
i.e. the distance between the centres is less than sum of
their radii and greater than difference of their radii.
In this case two direct common tangents are real and
distinct while the transverse tangents are imaginary.
How to find transverse common tangents Let
Direct common
tangents
D
equation of common tangent through T ( g, d) is
y - d = M(x - g )
…(ii)
Now, length of ^ from C 1 or C 2 on Eq. (i) = r1 or r2
then, we get two values of M.
C2
Substituting the values of M in Eq. (i), we get two
transverse common tangents.
C1
Remark
In this case circles neither cut nor touch i.e. Number of
solutions of two circles is zero.
Case II : When |C 1 C 2 | = r1 + r2
i.e. the distance between the centres is equal to the sum of
their radii.
In this case two direct common tangents are real and
distinct while the transverse tangents are coincident.
Remark
In this case circles cuts at two points i.e. Number of solutions of
two circles is two.
Case IV : When | C 1 C 2 | = |r1 - r2 |, i.e. the distance
between the centres is equal to the difference of their
radii.
Chap 04 Circle
Tangent at the
point of contact
P
r2
C1
287
Common Tangents to two Circles
If two circles with centres C 1 and C 2 and their radii are r1
and r2 , then
Condition
C2
Figure
Number of
common tangents
(i)
| C1C 2 | > r1 + r 2
4
(ii)
| C1C 2 | = r1 + r 2
3
(iii)
| r1 - r 2 | <| C1C 2 |< r1 + r 2
2
(iv)
| C1C 2 | = | r1 - r 2 |
1
(v)
| C1C 2 | <| r1 - r 2 |
0
r1
In this case two tangents are real and coincident while the
other two tangents are imaginary.
If circles are represented by S 1 = 0 and S 2 = 0, then
equation of common tangent is S 1 - S 2 = 0 .
Remark
If circles touch each other externally, i.e.| C1C2| = r1 + r2, then
equation of tangent at the point of contact is
S1 - S2 = 0
Tangent at the
point of contact
r2
C2
P
r1
C1
In this case circles touch at one point.
i.e. Number of solutions of two circles is one.
Case V : When | C 1 C 2 | < | r1 - r2 | , i.e. the distance
between the centres is less than the difference of their
radii.
Length of External Common Tangent and
Internal Common Tangent to Two Circles
Length of external common tangent Lex = d 2 - (r1 - r2 ) 2
and length of internal common tangent
A
Lex
r1–r2
D
r2
C1
C2
r1
d
r1
d
C1
r1+ r2
r1
A¢
B
r2 r
2
C2
Lin
r2
Lin
A¢¢
In this case, all the four common tangents are imaginary.
Remark
In this case circles neither cut nor touch each other i.e. Number
of solution of two circles is zero.
Lin = d 2 - (r1 + r2 ) 2 (Applicable only when d > r1 + r2 )
where, d is the distance between the centres of two circles
and r1 ,r2 are the radii of two circles, when |C 1 C 2 | = d .
288
Textbook of Coordinate Geometry
Angle between Direct Common Tangents (DCT)
and Transverse Common Tangents (TCT)
Case I : If d > r1 + r2 , then
æ |r - r | ö
Angle between DCT = 2 sin -1 ç 1 2 ÷
è d ø
æ r +r ö
And angle between TCT = 2 sin -1 ç 1 2 ÷
è d ø
and the point C divide C 2C 1 in the ratio 3 : 1 (internally)
æ 3(1) + 1( -3) 3(3) + 1(1) ö
then coordinates of C are ç
,
÷ or
3+1 ø
è 3+1
(0, 5/2)
Direct tangents : Any line through (3, 4) is
y - 4 = m( x - 3 )
…(i)
Þ
mx - y + 4 - 3m = 0
Apply the usual condition of tangency to any of the circle
m - 3 + 4 - 3m
= ±1
m2 + 1
Case II : If d = r1 + r2 , then
æ |r - r | ö
angle between DCT = 2 sin -1 ç 1 2 ÷
è r1 + r2 ø
and angle between TCT = p
Case III : If |r1 - r2 | < d < r1 + r2 , then
æ |r - r | ö
angle between DCT = 2 sin -1 ç 1 2 ÷
è d ø
Here, transverse common tangents are not possible .
Case IV : If d = |r1 - r2 |
Angle between DCT = p
Here, transverse common tangents are not possible.
Case V : If d <|r1 - r2 |
Here, tangents are not possible.
y Example 64. Find all the common tangents to the
circles
x 2 + y 2 - 2 x - 6 y + 9 = 0 and x 2 + y 2 + 6 x - 2 y + 1 = 0.
( x - 1) 2 + ( y - 3) 2 = 1
and
Þ
2
…(i)
2
x + y + 6x - 2y + 1 = 0
( x + 3) 2 + ( y - 1) 2 = 9
…(ii)
Centres and radii of circles Eq. (i) and Eq. (ii) are
D
1
3
C
3
(1,
C1
)
C2 (– 3, 1)
Þ
( -2m + 1)2 = m 2 + 1
Þ
3m 2 - 4m = 0
Þ
m = 0, m = 4 / 3
\ Equations of direct common tangents are
y = 4 and 4 x - 3y = 0
Transverse tangents : Any line through C (0, 5 / 2) is
y - 5 / 2 = mx
or
…(ii)
mx - y + 5 / 2 = 0
Apply the usual condition of tangency to any of the circle
m ×1 - 3 + 5 / 2
\
= ±1
m2 + 1
1
- m = m2 + 1
4
3
0 ×m2 - m - = 0
Þ
4
\
m = ¥ and m = - 3 / 4
Hence, equations of transverse tangents are
x = 0 and 3x + 4y - 10 = 0
Þ
Sol. The given circles are
x 2 + y 2 - 2x - 6x + 9 = 0
Þ
Hence, the circles do not intersect to each other.
The direct common tangents meet AB produced at D, then
point D will divide C 2C 1 in the ratio 3 : 1 (externally).
æ 3(1) - 1( -3) 3(3) - 1(1) ö
Coordinates of D are ç
,
÷ or (3, 4)
3-1 ø
è 3-1
y Example 65. Show that the common tangents to the
circles x 2 + y 2 - 6 x = 0 and x 2 + y 2 + 2x = 0 form an
equilateral triangle.
Sol. The given circles are
x 2 + y 2 - 6x = 0
or
and
A
or
and
\
C 1 (1, 3), r1 = 1
C 2 ( -3, 1),r 2 = 3 respectively.
C 1C 2 = (16 + 4 ) = 2 5
\
C 1C 2 > r1 + r 2
m2 +
( x - 3) 2 + ( y - 0) 2 = 9
2
…(i)
2
x + y + 2x = 0
( x + 1) 2 + ( y - 0) 2 = 1
…(ii)
Centres and radii of circles Eqs. (i) and (ii) are C 1(3, 0), r1 = 3
and C 2 ( -1, 0), r 2 = 1, respectively.
Q
C 1C 2 = [3 - ( -1)]2 + 0 = 4
\
C 1C 2 = r1 + r 2
Chap 04 Circle
289
Hence, the two circles touch each other externally,
therefore, there will be three common tangents. Equation of
the common tangent at the point of contact is S1 - S 2 = 0
Þ
( x 2 + y 2 - 6x ) - ( x 2 + y 2 + 2x ) = 0
y Example 66. Find the number of common tangents
to the circles x 2 + y 2 - 8 x + 2y + 8 = 0 and
x 2 + y 2 - 2x - 6 y - 15 = 0.
Þ
\
Sol. For x 2 + y 2 - 8x + 2y + 8 = 0
- 8x = 0
x =0
C 1 º ( 4, - 1), r1 = (16 + 1 - 8) = 3
2
and for x + y 2 - 2x - 6y - 15 = 0
L
P
C 2 º (1,3), r 2 = (1 + 9 + 15) = 5
3
M
1
Q
1
(3, 0)
C1
3
C2
(–1, 0)
= ( 4 - 1) 2 + ( - 1 - 3) 2 = 5
and
R
Let the coordinates of Q be (h , k ), then
QC 2 C 2 M 1
=
=
3
QC 1
C 1L
\
QC 2 : QC 1 = 1 : 3
\
h=
1 × ( 3) - 3 × ( - 1)
= - 3 and k = 0
1- 3
\
Q º ( - 3, 0)
Equation of line passing through Q( -3, 0) is
y - 0 = m( x + 3 )
or
…(iii)
mx - y + 3m = 0
where, m is the slope of direct tangents since Eq. (iii) is the
common tangent (direct) of the circles Eqs. (i) and (ii), then
Length of perpendicular from centre of Eq. (ii) i.e. (–1, 0) to
the Eq. (iii) = radius of circle Eq. (ii)
| -m - 0 + 3m |
Þ
= 1 or 4m 2 = m 2 + 1
2
m +1
Þ
Now, | C 1C 2 | = Distance between centres
3m 2 = 1
\
m=±
1
3
From Eq. (iii), common tangents are (direct)
x
x
y=
+ 3 and y = - 3
3
3
Hence, all common tangents are
…(iv)
x =0
x
…(v)
y=
+ 3
3
x
…(vi)
and
y=- 3
3
Let P,Q,R be the point of intersection of lines Eqs.(iv), (v);
(v), (vi) and (iv), (vi) respectively, then
P º (0, 3); Q º ( -3, 0) and R º (0, - 3)
Now, PQ = QR = RP = 2 3
Hence, DPQR is an equilateral triangle thus common
tangents form an equilateral triangle.
r1 + r 2 = 3 + 5 = 8
| r1 - r 2 | = |3 - 5| = 2
or
| r1 - r 2 | <| C 1C 2 | < r1 + r 2
Hence, the two circles intersect at two distinct points.
Therefore, two tangents can be drawn.
y Example 67. Find the lengths of external and
internal common tangents and also find the angle
between external common tangents and internal
common tangents of the circles
x 2 + y 2 + 2x - 8 y + 13 = 0 and x 2 + y 2 - 8 x - 2y + 8 = 0.
Sol. The given circles are x 2 + y 2 + 2x - 8y + 13 = 0
( x + 1) 2 + ( y - 4 ) 2 = 22
Þ
and
2
x + y - 8x - 2y + 8 = 0
( x - 4 ) 2 + ( y - 1) 2 = 32
Þ
…(i)
2
…(ii)
Centres and radii of circles Eqs. (i) and (ii) are C 1( -1, 4 ),
r1 = 2 and C 2 ( 4,1), r 2 = 3 respectively.
\
| C 1C 2 | = d = (25 + 9 ) = 34
Þ
d > r1 + r 2
Hence, the circles do not intersect to each other.
\
Lex = d 2 - (r1 - r 2 )2 = 34 - 1 = 33
and
Lin = d 2 - (r1 + r 2 )2 = (34 - 25) = 3
Angle between external common tangents
æ 1 ö
æ |r - r | ö
= 2sin -1 ç 1 2 ÷ = 2sin -1 ç
÷
è d ø
è 33 ø
and angle between internal common tangent
æ 5 ö
ær + r ö
= 2sin -1 ç 1 2 ÷ = 2sin -1 ç
÷
è d ø
è 33 ø
Common Chord of Two Circles
The chord joining the points of intersection of two given
circles is called their common chord.
Theorem : The equation of common chord of two circles
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
290
Textbook of Coordinate Geometry
and
S ¢ º x 2 + y 2 + 2g ¢ x + 2 f ¢y + c ¢ = 0
is
2 x ( g - g ¢ ) + 2y ( f - f ¢ ) + c - c ¢ = 0
i.e.
S - S¢ = 0
Proof : Q S = 0 and S ¢ = 0
P
be two intersecting circles.
C2
Then, S - S ¢ = 0 or
M
C1
2 x ( g - g ¢ ) + 2y ( f - f ¢ ) + c - c ¢ = 0
Q
S´ = 0
is a first degree equation in
S=0
x and y.
So, it represent a straight line. Also, this equation satisfied
by the intersecting points of two given circles S = 0 and
S ¢ = 0.
Hence, S - S ¢ = 0 represents the common chord of circles
S = 0 and S ¢ = 0
Length of common chord :
We have,
(QM is mid-point of PQ)
PQ = 2( PM )
= 2 {(C 1 P ) 2 - (C 1 M ) 2 }
where, C 1 P = radius of the circle (S = 0)
and C 1 M = length of perpendicular from C 1 on common
chord PQ.
Corollary 1 : The common chord PQ of two circles
becomes of the maximum length when it is a diameter of
the smaller one between them.
Corollary 2 : Circle on the common chord a diameter,
then centre of the circle passing through P and Q lie on
the common chord of two circles i.e.
S - S¢ = 0
Corollary 3 : If the length of common chord is zero, then
the two circles touch each other and the common chord
becomes the common tangent to the two circles at the
common point of contact.
y Example 68. Prove that the length of the common
chord of the two circles :
(x - a) 2 + (y - b) 2 = c 2
and
2
2
2
2
or
Þ
Þ
(2x - a - b )( -a + b ) + (2y - b - a )( -b + a ) = 0
2x - a - b - 2y + b + a = 0
x -y =0
(a, b)
C1
P
M
C2 ( b, a)
Q
Now, C 1M = Length of perpendicular from C 1 (a, b ) on
|a - b |
PQ ( x - y = 0) =
2
and
C 1P = radius of the circle Eq. (i) = c
\ In DPC 1M , PM = ( PC 1 )2 - (C 1M )2
= c2 \
(a - b )2
2
(a - b )2
2
PQ = 2PM = 2 c 2 -
= 4c 2 - 2(a - b )2
Also, when the circles touch, then chord PQ becomes the
tangent and PQ = 0.
\ The condition of tangency is 4c 2 - 2(a - b )2 = 0.
i.e.
2c 2 = ( a - b ) 2
Family of Circles
1. The equation of the family of circles passing through the
point of intersection of two given circles S = 0 and S¢ = 0
is given as
S + lS ¢ = 0 (where, l is a parameter, l ¹ -1)
2
(x - b) + (y - a) = c is 4c - 2(a - b) .
Find also the condition when the given circles touch.
Sol. The equation of circles are
S1 º ( x - a )2 + (y - b )2 - c 2 = 0
and
2
2
2
S 2 º ( x - b ) + (y - a ) - c = 0
then equation of common chord is S1 - S 2 = 0
Þ
( x - a )2 - ( x - b )2 + (y - b )2 - (y - a )2 = 0
…(i)
…(ii)
S=0
S+λS′ = 0
S′ = 0
Chap 04 Circle
2. The equation of the family of circles passing through the
point of intersection of circle S = 0 and a line L = 0 is
given as
(where, l is a parameter)
S + lL = 0
and if m is infinite, the family of circles is
( x - x 1 ) 2 + (y - y 1 ) 2 + l( x - x 1 ) = 0
(where, l is a parameter)
S + λL = 0
S=0
291
(x1, y1)
y – y1 = m (x – x1)
L=0
6. Equation of the circles given in diagram are
( x - x 1 )( x - x 2 ) + (y - y 1 ) (y - y 2 )
± cot q {( x - x 1 )(y - y 2 ) - ( x - x 2 )(y - y 1 )} = 0
3. The equation of the family of circles touching the circle
S = 0 and the line L = 0 at their point of contact P is
(where, l is a parameter)
S + lL = 0
θ
(x1, y1)
S+
λL
=
0
P
(x2, y2)
S=0
θ
L=0
4. The equation of a family of circles passing through two
given points P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) can be written in
the form
y Example 69. Find the equation of the circle
passing through (1, 1) and the points of intersection
of the circles
x + y 2 + 13 x - 3 y = 0 and 2 x 2 + 2 y 2 + 4x - 7 y - 25 = 0.
2
P (x1, y1 )
Q (x2, y2 )
( x - x 1 )( x - x 2 ) + (y - y 1 )(y - y 2 )
x
+ l x1
x2
y
y1
y2
1
1 =0
1
(where, l is a parameter)
5. The equation of family of circles which touch
y - y 1 = m( x - x 1 ) at ( x 1 , y 1 ) for any finite m is
( x - x 1 ) 2 + (y - y 1 ) 2
+ l {(y - y 1 ) - m( x - x 1 )} = 0
Sol. The given circles are
…(i)
x 2 + y 2 + 13x - 3y = 0
2
and
2x + 2y 2 + 4 x - 7y - 25 = 0
7
25
or
…(ii)
=0
x 2 + y 2 + 2x - y 2
2
Equation of any circle passing through the point of
intersection of the circles Eqs. (i) and (ii) is
7
25 ö
æ
( x 2 + y 2 + 13x - 3y ) + l ç x 2 + y 2 + 2x - y - ÷ = 0 …(iii)
è
2
2ø
Its passes through (1, 1), then
7 25 ö
æ
(1 + 1 + 13 - 3) + l ç1 + 1 + 2 - - ÷ = 0
è
2 2ø
Þ
12 + l ( -12) = 0 \ l = 1
Substituting the value of l in Eq. (iii), the required
equation is
7
25
=0
x 2 + y 2 + 13x - 3y + x 2 + y 2 + 2x - y 2
2
13
25
Þ
2x 2 + 2y 2 + 15x - y =0
2
2
Þ
4 x 2 + 4y 2 + 30x - 13y - 25 = 0
292
Textbook of Coordinate Geometry
y Example 70. Find the equation of the circle passing
through the point of intersection of the circles
x 2 + y 2 - 6 x + 2y + 4 = 0, x 2 + y 2 + 2x - 4 y - 6 = 0
and with its centre on the line y = x .
y Example 72. Find the equation of the circle passing
through points of intersection of the circle
x 2 + y 2 - 2x - 4 y + 4 = 0 and the line x + 2y = 4 which
touches the line x + 2y = 0.
Sol. Equation of any circle through the points of intersection
of given circles is
( x 2 + y 2 - 6x + 2y + 4 ) + l ( x 2 + y 2 + 2x - 4y - 6) = 0
Sol. Equation of any circle through points of intersection of
the given circle and the line is
( x 2 + y 2 - 2x - 4y + 4 ) + l( x + 2y - 4 ) = 0
Þ
x 2 (1 + l ) + y 2 (1 + l ) - 2x (3 - l ) + 2y (1 - 2l )
or
+ ( 4 - 6l ) = 0
x
y
(
l
)
(
l
)
(
l
)
2
3
2
1
2
4
6
…(i)
or x 2 + y 2 =0
+
+
(1 + l )
(1 + l )
(1 + l)
ì 3 - l 2l - 1ü
Its centre í
,
ý lies on the line y = x
î1 + l 1 + l þ
2l - 1 3 - l
then
Þ l ¹ -1
=
1+ l 1+ l
x 2 + y 2 + ( l - 2)x + (2l - 4 )y + 4(1 - l ) = 0
It will touch the line x + 2y = 0, then solution of Eq. (i) and
x = - 2y be unique.
Hence, the roots of the equation
( -2y )2 + y 2 + ( l - 2)( -2y ) + (2l - 4 ) y + 4 (1 - l ) = 0
or
5y 2 + 4(1 - l ) = 0
must be equal.
Then, 0 - 4 × 5 × 4(1 - l ) = 0 or 1 - l = 0 or l = 1
From Eq. (i), the required circle is x 2 + y 2 - x - 2y = 0
\
2l - 1 = 3 - l or 3l = 4
\
l = 4 /3
\Substituting the value of l = 4 / 3 in Eq. (i), we get the
required equation is
7 x 2 + 7y 2 - 10x - 10y - 12 = 0
y Example 73. Find the circle whose diameter is the
common chord of the circles x 2 + y 2 + 2x + 3y + 1 = 0
and x 2 + y 2 + 4 x + 3y + 2 = 0.
y Example 71. Find the equation of the circle
passing through the points of intersection of the
circles x 2 + y 2 - 2x - 4 y - 4 = 0 and
Sol. Given circles are
S º x 2 + y 2 + 2x + 3y + 1 = 0
and
x 2 + y 2 - 10x - 12y + 40 = 0 and whose radius is 4.
Sol. Equation of the any circle through the points of
intersection of given circles is
( x 2 + y 2 - 2x - 4y - 4 )
+ l( x 2 + y 2 - 10x - 12y + 40) = 0
Þ
x 2 ( 1 + l ) + y 2 ( 1 + l ) - 2x ( 1 + 5l )
or
- 2y(2 + 6l ) - 4 + 40l = 0
( 1 + 5l )
(2 + 6l ) ( 40l - 4 )
2
2
- 2y
=0
x + y - 2x
+
(1 + l )
(1 + l )
(1 + l )
…(i)
S ¢ º x 2 + y 2 + 4 x + 3y + 2 = 0
Hence, their common chord is S - S ¢ = 0
…(i)
Þ
- 2x - 1 = 0 or 2x + 1 = 0
Now, the required circle must pass through the point of
intersection of S and S¢ .
Hence, its equation is
S + lS ¢ = 0
2
2
Þ
( x + y + 2x + 3y + 1)
+ l ( x 2 + y 2 + 4 x + 3y + 2) = 0
Þ
or
x 2 ( 1 + l ) + y 2 ( 1 + l ) + 2x ( 1 + 2l )
+ 3y (1 + l ) + (1 + 2l ) = 0
x 2 + y 2 + 2x
Its radius
2
2
æ 40l - 4 ö
æ 2 + 6l ö
æ 1 + 5l ö
÷ =4
÷ -ç
÷ +ç
ç
è 1+ l ø
è1+ l ø
è1+ l ø
Þ
Þ
(1 + 5l )2 + (2 + 6l )2 - ( 40l - 4 )(1 + l )
(1 + l )2
(given)
= 16
5l2 - 34 l - 7 = 0
or
( l - 7 ) ( 5l + 1) = 0
1
\
l = 7 or l = 5
Substituting the values of l in Eq. (i), the required circles are
2x 2 + 2y 2 - 18x - 22y + 69 = 0
and
x 2 + y 2 - 2y - 15 = 0
…(i)
( 1 + 2l )
( 1 + 2l )
= 0 …(ii)
+ 3y +
(1 + l )
(1 + l )
æ 1 + 2l
3ö
Its centre is ç ,- ÷
2ø
è 1+ l
But from Eq. (i), 2x + 1 = 0 is a diameter of this circle.
Hence, its centre must lie on this line
æ 1 + 2l ö
-2 ç
\
÷ +1=0
è1+ l ø
Þ
Þ
-2 - 4 l + 1 + l = 0
- 1 - 3l = 0
1
3
Hence, from Eq. (ii), the required circle is
2x 2 + 2y 2 + 2x + 6y + 1 = 0
\
l=-
Chap 04 Circle
Four concyclic points lie on a circle, then Eq. (i) represents
a circle. Then,
coefficient of x 2 = coefficient of y 2 and coefficient of
xy = 0
Þ
a + la ¢ = b + lb ¢
or
…(ii)
( a - b ) = - l( a ¢ - b ¢ )
and
2( h + h ¢ l ) = 0
h
or
…(iii)
l=h¢
Substituting the value of l from Eq. (iii) in Eq. (ii), then
a - b a¢ - b¢
=
h
h¢
y Example 74. If two curves, whose equations are
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 and
a ¢ x 2 + 2h ¢ xy + b ¢ y 2 + 2g ¢ x + 2 f ¢ y + c ¢ = 0 intersect in
a - b a¢ - b ¢
four concyclic points, prove that
=
h
h¢
Sol. The equation of family of curves passing through the
points of intersection of two curves is
(ax 2 + 2hxy + by 2 + 2gx + 2 fy + c )
+ l (a ¢ x 2 + 2h ¢ xy + b ¢ y 2 + 2g ¢ x + 2 f ¢ y + c ¢ ) = 0
or
293
(a + la ¢ ) x 2 + 2xy (h + h ¢ l ) + (b + lb ¢ )y 2
+ 2x ( g + l g ¢ ) + 2y ( f + lf ¢ ) + (c + l c ¢ ) = 0 …(i)
Exercise for Session 6
1.
Circles x 2 + y 2 - 2x - 4y = 0 and x 2 + y 2 - 8y - 4 = 0
(a) touch each other internally
(c) cuts each other at two points
2.
(b) touch each other externally
(d) None of these
The number of common tangents that can be drawn to the circles x 2 + y 2 - 4x - 6y - 3 = 0 and
x 2 + y 2 + 2x + 2y + 1 = 0 is
(a) 1
3.
(b) 2
2
2
(b) ab > 0, c < 0
(d) ab < 0, c < 0
The condition that the circle ( x - 3)2 + ( y - 4)2 = r 2 lies entirely within the circle x 2 + y 2 = R 2 is
(b) R 2 + r 2 < 49
(d) R - r > 5
(a) R + r £ 7
(c) R 2 - r 2 < 25
5.
The circles whose equations are x 2 + y 2 + c 2 = 2ax and x 2 + y 2 + c 2 - 2by = 0 will touch one another
externally, if
(a)
(c)
6.
7.
1
b2
1
a2
+
+
1
c2
1
b2
=
=
1
(b)
a2
1
(d)
c2
1
c2
1
b2
+
+
1
a2
1
c2
=
=
1
b2
2
a2
Two circles with radii r1 and r2, r1 > r2 ³ 2, touch each other externally. If q be the angle between the direct
common tangents, then
ær + r ö
(a) q = sin-1 ç 1 2 ÷
è r1 - r2 ø
ær - r ö
(b) q = 2 sin-1 ç 1 2 ÷
è r1 + r2 ø
ær - r ö
(c) q = sin-1 ç 1 2 ÷
è r1 + r2 ø
(d) None of these
The circles x 2 + y 2 - 10x + 16 = 0 and x 2 + y 2 = r 2 intersect each other in two distinct points if
(a) r < 2
8.
(d) 4
2
If one of the circles x + y + 2ax + c = 0 and x + y + 2bx + c = 0 lies within the other, then
(a) ab > 0, c > 0
(c) ab < 0, c > 0
4.
(c) 3
2
(b) r > 8
(c) 2 < r < 8
(d) 2 £ r £ 8
If the circle x 2 + y 2 + 4x + 22y + c = 0 bisects the circumference of the circle x 2 + y 2 - 2x + 8y - d = 0 , then
c + d is equal to
(a) 40
(b) 50
(c) 60
(d) 70
294
Textbook of Coordinate Geometry
9.
Two circles x 2 + y 2 = 6 and x 2 + y 2 - 6x + 8 = 0 are given. Then, the equation of the circle through their points
of intersection and the point (1, 1) is
(a) x 2 + y 2 - 6x + 4 = 0
2
(b) x 2 + y 2 - 3x + 1 = 0
2
(d) x 2 + y 2 - 2x + 1 = 0
(c) x + y - 4x + 2 = 0
10.
11.
The equation of the circle described on the common chord of the circles x 2 + y 2 + 2x = 0 and x 2 + y 2 + 2y = 0
as diameter is
(a) x 2 + y 2 + x - y = 0
(b) x 2 + y 2 - x + y = 0
(c) x 2 + y 2 - x - y = 0
(d) x 2 + y 2 + x + y = 0
The equation of the diameter of the circle 3( x 2 + y 2 ) - 2x + 6y - 9 = 0 which is perpendicular to the line
2x + 3y = 12 is
(a) 3x - 2y + 3 = 0
(c) 3x - 2y + 1 = 0
12.
If the curves ax 2 + 4xy + 2y 2 + x + y + 5 = 0 and ax 2 + 6xy + 5y 2 + 2x + 3y + 8 = 0 intersect at four concyclic
points, then the value of a is
(a) -6
13.
(b) 3x - 2y - 3 = 0
(d) 3x - 2y - 1 = 0
(b) -4
(c) 4
(d) 6
Find the equation of the circle passing through the points of intersection of x 2 + y 2 + 13x - 3y = 0 and
2x 2 + 2y 2 + 4x - 7y - 25 = 0 and the point (1, 1).
14.
Show that the common chord of the circles x 2 + y 2 - 6x - 4y + 9 = 0
and
x 2 + y 2 - 8x - 6y + 23 = 0
pass through the centre of the second circle and find its length.
15.
Prove that the circles x 2 + y 2 + 2ax + 2by = 0 and x 2 + y 2 + 2a1x + 2b1y = 0 touch each other, if ab1 = a1 b .
16.
Find the equations of common tangents to the circles x 2 + y 2 - 24x + 2y + 120 = 0 and
x 2 + y 2 + 20x - 6y - 116 = 0.
.
Session 7
Angle of Intersection of Two Circles, Radical Axis,
Radical Centre, Co-axial System of Circles,
Limiting Point, Image of the Circle by the Line
Mirror
X¢
Angle of Intersection of
Two Circles
Let the two circles
S ¢ º x 2 + y 2 + 2g 1 x + 2 f 1y + c 1 = 0
intersect each other at the points P and Q.The angle q
between two circles S = 0 and S ¢ = 0 is defined as the angle
between the tangents to the two circles at the point of
intersection.
A´
S=
B´
θ
P
0
C2
θ
cosq =
r12 + r22 - d 2
2r1 r2
Then, from Eq. (i), 0 =
A
B
Q
…(i)
r12 + r22 - d 2
2r1 r2
or
r12 + r22 - d 2 = 0 or r12 + r22 = d 2
Þ
g 2 + f 2 - c + g 12 + f 12 - c 1 = g 2 + f 2 + g 12
+ f 12 - 2 gg 1 - 2 ff 1
d
C1
(\ a = 180° - q )
Orthogonal Intersection of Circles
If the angle between the circles is 90°, i.e. q = 90°, then the
circles are said to be orthogonal circles or we say that
the circles cut each other orthogonally.
S´ = 0
r2
α
r1
æ r 2 + r22 - d 2 ö
÷÷
cos (180° - q ) = çç 1
2r1 r2
è
ø
\
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
and
or
or
2 gg 1 + 2 ff 1 = c + c 1
Remark
C 1 and C 2 are the centres of circles
S = 0 and S ¢ = 0, then
C 1 º ( -g, - f ) and C 2 º ( -g 1 , - f 1 )
and radii of circles S = 0 and S ¢ = 0 are
r1 = ( g 2 + f 2 - c ) and r2 = ( g 12 + f 12 - c 1 )
Let
d = |C 1 C 2 | = Distance between their centres
Equation of a circle cutting the three circles
x 2 + y 2 + 2gi x + 2fi y + ci = 0 ( i = 1, 2, 3) orthogonally is
x 2 + y2 x
- c1 g1
- c2 g 2
- c3 g 3
y 1
f1 -1
f2 -1
f3 -1
y Example 75. Find the angle between the circles
S : x 2 + y 2 - 4x + 6 y + 11 = 0
= ( -g + g 1 ) 2 + ( - f + f 1 ) 2
and S ¢ : x 2 + y 2 - 2 x + 8y + 13 = 0
= ( g 2 + f 2 + g 12 + f 12 - 2 gg 1 - 2 ff 1 )
Sol. Centres and radii of circles S and S¢ are
C 1(2, - 3), r1 = 2, C 2 (1, - 4 ), r 2 = 2.
æ r 2 +r 2 -d 2 ö æQa + q + 90° + 90° ö
÷÷ ç
Now, in DC1 PC 2 , cosa = çç 1 2
÷
= 360° ø
è 2r1 r2 ø è
Distance between centres, d = | C 1C 2 |
= ( 2 - 1) 2 + ( - 3 + 4 ) 2 = 2
296
Textbook of Coordinate Geometry
If angle between the circles is q, then
y Example 78. Find the equations of the two circles
which intersect the circles
x 2 + y 2 - 6 y + 1 = 0 and x 2 + y 2 - 4y + 1 = 0
r 2 + r 22 - d 2
cos q = 1
2r1r 2
cos q =
\
2+ 4 -2
2 × 2 ×2
=
1
orthogonally and touch the line 3 x + 4y + 5 = 0.
2
Sol. Let the required circle be
q = 45°
y Example 76. Show that the circles
x 2 + y 2 - 6 x + 4 y + 4 = 0 and
x 2 + y 2 + x + 4 y + 1 = 0 cut orthogonally.
and
2
x + y + 2g 1x + 2 f 1y + c 1 = 0
then,
g = - 3, f = 2, c = 4
1
and
g 1 = , f 1 = 2, c 1 = 1
2
Then, given circles cut orthogonally, if
We have,
and given circles are x 2 + y 2 - 6y + 1 = 0
…(ii)
2
2
…(iii)
x + y - 4y + 1 = 0
Since, Eq. (i) cuts Eq. (ii) and Eq. (iii) orthogonally
\
2g ´ 0 + 2 f ´ ( - 3) = c + 1
or
-6 f = c + 1
and 2g ´ 0 + 2 f ´ ( -2) = c + 1
or
-4 f = c + 1
Solving, Eqs. (iv) and (v), we get
f = 0 and c = - 1
From Eq. (i),
x 2 + y 2 + 2gx - 1 = 0
…(iv)
…(v)
…(vi)
2
2gg 1 + 2 ff 1 = c + c 1
1
2 ´ ( - 3) ´ + 2 ´ 2 ´ 2 = 4 + 1
2
Þ
-3 + 8 = 5 or 5 = 5.
Hence, the given circles cut each other orthogonally.
y Example 77. Find the equation of the circle which
cuts the circle x 2 + y 2 + 5x + 7 y - 4 = 0 orthogonally,
has its centre on the line x = 2 and passes through the
point (4, - 1) .
Sol. Let the required circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
…(i)
and
Sol. Comparing the given circles by general equation of circles
x 2 + y 2 + 2gx + 2 fy + c = 0
2
x 2 + y 2 + 2gx + 2 fy + c = 0
…(i)
Since, ( 4, - 1) lie on Eq. (i), then
…(ii)
17 + 8g - 2 f + c = 0
Centre of Eq. (i) is (- g , - f )
Since, centre lie on x = 2 then - g = 2
…(iii)
\
g = -2
From Eq. (ii), 1 - 2 f + c = 0
…(iv)
and given circle is
…(v)
x 2 + y 2 + 5x + 7y - 4 = 0
Given the circles Eqs. (i) and (v) cut each other
orthogonally,
7
5
\
2g ´ + 2 f ´ = c - 4
2
2
or
5g + 7 f = c - 4
[from Eq. (iii)]
-10 + 7 f = c - 4
or
…(vi)
-6 + 7 f - c = 0
Solving Eqs. (iv) and (vi), we get
f = 1 and c = 1
Substituting the values of g, f, c in Eq. (i), we get
x 2 + y 2 - 4 x + 2y + 1 = 0
centre and radius of Eq. (vi) are ( -g , 0) and ( g + 1),
respectively.
Since, 3x + 4y + 5 = 0 is tangent of Eq. (vi), then length of
perpendicular from ( -g , 0) to this line = radius of circle
| - 3g + 0 + 5|
or
= ( g 2 + 1)
(9 + 16)
| - 3g + 5| = 5 ( g 2 + 1)
or
( -3g + 5)2 = 25( g 2 + 1)
or
9 g 2 + 25 - 30g = 25g 2 + 25
or
16g 2 + 30g = 0
15
8
Equations of circles are from Eq. (vi),
\
g = 0 and g = -
15
x -1=0
4
x 2 + y 2 - 1 = 0 and 4 x 2 + 4y 2 - 15x - 4 = 0.
x 2 + y 2 - 1 = 0 and x 2 + y 2 -
or
y Example 79. Prove that the two circles, which
pass through (0, a) and (0, - a ) and touch the line
y = mx + c , will cut orthogonally, if c 2 = a 2 (2 + m 2 )
Sol. Let the equation of the circles be
x 2 + y 2 + 2gx + 2 fy + d = 0
…(i)
Since, these circles pass through (0, a ) and (0, - a ), then
…(ii)
a 2 + 2 fa + d = 0
and
…(iii)
a 2 - 2 fa + d = 0
2
Solving, Eq. (ii) and Eq. (iii), we get f = 0 and d = - a .
Substituting these values of f and d in Eq. (i), we obtain
…(iv)
x 2 + y 2 + 2gx - a 2 = 0
Chap 04 Circle
297
or
c = -4
Substituting the values of g , f , c in Eq. (i), then required
circle is
x 2 + y 2 - 16x - 18y - 4 = 0
Now, y = mx + c touch this circle, therefore, length of the
perpendicular from the centre = radius
| -mg - 0 + c |
= ( g 2 + a2 )
2
1+m
(c - mg )2 = (1 + m 2 )( g 2 + a 2 )
or
Radical Axis
g 2 + 2mcg + a 2 (1 + m 2 ) - c 2 = 0
Let g 1, g 2 are the roots of this equation
\
g 1g 2 = a 2 (1 + m 2 ) - c 2
…(v)
Now, the equations of the two circles represented by
Eq. (iv) are
x 2 + y 2 + 2g 1x - a 2 = 0
and
2
2
2
…(vi)
S ¢ º x + y + 2g 1 x + 2 f 1y + c 1 = 0
Þ
On squaring,
x 12 + y 12 + 2 gx 1 + 2 fy 1 + c
= x 12 + y 12 + 2 g 1 x 1 + 2 f 1 y 1 + c 1
y Example 80. Find the equation of the circle which
cuts orthogonally each of the three circles given below
:
x 2 + y 2 - 2 x + 3 y - 7 = 0, x 2 + y 2 + 5 x - 5 y + 9 = 0
and x 2 + y 2 + 7 x - 9x + 29 = 0.
Þ
2 ( g - g 1 ) x 1 + 2( f - f 1 ) y 1 + c - c 1 = 0
\ Locus of P ( x 1 , y 1 ) is
2 ( g - g 1 ) x + 2( f - f 1 )y + c - c 1 = 0
P (x1, y1)
…(i)
Since, it is orthogonal to three given circles respectively,
therefore
3
2g ´ ( - 1) + 2 f ´ = c - 7
2
or
…(ii)
- 2g + 3 f = c - 7
5
5
æ ö
2g ´ + 2 f ´ ç - ÷ = c + 9
è 2ø
2
5g - 5 f = c + 9
7
æ 9ö
2g ´ + 2 f ´ ç - ÷ = c + 29
è 2ø
2
…(iii)
or
7 g - 9 f = c + 29
Subtracting, Eq. (ii) from Eq. (iii),
7 g - 8 f = 16
and subtracting Eq. (iii) from Eq. (iv),
2g - 4 f = 20
Solving Eq. (v) and Eq. (vi), we get
g = - 8 and f = - 9
Putting the values of g and f in Eq. (iii)
-40 + 45 = c + 9
Þ
5=c +9
…(iv)
and
…(ii)
( x 12 + y 12 + 2 gx 1 + 2 fy 1 + c )
which is the required condition.
or
…(i)
= ( x 12 + y 12 + 2 g 1 x 1 + 2 f 1 y 1 + c 1 )
c 2 = a 2 (2 + m 2 )
Sol. Let the required circle be
x 2 + y 2 + 2gx + 2 fy + c = 0
2
Let P ( x 1 , y 1 ) be a point such that
| PA| = | PB|
From Eqs. (v) and (vi),
-a 2 = a 2 ( 1 + m 2 ) - c 2
or
2
and
x + y + 2g 2 x - a = 0.
g 1g 2 = - a 2
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
Consider,
These two circles will be cuts orthogonal, if
2g 1g 2 + 0 = - a 2 - a 2
or
The radical axis of two circles is the locus of a point which
moves such that the lengths of the tangents drawn from it
to the two circles are equal.
…(v)
…(vi)
B
A
C2
C1
S=0
S´ = 0
P (x1, y1)
B
A
C1
C2
which is the required equation of radical axis of the given
circles. Clearly this is a straight line.
298
Textbook of Coordinate Geometry
Some Properties of the Radical Axis
(i) The radical axis and common chord are
identical : Since, the radical axis and common chord
of two circles S = 0 and S ¢ = 0 are the same straight
line S - S ¢ = 0, they are identical. The only difference
is that the common chord exists only if the circles
intersect in two real points, while the radical axis
exists for all pair of circles irrespective of their
position.
\
m 1m 2 = - 1
Hence, C 1 C 2 and radical axis are perpendicular to
each other.
(iii) The radical axis bisects common tangents of two
circles : Let AB be the common tangent. If it meets
the radical axis LM in M, then MA and MB are two
tangents to the circles. Hence, MA = MB since lengths
of tangents are equal from any point on radical axis.
Hence, radical axis bisects the common tangent AB.
Common tangent
Radical axis
A
C1
C1
C2
C2
C1
B
L
C2
Touching circles
Non-intersecting circles
Common chord
T
T
C2
C1
M
C1
A B
C2
C1 C
2
A
Intersecting circles
If the two circles touch each other externally or
internally, then A and B coincide. In this case the
common tangent itself becomes the radical axis.
(iv) The radical axis of three circles taken in pairs
are concurrent : Let the equations of three circles be
The position of the radical axis of the two circles
geometrically is shown below:
P (x1, y1)
B
A
S 1 º x 2 + y 2 + 2g 1 x + 2 f 1y + c 1 = 0
R
2
C1
Q
C2
S= 0
S´ = 0
From Euclidian geometry
( PA ) 2 = PR × PQ = ( PB ) 2
(ii) The radical axis is perpendicular to the straight
line which joins the centres of the circles :
Consider, S º x 2 + y 2 + 2 gx + 2 fy + c = 0
and
S 1 º x 2 + y 2 + 2g 1 x + 2 f 1y + c 1 = 0
…(i)
…(ii)
Since, C 1 º ( -g, - f ) and C 2 º ( -g 1 , - f 1 ) are the
centres of the circles Eqs. (i) and (ii), then slope of
-f + f f - f1
(say)
= m1
C 1C 2 = 1
=
-g 1 + g g - g 1
Equation of the radical axis is
2( g - g 1 ) x + 2 ( f - f 1 )y + c - c 1 = 0
(g - g1 )
= m2
Slope of radical axis is f - f1
(say)
2
…(i)
S 2 º x + y + 2g 2 x + 2 f 2y + c 2 = 0
…(ii)
S 3 º x 2 + y 2 + 2g 3 x + 2 f 3y + c 3 = 0
…(iii)
The radical axis of the above three circles taken in
pairs are given by
S 1 - S 2 º 2 x ( g 1 - g 2 ) + 2y ( f 1 - f 2 )
+ c 1 - c 2 = 0 …(iv)
S 2 - S 3 º 2 x ( g 2 - g 3 ) + 2y ( f 2 - f 3 )
+ c 2 - c 3 = 0 …(v)
S 3 - S 1 º 2 x ( g 3 - g 1 ) + 2y ( f 3 - f 1 )
+ c 3 - c 1 = 0 …(vi)
Adding Eqs. (iv), (v) and (vi), we find LHS vanished
identically. Thus, the three lines are concurrent.
(v) If two circles cut a third circle orthogonally, the
radical axis of the two circles will pass through
the centre of the third circle.
OR
The locus of the centre of a circle cutting two
given circles orthogonally is the radical axis of
the two circles.
Chap 04 Circle
Let
S 1 º x 2 + y 2 + 2g 1 x + 2 f 1y + c 1 = 0
…(i)
S 2 º x 2 + y 2 + 2g 2 x + 2 f 2y + c 2 = 0
…(ii)
S 3 º x 2 + y 2 + 2g 3 x + 2 f 3y + c 3 = 0
…(iii)
Since, Eqs. (i) and (ii) both cut Eq. (iii) orthogonally
\
2g 1 g 3 + 2 f 1 f 3 = c 1 + c 3
and
2g 2 g 3 + 2 f 2 f 3 = c 2 + c 3
Subtracting, we get
…(iv)
2g 3 ( g 1 - g 2 ) + 2 f 3 ( f 1 - f 2 ) = c 1 - c 2
Now, radical axis of Eqs. (i) and (ii) is
S1 - S2 = 0
or 2 x ( g 1 - g 2 ) + 2y ( f 1 - f 2 ) + c 1 - c 2 = 0
Since, it will pass through the centre of Eq. (iii) circle
\ -2 g 3 ( g 1 - g 2 ) - 2 f 3 ( f 1 - f 2 ) + c 1 - c 2 = 0
or 2 g 3 ( g 1 - g 2 ) + 2 f 3 ( f 1 - f 2 ) = c 1 - c 2
…(v)
which is true by Eq. (iv),
Radical axis need not always pass through the mid-point of the
line joining the centres of the two circles.
y Example 81. If two circles x 2 + y 2 + 2gx + 2 fy = 0
and x 2 + y 2 + 2g ¢ x + 2 f ¢ y = 0 touch each other, then
f ¢ g = fg ¢.
Sol. If two circles touch each other, then their radical axis is
their common tangent.
\ Radical axis of two circles is
( x 2 + y 2 + 2gx + 2 fy ) - ( x 2 + y 2 + 2g ¢ x + 2 f ¢ y ) = 0
or
or
…(i)
2
If this touches the circle x + y + 2gx + 2 fy = 0, then the
length of perpendicular from its centre ( - g , - f ) to (i)
=radius ( g 2 + f 2 ) of the circle
or
( g - g ¢ )2 + ( f - f ¢ )2 = ( g 2 + f 2 ) ± ( g ¢ 2 + f ¢ 2 )
| - g(g - g ¢ ) - f ( f - f ¢ ) |
( g - g ¢ )2 + ( f - f ¢ )2
+ g ¢ 2 + f ¢ 2 - 2gg ¢ - 2 ff ¢ )
On squaring, we have
g2 + f
2
+ g ¢ 2 + f ¢ 2 - 2gg ¢ - 2 ff ¢
= g2 + f
2
+ g ¢2 + f ¢2± 2 ( g 2 + f 2 ) ( g ¢2 + f ¢2 )
or ( gg ¢ + ff ¢ ) = ± ( g 2 + f 2 )( g ¢ 2 + f ¢ 2 )
Again, on squaring both sides, we get
g 2 g ¢ 2 + f 2 f ¢ 2 + 2gg ¢ ff ¢ = g 2 g ¢ 2
+ g 2 f ¢2 + f 2g ¢2 + f 2 f ¢2
or
g 2 f ¢ 2 + f 2 g ¢ 2 - 2gg ¢ ff ¢ = 0
or
( gf ¢ - g ¢ f )2 = 0
gf ¢ - g ¢ f = 0
gf ¢ = g ¢ f
y Example 82. A and B are two fixed points and P
moves so that PA = nPB. Show that locus of P is a
circle and for different values of n all the circles have a
common radical axis.
Sol. Let A º (a , 0), B º ( -a , 0) and P º (h , k )
\
PA = (h - a )2 + k 2
PB = (h + a )2 + k 2
Since,
Þ
PA = nPB
( PA )2 = n 2 ( PB )2
{(h - a )2 + k 2 } = n 2 {(h + a )2 + k 2 }
Þ
(h 2 + k 2 - 2ah + a 2 ) = n 2 (h 2 + k 2 + 2ah + a 2 )
Þ
(1 - n 2 )h 2 + (1 - n 2 )k 2 - 2ah (1 + n 2 )
+ (1 - n 2 )a 2 = 0
= g2 + f
2
or
2
2
2
= (g + f ) { (g - g ¢ ) + ( f + f ¢ ) }
2
or ( g + f ) + ( gg ¢ + ff ¢ )2 - 2( g 2 + f 2 )( gg ¢ + ff ¢ )
2 2
2
2
2
2
2
2
= ( g + f ){( g + f ) + ( g ¢ + f ¢ ) - 2 ( gg ¢ + ff ¢ )}
h 2 + k 2 - 2ah
\ Locus of P is
On simplifying,
2gg ¢ ff ¢ = g 2 f ¢ 2 + f 2 g ¢ 2
( gf ¢ - g ¢ f )2 = 0
gf ¢ = g ¢ f
(1 + n 2 )
(1 - n 2 )
+ a2 = 0
æ1 + n2 ö
÷ 2ax + a 2 = 0
x 2 + y 2 - çç
2÷
è1 - n ø
which is a circle. For different values of n.
If two different values of n are n1 and n 2 , then circles are
or ( gg ¢ + ff ¢ )2 = ( g 2 + f 2 )( g ¢ 2 + f ¢ 2 )
or
2
= ( g 2 + f 2 ) ± ( g ¢2 + f ¢2 )
{( -( g 2 + f 2 ) + gg ¢ + ff ¢ )} 2
2
or
(g 2 + f
or
or
2x ( g - g ¢ ) + 2y ( f - f ¢ ) = 0
x ( g - g ¢ ) + y( f - f ¢ ) = 0
2
i.e.
Aliter : If two circles touch each other, then
distance between their centres = sum or difference of their
radii
or
or
Remark
299
and
æ 1 + n12 ö
2
x2 + y2 - ç
÷ 2ax + a = 0
è 1 - n12 ø
…(i)
æ 1 + n 22 ö
2
x2 + y2 - ç
÷ 2ax + a = 0
è 1 - n 22 ø
…(ii)
300
Textbook of Coordinate Geometry
\ Radical axis of Eqs. (i) and (ii) is
ìï 1 + n 22
1 + n12 ü
2ax í
ý=0
2
1 - n12 þ
ïî 1 - n 2
or
x = 0 or Y-axis.
Hence, for different values of n the circles have a common
radical axis.
y Example 83. Show that the difference of the
squares of the tangents to two coplanar circles from
any point P in the plane of the circles varies as the
perpendicular from P on their radical axis. Also,
prove that the locus of a point such that the
difference of the squares of the tangents from it to
two given circles is constant is a line parallel to their
radical axis.
Sol. Let the two circles be
S1 º x 2 + y 2 + 2g 1x + 2 f 1y + c 1 = 0
2
and
…(i)
2
…(ii)
S 2 º x + y + 2g 2 x + 2 f 2y + c 2 = 0
and let P = (h , k )
\ Radical axis of Eqs. (i) and (ii) is
2( g 1 - g 2 )x + 2( f 1 - f 2 )y + c 1 - c 2 = 0 …(iii)
Let length of tangents from P (h , k ) on Eqs. (i) and (ii) are l1
and l 2 , then
l1 = S1 = (h 2 + k 2 + 2g 1h + 2 f 1k + c 1 )
l 2 = S 2 = (h 2 + k 2 + 2g 2h + 2 f 2k + c 2 )
and
According to the question,
l12 - l 22 = 2( g 1 - g 2 )h + 2( f 1 - f 2 ) k + c 1 - c 2
…(iv)
Let p be the perpendicular distance from P (h , k ) on Eq. (iii),
| 2( g 1 - g 2 )h + 2( f 1 - f 2 )k + c 1 - c 2 |
…(v)
\
p=
4( g 1 - g 2 ) 2 + 4( f 1 - f 2 ) 2
From Eqs. (iv) and (v), we get
| l12 - l 22 |
p=
2 ( g 1 - g 2 )2 + ( f 1 - f 2 )2
or
| l12
\
| l12 - l 22 | µ p
p
l 22
|
…(i)
…(ii)
…(iii)
S1 = 0
S2 = 0
S3 =0
= 2 ( g 1 - g 2 )2 + ( f 1 - f 2 )2 = constant
Locus of P (h , k ) in Eq. (iv) is
2( g 1 - g 2 )x + 2( f 1 - f 2 )y + c 1 - c 2 = (l 12 - l 22 )
a line which is parallel to Eq. (iii).
L
O
M
N
S3 = 0
Let OL, OM and ON be radical axes of the pair sets of
circles
{S 1 = 0, S 2 = 0 }, {S 3 = 0, S 1 = 0}
and
{S 2 = 0, S 3 = 0 } respectively.
Equations of OL, OM and ON are respectively
…(iv)
S1 - S2 = 0
…(v)
S 3 - S1 = 0
…(vi)
S2 - S 3 = 0
Let the straight lines Eqs. (iv) and (v) i.e. OL and OM meet
in O. The equation of any straight line passing through O is
(S 1 - S 2 ) + l (S 3 - S 1 ) = 0
where l is any constant.
For l = 1 this equation becomes
S2 - S 3 = 0
which is, by Eq. (vi), equation of ON.
Thus, the third radical axis also passes through the point
where the Eqs. (iv) and (v) meet. In the above figure O is
the radical centre.
Properties of Radical Centre
1. Coordinates of radical centre can be found by solving
the equations
S1 = S2 = S 3 = 0
2. The radical centre of three circles described on
the sides of a triangle as diameters is the
orthocentre of the triangle :
A
F
Radical Centre
The radical axes of three circles, taken in pairs, meet in a
point, which is called their radical centre. Let the three
circles be
S2 = 0
S1 = 0
B
I
D
E
C
Chap 04 Circle
Draw perpendicular from A on BC.
\
ÐADB = ÐADC = p / 2
Therefore, the circles whose diameters are AB and
AC passes through D and A. Hence, AD is their
radical axis. Similarly, the radical axis of the circles
on AB and BC as diameters is the perpendicular line
from B on CA and radical axis of the circles on BC
and CA as diameters is the perpendicular line from C
on AB. Hence, the radial axis of three circles meet in
a point. This point I is radical centre but here radical
centre is the point of intersection of altitudes i.e.
AD, BE and CF. Hence, radical centre = orthocentre.
3. The radical centre of three given circles will be the
centre of a fourth circle which cuts all the three
circles orthogonally and the radius of the fourth circle
is the length of tangent drawn from radical centre of
the three given circles to any of these circles.
Let the fourth circle be ( x - h ) 2 + (y - k ) 2 = r 2 , where
(h, k ) is centre of this circle and r be the radius. The
centre of circle is the radical centre of the given
circles and r is the length of tangent from (h, k) to any
of the given three circles.
y Example 84. Find the radical centre of circles
x 2 + y 2 + 3x + 2y + 1 = 0 , x 2 + y 2 - x + 6 y + 5 = 0 and
x 2 + y 2 + 5x - 8 y + 15 = 0. Also, find the equation of
the circle cutting them orthogonally.
Sol. : Given circles are
S1 º x 2 + y 2 + 3x + 2y + 1 = 0
S 2 º x 2 + y 2 - x + 6y + 5 = 0
S 3 º x 2 + y 2 + 5x - 8y + 15 = 0
Equations of two radical axes are
S1 - S 2 º 4 x - 4y - 4 = 0 or x - y - 1 = 0
and S 2 - S 3 º - 6x + 14y - 10 = 0 or 3x - 7 x + 5 = 0
Solving them the radical centre is (3, 2) also, if r is the
length of the tangent drawn from the radical centre (3, 2) to
any one of the given circles, say S1, we have
Hence, (3, 2) is the centre and 27 is the radius of the circle
intersecting them orthogonally.
\ Its equation is
( x - 3)2 + (y - 2)2 = r 2 = 27 or x 2 + y 2 - 6x - 4y - 14 = 0
Aliter :
Let x 2 + y 2 + 2gx + 2 fy + c = 0 be the
equation of the circle cutting the given circles orthogonally.
\
æ3ö
2g ç ÷ + 2 f ( 1) = c + 1
è2ø
or
3g + 2 f = c + 1
æ 1ö
2g ç - ÷ + 2 f ( 3) = c + 5
è 2ø
or
and
…(ii)
-g + 6f = c + 5
æ5ö
2g ç ÷ + 2 f ( -4 ) = c + 15
è2ø
or
5g - 8 f = c + 15
Solving, Eqs. (i), (ii) and (iii), we get
g = - 3, f = - 2 and c = - 14
\ Equation of required circle is
x 2 + y 2 - 6x - 4y - 14 = 0
…(iii)
y Example 85. Find the radical centre of three
circles described on the three sides 4 x - 7 y + 10 = 0,
x + y - 5 = 0 and 7 x + 4 y - 15 = 0 of a triangle as
diameters.
Sol. Since, the radical centre of three circles described on the
sides of a triangle as diameters is the orthocentre of the
triangle.
Radical centre = orthocentre
\
Given sides are
…(i)
4 x - 7y + 10 = 0
…(ii)
x +y -5=0
…(iii)
7 x + 4y - 15 = 0
Since, lines Eqs. (i) and (iii) are perpendiculars the point of
intersection of Eqs. (i) and (iii) is (1, 2), the orthocentre of
the triangle. Hence, radical centre is (1, 2).
y Example 86. Prove that if the four points of
intersection of the circles x 2 + y 2 + ax + by + c = 0 and
x 2 + y 2 + a ¢ x + b ¢ y + c ¢ = 0 by the lines Ax + By + C = 0
and A ¢ x + B ¢ y + C ¢ = 0 respectively are concyclic, then
a - a¢ b - b ¢ c - c ¢
A
B
C =0
A¢
B¢
C¢
Sol. The given circles and given lines are
S1 º x 2 + y 2 + ax + by + c = 0
S2 º x 2 + y 2 + a¢ x + b¢y + c ¢ = 0
L1 º Ax + By + C = 0
r = S1 = 32 + 22 + 3 × 3 + 2 × 2 + 1 = 27
L2 º A ¢ x + B¢y + C ¢ = 0
S1 = 0
S2 = 0
N
P
R
S
L1 = 0 Q
L2 = 0
S=0
…(i)
301
S1 – S2 = 0
302
Textbook of Coordinate Geometry
λ(
S 1+
)=0
S 1–S 2
S2 = 0
Radical axis of S1 = 0 and S 2 = 0 is
S1 = 0
S1 - S 2 = 0
Þ
The radical axis of S1 = 0 and S = 0 is L1 = 0
and
or
Ax + By + C = 0
…(iii)
and
radical axis of S 2 = 0 and S = 0 is L 2 = 0
or
A ¢ x + B¢y + C ¢ = 0
…(iv)
Since, the radical axes of any three circles taken in pairs are
concurrent. (i.e. lines Eqs. (ii), (iii) and (iv) are concurrent).
a - a¢ b - b¢ c - c ¢
we have
A
B
C =0
A¢
B¢
C¢
Co-axial System of Circles
S +λP = 0
A system (or a family) of circles, every pair of which have
the same radical axis, are called co-axial circles.
(1) The equation of a system of co-axial circles, when the
equation of the radical axis and of one circle of the
system are
S +λP = 0
S +λP = 0
S=0
P=0
P º lx + my + n = 0
and
2
2
S º x + y + 2 gx + 2 fy + c = 0
respectively, is
S + lP = 0 (l is an arbitrary constant)
(2) The equation of a co-axial system of circles, where
the equation of any two circles of the system are
S1 + λS2 = 0
S2 = 0
S 2 º x 2 + y 2 + 2g 2 x + 2 f 2y + c 2 = 0
respectively is
( l ¹ -1)
S 1 + l(S 1 - S 2 ) = 0
or
S 2 + l1 (S 1 - S 2 ) = 0
( l1 ¹ - 1)
Other formS 1 + lS 2 = 0
( l ¹ -1)
(3) The equation of a system of co-axial circles in the
simplest form is
x 2 + y 2 + 2 gx + c = 0
where, g is variable and c, a constant.
The common radical axis is the Y-axis
(since centre on X-axis) and the equation of a system
of other co-axial circles in the simplest form is
x 2 + y 2 + 2 fy + c = 0
where, f is variable and c, a constant
(since centre on Y-axis). The common radical axis is
the X-axis.
y Example 87. Find the equation of the system of
circles co-axial with the circles
x 2 + y 2 + 4x + 2 y + 1 = 0
and x 2 + y 2 - 2 x + 6 y - 6 = 0
Also, find the equation of that particular circle
whose centre lies on the radical axis.
Sol. Given circles are
S1 º x 2 + y 2 + 4 x + 2y + 1 = 0
and
S 2 º x 2 + y 2 - 2x + 6y - 6 = 0
\ Radical axis is
S1 - S 2 = 0
i.e.
6x - 4y + 7 = 0
Now, system of co-axial circle is
S1 + l (S1 - S 2 ) = 0
Þ
Þ
S1 = 0
S1 − S2 = 0
S 1 º x 2 + y 2 + 2g 1 x + 2 f 1y + c 1 = 0
…(ii)
(a - a ¢ )x + (b - b ¢ )y + c - c ¢ = 0
S1 +λ (S1–S2) = 0
Let S1 = 0 meet L1 = 0 at two points P and Q and S 2 = 0 meet
L 2 = 0 at two points R and S.
Further P,Q, R and S are given to be concyclic. Let the circle
through them is
…(i)
x 2 + y 2 + 2gx + 2 fy + l = 0
…(i)
( x 2 + y 2 + 4 x + 2y + 1) + l(6x - 4y + 7 ) = 0
x 2 + y 2 + 2x (2 + 3l ) + 2y (1 - 2l ) + 1 + 7 l = 0 …(ii)
Its centre [ -(2 + 3l ), - (1 - 2l )] lies on Eq. (i)
\
6 ´ - ( 2 + 3l ) - 4 ´ - ( 1 - 2l ) + 7 = 0
or
-12 - 18l + 4 - 8l + 7 = 0
or
-26 l - 1 = 0
Chap 04 Circle
1
26
Substituting the value of l in Eq. (ii), the equation of circle
is
7
2ö
3ö
æ
æ
=0
x 2 + y 2 + 2x ç2 - ÷ + 2y ç1 + ÷ + 1 ø
è
ø
è
26
26
26
\
Þ
l=-
Sol. Let the equations of the circles be x 2 + y 2 + 2g i x + c = 0,
i = 1,2, 3. Since, all the three circles are fixed
g 1, g 2 and g 3 are constants.
Let P (h , k ) be any point on the first circle, so that
…(i)
h 2 + k 2 + 2g 1h + c = 0
Let PQ and PR be the tangents from P on the other two
circles
2
= S ( g 2 - g 3 ) (h 2 + k 2 + 2g 1h + c )
= (h 2 + k 2 + c ) S ( g 2 - g 3 ) + 2h S g 1( g 2 - g 3 )
= (h 2 + k 2 + c )( g 2 - g 3 + g 3 - g 1 + g 1 - g 2 )
+ 2h { g 1( g 2 - g 3 ) + g 2 ( g 3 - g 1 ) + g 3 ( g 1 - g 2 )}
= (h 2 + k 2 + c )(0) + 2h (0) = 0
PQ = (h + k + 2g 2h + c )
and
PR = (h 2 + k 2 + 2g 3h + c )
( PQ )2
( PR )2
=
Limiting Point
Limiting points of system of co-axial circles are the
centres of the point circles belonging to the family (Circles
whose radii are zero are called point circles).
1. Limiting points of the co-axial system
Let the circle is
x 2 + y 2 + 2 gx + c = 0
( g 2 - c ) , respectively. Let
h 2 + k 2 + 2g 3h + c
g2 - c = 0
[from Eq. (i)]
g 2 - g1
= constant
g 3 - g1
\
( c , 0 ) and ( - c , 0 )
y Example 89. If A, B, C be the centres of three
co-axial circles and t 1 , t 2 , t 3 be the lengths of the
tangents to them from any point, prove that
¾¾
¾¾
BC × t 12 + CA × t 22 + AB t 32 = 0
Sol. Let the equations of three circles are
x 2 + y 2 + 2g i x + c = 0, i = 1, 2, 3, .
According to the question
A º ( - g 1, 0) , B º ( - g 2 , 0), C º ( - g 3 , 0)
Let any point be P (h , k )
\
and
Clearly the above limiting points are real and
distinct, real and coincident or imaginary according
as c > , = , < 0
2. System of co-axial circles whose two
limiting point are given :
Let (a, b) and (g , d) be the two given limiting points.
Then, the corresponding point circles with zero radii
are
( x - a ) 2 + (y - b) 2 = 0
t 1 = h 2 + k 2 + 2g 1h + c
and
( x - g ) 2 + (y - d)2 = 0
t 2 = h 2 + k 2 + 2g 2h + c
or
x 2 + y 2 - 2ax - 2by + a 2 + b2 = 0
t 3 = h 2 + k 2 + 2g 3h + c
and
¾¾
The equation of co-axial system is
¾¾
( x 2 + y 2 - 2ax - 2by + a 2 + b2 )
AB = ( g 1 - g 2 )
BC = ( g 2 - g 3 )
and
g =± c
Thus, we get the two limiting points of the given
co-axial system as
because g 1, g 2 , g 3 are constants.
¾¾
¾¾
CA = ( g 3 - g 1 )
…(i)
where, g is variable and c is constant.
\ Centre and the radius of Eq. (i) are ( -g, 0 ) and
h 2 + k 2 + 2g 2h + c
-2g 1h + 2g 2h
=
-2g 1h + 2g 3h
=
which proves the result.
2
\
¾¾
¾¾
Now, BC × t 12 + CA × t 22 + AB t 32
26( x 2 + y 2 ) + 98x + 56y + 9 = 0
y Example 88. Prove that the tangents from any point
of a fixed circle of co-axial system to two other fixed
circles of the system are in a constant ratio.
\
¾¾
303
x 2 + y 2 - 2 gx - 2 dy + g 2 + d2 = 0
+l( x 2 + y 2 - 2 gx - 2 dy + g 2 + d2 ) = 0
where, l ¹ -1 is a variable parameter.
304
Textbook of Coordinate Geometry
Þ x 2 (1 + l) + y 2 (1 + l) - 2 x (a + gl)
- 2y(b + dl) + (a 2 + b2 ) + l( g 2 + d2 ) =0
x2 +y2 -
or
2( a + gl)
(b + dl)
x -2
y
(1 + l)
(1 + l)
…(i)
=
+
(1 + l) 2
( b + dl ) 2
(1 + l) 2
( a 2 + b2 ) + l( g 2 + d2 )
=0
(1 + l)
After solving, find l . Substituting value of l in Eq. (i),
we get the limiting point of co-axial system.
y Example 90. Find the coordinates of the limiting
points of the system of circles determined by the two
circles
x 2 + y 2 + 5 x + y + 4 = 0 and x 2 + y 2 + 10x - 4y - 1 = 0
Sol. The given circles are
S1 º x 2 + y 2 + 5x + y + 4 = 0
S 2 º x 2 + y 2 + 10x - 4y - 1 = 0
and
\ Equation of the co-axial system of circles is S1 + lS 2 = 0
( x 2 + y 2 + 5x + y + 4 )
or
2
2
+ l( x + y + 10x - 4y - 1) = 0
2
or
or
2
x ( 1 + l ) + y ( 1 + l ) + 5x ( 1 + 2l )
+ y (1 - 4 l ) + ( 4 - l ) = 0
(1 - 4 l )
(4 - l)
5 ( 1 + 2l )
=0
x +y +
x+
y+
(1 + l )
(1 + l )
(1 + l )
2
2
The centre of this circles is
æ - 5( 1 + 2l ) ( 1 - 4 l ) ö
,ç
÷
2 (1 + l ) ø
è 2 (1 + l )
Radius =
25 (1 + 2l )
4 (1 + l )2
2
2
+
(1 - 4 l )
2
4 (1 + l )2
-
…(i)
(4 - l)
=0
(1 + l )
2
or 25 (1 + 2l ) + (1 - 4 l ) - 4 ( 4 - l ) (1 + l ) = 0
or
or
or
25 ( 4 l2 + 4 l + 1) + (16l2 - 8l + 1) - 4 ( - l2 + 3l + 4 ) = 0
or
or
\
x 2 (1 + l ) + y 2 (1 + l ) + 3x + 4y + 25 = 0
3
4
25
x+
y+
=0
(1 + l )
(1 + l )
(1 + l )
x2 + y2 +
æ
3
2 ö
Its centre = ç ,÷
è 2(1 + l ) (1 + l ) ø
Radius of Eq. (i) can be zero for limiting point, then
9
4
25
+
=0
2
2
(1 + l )
4( 1 + l )
(1 + l )
9 + 16 - 100(1 + l ) = 0
1
3
or l = Þ
1 +l =
4
4
æ
-3
-2 ö
From Eq. (ii),
,
ç
÷
è 2(1 - 3 / 4 ) (1 - 3 / 4 ) ø
or ( -6, - 8) is the other limiting point of the system.
y Example 92. Prove that the limiting points of the
system
x 2 + y 2 + 2 gx + c + l(x 2 + y 2 + 2 fx + k ) = 0
c
k
subtend a right angle at the origin, if 2 + 2 = 2 .
g
f
Sol. The given circle is
x 2 + y 2 + 2gx + c + l( x 2 + y 2 + 2 fy + k ) = 0
or (1 + l )x 2 + (1 + l )y 2 + 2gx + 2 fyl + c + kl = 0
or x 2 + y 2 +
2g
2 fl
c + kl
=0
x+
y+
(1 + l )
(1 + l )
(1 + l )
æ -g - f l ö
Its centre is ç
,
÷
è1 + l 1 + l ø
B
90°
O (0, 0)
…(i)
…(ii)
( 6l + 1) ( 2l + 1) = 0
l=-
…(i)
…(ii)
120l2 + 80l + 10 = 0 or 12l2 + 8l + 1 = 0
1
1
and 6
2
Substituting these values of l in Eq. (i), we get the points
(–2, –1) and (0, – 3) which are the required limiting points.
i.e.
belongs to the system of co-axial circles of which one
member is
x 2 + y 2 + 3x + 4y + 25 = 0
Hence, the equation of the whole system is
( x 2 + y 2 + 3x + 4y + 25) + l( x 2 + y 2 ) = 0
For limiting point,
Radius
( a + gl) 2
Equation of circle with origin as limiting point is
( x - 0)2 + (y - 0)2 = 0 or x 2 + y 2 = 0
Sol.
( a 2 + b2 ) + l( g 2 + d2 )
+
=0
(1 + l)
æ ( a + gl) (b + dl) ö
Centre of this circle is ç
,
÷
è (1 + l) (1 + l) ø
y Example 91. If the origin be one limiting point of a
system of co-axial circles of which
x 2 + y 2 + 3x + 4 y + 25 = 0 is a member, find the
other limiting point.
A
305
Chap 04 Circle
( x - 1) 2 + ( y - 2) 2 = 0
Radius of circle Eq. (i) is =0
g2
Þ
or
(1 + l )
2
l (f
2
2
+
f 2 l2
(1 + l )
2
-
(c + kl )
=0
(1 + l )
x 2 + y 2 - 2x - 4y + 5 = 0
or
2
and
2
- k ) - l( k + c ) + g - c = 0
¼(i)
2
( x - 4 ) + ( y - 3) = 0
2
x + y 2 - 8x - 6y + 25 = 0
or
¼(ii)
which is a quadratic in l. Let roots be l1 and l 2 .
g2 - c
k +c
and l1l 2 = 2
\
l1 + l 2 = 2
f -k
f -k
Therefore, the corresponding system of co-axial circles is
( x 2 + y 2 - 2x - 4y + 5)
then limiting points are
æ - g - f l1 ö
,
Aç
÷
è 1 + l1 1 + l1 ø
It passes through origin, then
5 + 25l = 0
1
\
l=5
Substituting the value of l in Eq. (iii), the required circle is
5( x 2 + y 2 - 2x - 4y + 5)
2
2
- ( x + y - 8x - 6y + 25) = 0
+ l( x 2 + y 2 - 8x - 6y + 25) = 0
[from Eq. (ii)]
æ -g - f l2 ö
and B ç
,
÷
è1 + l2 1 + l2 ø
But given that AB subtend a right angle at the origin.
\
Þ
Slope of OA ´ Slope of OB = - 1
æ - f l1 ö æ - f l 2 ö
÷
ç
÷ ç
ç 1 + l1 ÷ ´ ç 1 + l 2 ÷ = -1
ç -g ÷ ç -g ÷
ç
÷ ç
÷
è 1 + l1 ø è 1 + l 2 ø
f l1 f l 2
´
= -1
g
g
or
or
or
f
2
(g 2 - c )
(f 2 -k)
2
+g =0
2g 2 f 2 - cf 2 - kg 2 = 0
or
2=
c
g
4 x 2 + 4y 2 - 2x - 14y = 0
or
2x 2 + 2y 2 - x - 7y = 0
Image of the Circle by
the Line Mirror
f 2 l1l 2 + g 2 = 0
or
or
2
+
k
f
2
y Example 93. Find the radical axis of co-axial system
of circles whose limiting points are (–1,2) and (2, 3).
Sol. Equations of circles with limiting points are (–1, 2) and
(2,3) are
( x + 1) 2 + ( y - 2) 2 = 0
or
x 2 + y 2 + 2x - 4y + 5 = 0
and
( x - 2) 2 + ( y - 3) 2 = 0
or
x 2 + y 2 - 4 x - 6y + 13 = 0
¼(i)
Let the circle be x 2 + y 2 + 2 gx + 2 f y +c = 0 and line mirror
is lx + my +n = 0 in this condition, radius of circle remains
unchanged but centres changes. Let the centre of imaged
circle be ( x 1 , y 1 ) .
x 1 - ( - g ) y 1 - ( - f ) - 2 ( -lg -mf +c )
Then,
=
=
l
m
(l 2 +m 2 )
we get, x 1 =
and
y1 =
(l 2 g -m 2 g + 2ml f - 2 nl )
(l 2 +m 2 )
(m 2 f - l 2 f + 2ml g - 2mn )
(l 2 +m 2 )
¼(ii)
respectively.
\ Radical axis of circles Eqs. (i) and (ii) is
n=
0
Sol. Equations of circles whose limiting points are (1,2) and
(4,3) are
C1
imaged circle
+
y Example 94. Find the equation of the circle which
passes through the origin and belongs to the co-axial
of circles whose limiting points are (1,2) and (4,3).
r
my
6x + 2y - 8 = 0
3x + y - 4 = 0
(– g, – f)
lx +
- ( x 2 + y 2 - 4 x - 6y + 13) = 0
C2
r
( x 2 + y 2 + 2x - 4y + 5)
or
or
¼(iii)
given circle
\ Required imaged circle is ( x - x 1 ) 2 +(y - y 1 ) 2 = r 2
where, r = ( g 2 + f 2 - c )
306
Textbook of Coordinate Geometry
x 1 - ( -8) y1 - 12
=
4
7
-2 ( 4( -8) + 7(12) + 13)
=
(42 + 72 )
y Example 95. Find the equation of the image of
the circle x 2 + y 2 + 16 x - 24 y + 183 = 0 by the line
mirror 4 x + 7 y + 13 = 0.
Sol. The given circle and line are
x 2 + y 2 + 16x - 24y + 183 = 0
x 1 + 8 y1 - 12
=
= -2
4
7
\ x 1 = - 16 , y1 = - 2
\ Equation of the imaged circle is ( x + 16)2 + (y + 2)2 = 52
Þ
¼(i)
and
4 x + 7y + 13 = 0
¼(ii)
Centre and radius of circle Eq. (i) are (-8,12) and 5,
respectively. Let the centre of the imaged circle be ( x 1,y1 ).
Hence, ( x 1,y1 ) be the image of the point ( -8, 12) with respect
to the line 4 x + 7y + 13 = 0, then
x 2 + y 2 + 32 x + 4y + 235 = 0
or
Exercise for Session 7
1.
The circles x 2 + y 2 + x + y = 0 and x 2 + y 2 + x - y = 0 intersect at an angle of
(a) p / 6
2.
(b) p / 4
(b) 2
2
(c) 3
2
2
3
2
(b) -2 or -
3
2
(c) 2 or
2
(d) -2 or
3
2
(b) 2ax + 2by - (a 2 + b 2 + 4) = 0
2
(d) 2ax - 2by - (a 2 + b 2 + 4) = 0
(c) 2ax - 2by + (a + b + 4) = 0
6.
3
2
If a circle passes through the point (a,b ) and cuts the circle x 2 + y 2 = 4 orthogonally, then the locus of its centre
is
(a) 2ax + 2by + (a 2 + b 2 + 4) = 0
5.
(d) 4
2
If the circles x + y + 2x + 2ky + 6 = 0 and x + y + 2ky + k = 0 intersect orthogonally, k is
(a) 2 or -
4.
(d) p / 2
If the circles of same radius a and centres at (2, 3) and (5, 6) cut orthogonally, then a equals to
(a) 1
3.
(c) p / 3
The locus of the centre of the circle which cuts orthogonally the circle x 2 + y 2 - 20x + 4 = 0 and which touches
x = 2 is
(a) x 2 = 16y
(b) x 2 = 16y + 4
(c) y 2 = 16x
(d) y 2 = 16x + 4
The equation of a circle which cuts the three circles x 2 + y 2 - 3x - 6y + 14 = 0, x 2 + y 2 - x - 4y + 8 = 0
and x 2 + y 2 + 2x - 6y + 9 = 0 orthogonally is
(a) x 2 + y 2 - 2x - 4y + 1 = 0
2
(b) x 2 + y 2 + 2x + 4y + 1 = 0
2
(d) x 2 + y 2 - 2x - 4y - 1 = 0
(c) x + y - 2x + 4y + 1 = 0
7.
The equation of radical axis of the circles x 2 + y 2 + x - y + 2 = 0 and 3x 2 + 3y 2 - 4x - 12 = 0 is
(a) 2x 2 + 2y 2 - 5x + y - 14 = 0
(b) 7x - 3y + 18 = 0
(c) 5x - y + 14 = 0
8.
The radical centre of the circles x + y = 1, x + y + 10y + 24 = 0 and x 2 + y 2 - 8x + 15 = 0 is
(a) (2, 5 / 2)
(c) (-2, - 5 / 2)
9.
(d) None of these
2
2
2
2
(b) (-2, 5 / 2)
(d) (2, - 5 / 2)
If (1,2) is a limiting point of the co-axial system of circles containing the circle x 2 + y 2 + x - 5y + 9 = 0, then the
equation of the radical axis is
(a) x - 9y + 4 = 0
(c) x + 3y - 4 = 0
(b) 3x - y + 4 = 0
(d) 9x + y - 4 = 0
Chap 04 Circle
10.
The limiting points of the system of circles represented by the equation 2( x 2 + y 2 ) + lx +
3
(a) æç ± , 0ö÷
è 2 ø
9
(c) æç ± , 0ö÷
è 2 ø
11.
12.
307
9
= 0 are
2
9
(b) (0, 0) and æç , 0ö÷
è2 ø
(d) (±3, 0)
One of the limiting points of the co-axial system of circles containing the circles x 2 + y 2 - 4 = 0 and
x 2 + y 2 - x - y = 0 is
(a) ( 2, 2 )
(b) (- 2, 2 )
(c) (- 2 , – 2 )
(d) None of these
The point (2, 3) is a limiting point of a co-axial system of circles of which x 2 + y 2 = 9 is a member. The
coordinates of the other limiting point is given by
18 27
(a) æç , ö÷
è 13 13 ø
9 6
(b) æç , ö÷
è 13 13 ø
18 27
(c) æç ,- ö÷
è 13 13 ø
18
9
(d) æç - , - ö÷
è 13 13 ø
13.
Two circles are drawn through the points (a, 5a ) and (4a, a ) to touch the Y-axis. Prove that they intersect at
æ 40 ö
angle tan-1ç ÷ .
è9ø
14.
Find the equation of the circle which cuts orthogonally the circle x 2 + y 2 - 6x + 4y - 3 = 0, passes through (3,0)
and touches the axis of y.
15.
Tangents are drawn to the circles x 2 + y 2 + 4x + 6y - 19 = 0, x 2 + y 2 = 9 from any point on the line 2x + 3y = 5.
Prove that their lengths are equal.
16.
Find the coordinates of the point from which the lengths of the tangents to the following three circles be equal
3x 2 + 3y 2 + 4x - 6y - 1 = 0, 2x 2 + 2y 2 - 3x - 2y - 4 = 0 and 2x 2 + 2y 2 - x + y - 1 = 0
17.
Find the equation of a circle which is co-axial with the circles x 2 + y 2 + 4x + 2y + 1 = 0
3
x 2 + y 2 - x + 3y - = 0 and having its centre on the radical axis of these circles.
2
18.
Find the radical axis of a co-axial system of circles whose limiting points are (1, 2) and (3 , 4).
and
Shortcuts and Important Results to Remember
1 If the lines a1 x + b1 y + c1 = 0 and
a2 x + b2 y + c 2 = 0 cut
the X-axis and Y-axis in four concyclic points, then
a1 a2 = b1 b2 .
Remark
Equation of the circle circumscribing the triangle formed
by the lines ar x + br y + c r = 0, where r = 1, 2 , 3, is :
a12 + b12
a1 x + b1 y + c1
a22 + b22
a2 x + b2 y + c 2
a32 + b32
a3 x + b3 y + c 3
2 If two conic sections
a1 x 2 + 2 h1 xy + b1 y 2 + 2 g1 x + 2 f1 y + c1 = 0 and
a2 x 2 + 2 h2 xy + b2 y 2 + 2 g 2 x + 2 f2 y + c 2 = 0 will intersect
a -b
h
each other in four concyclic points, if 1 1 = 1 .
a2 - b2 h2
3 If the circle S1 = 0, bisects the circumference of the circle
S 2 = 0, then their common chord will be the diameter of
the circle S 2 = 0.
4 The radius of the director circle of a given circle is 2
times the radius of the given circle.
b1
a2
b2 = 0
a3
b3
10 Equation of circle circumscribing a quadrilateral whose
sides in order are represented by the lines
L1 = 0, L2 = 0, L3 = 0 and L4 = 0 is given by
L1 L3 + lL2 L4 = 0
5 The point of intersection of the tangents at the points
P (acos a , asin a ) and Q (acos b, asin b ) on the circle
x 2 + y 2 = a2 is
A
L3 = 0
æ
æ a + bö ö
æ a + bö
÷
÷ asin ç
ç acos çè
è 2 ø÷
ø
2
ç
÷
,
æ a - bö
æ a - bö ÷
ç
cos ç
÷
÷
ç cos çè
è 2 ø ÷ø
è
2 ø
D
L4 = 0
B
L2 = 0
6 If the tangent to the circle x 2 + y 2 = r 2 at the point (a, b)
meets the coordinates axes at the points A and B and O is
r4
the origin, then the area of the DOAB is
.
2 ab
7 The length of the common chord of the circles
x 2 + y 2 + ax + by + c = 0 and x 2 + y 2 + bx + ay + c = 0 is
1
(a + b)2 - 4c .
2
L1 = 0
C
provided coefficient of x 2 = coefficient of y 2 and
coefficient of xy = 0.
11 The locus of the middle point of a chord of a circle
subtending a right angle at a given point will be a circle.
12 The length of an equilateral triangle inscribed in the circle
x 2 + y 2 = a2 is a 3.
8 The length of the common chord of the circles
2 ab
.
( x - a)2 + y 2 = a2 and x 2 + ( y - b)2 = b2 is
a2 + b2
9 Family of circles circumscribing
a triangle whose sides are
given by L1 = 0, L2 = 0 and
L3 = 0 is given by
L1L2 + lL2 L3 + mL3 L1 = 0
provided coefficient of xy = 0
and coefficient of x 2 =
B
coefficient of y 2 .
a1
13 The distance between the chord of contact of tangents to
x 2 + y 2 + 2 gx + 2 fy + c = 0 from the origin and the point
|g 2 + f 2 - c|
.
(g, f ) is
2 (g 2 + f 2 )
A
14 The shortest chord of a circle passing through a point P
inside the circle is the chord whose middle point is P.
L3 = 0
L1 = 0
L2 = 0
C
15 The length of transverse common tangent < the length of
direct common tangent.
JEE Type Solved Examples :
Single Option Correct Type Questions
n
(3 − 4 λ )2 − (5 + 7 λ )(1 + λ ) = 0
⇒
This section contains 10 multiple choice examples.
Each example has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
⇒
9 − 16λ 2 − 24 λ − 5 − 5λ − 7 λ − 7 λ 2 = 0
⇒
Ex. 1 Two distinct chords drawn from the point ( p, q ) on
the circle x 2 + y 2 = px + qy , where pq ≠ 0, are bisected by
the X-axis. Then,
9 λ 2 − 36λ + 4 = 0
l
(a) | p | =|q |
λ=
∴
(b) p 2 = 8q 2 (c) p 2 < 8q 2 (d) p 2 > 8q 2
Sol. (d)
M
X
(l, 0)
Q (h
, –q
p +h
λ=
2
(a) ( x − 2)( x − 8 ) + (y − 4 )(y − 16 ) = 0
(b) ( x − 4 )( x − 8 ) + (y − 2)(y − 16 ) = 0
(c) ( x − 2)( x − 16 ) + (y − 4 )(y − 8 ) = 0
(d) ( x − 6 )( x − 8 ) + (y − 5)(y − 6 ) = 0
Sol. (a) Q
or
h + q = ph − q
⇒
h 2 − ph + 2q 2 = 0
2
then
2
f (3) = f (1) f (2) = 22
. 2 = 23
Hence,
p 2 − 4 ⋅ 1 ⋅ 2q 2 > 0
∴
p 2 > 8q 2
or
Ex. 2 The values of λ for which the circle
x 2 + y 2 + 6 x + 5 + λ( x 2 + y 2 − 8 x + 7 ) = 0 dwindles into a
point are
2
3
(b) 2 ±
2 2
4 2
4 2
(c) 2 ±
(d) 1 ±
3
3
3
Sol. (c) The given circle is
x 2 + y 2 + 6x + 5 + λ( x 2 + y 2 − 8x + 7 ) = 0
or
x 2 ( 1 + λ ) + y 2 ( 1 − λ ) + ( 6 − 8 λ ) x +( 5 + 7 λ ) = 0
⇒
5 + 7λ 
 6 − 8λ 
x2 +y2 + 
 =0
x + 
 1+ λ 
 1+ λ 
This will dwindle into a point circle, then radius of the
circle = 0
αn = f (n ) = 2n ∀n ∈ N
Ex. 4 Two circles of radii a and b touching each other
externally, are inscribed in the area bounded by
1
y = (1 − x 2 ) and the X-axis. If b = , then a is equal to
2
l
1
4
1
(c)
2
(a)
1
8
1
(d)
2
(b)
Sol. (a) Let the centres of circles be C 1 and C 2 , then
C 1 ≡ ( (1 − 2a ),a )
2
5 + 7λ 
3 − 4λ 
 =0
 +0− 

 1+ λ 
 1+ λ 
f (n ) = 2n
(α1,α 2 ) ≡ (2, 4 )
and
(α 3 ,α 4 ) ≡ (8,16)
Equation of circle in diametric form is
( x − 2)( x − 8) + (y − 4 )(y − 16) = 0
l
(a) 1 ±
f (2) = f (1). f (1) = 22
Now, in Eq. (i), x = 1,y = 2,then
for two distinct chords, B 2 − 4 AC > 0
or
f ( x + y ) = f ( x ). f (y )
Q
f ( 1) = 2
In Eq. (i), Put x = y = 1,
which lie on x 2 + y 2 = px + qy
2
4 2
3
Ex. 3 If f ( x + y ) = f ( x ) ⋅ f (y ) for all x and y, f (1) = 2
and α n = f (n ), n ∈N , then the equation of the circle having
(α 1 , α 2 ) and (α 3 , α 4 ) as the ends of its one diameter is
)
Suppose chord bisect at M( λ , 0 ), then other end point of chord
is (h, − q )
where,
λ =2±
l
P (p, q)
X′
36 ± (36)2 − 4.9.4
29
.
and
C 2 ≡ ( (1 − 2b ),b )
…(i)
310
Textbook of Coordinate Geometry
Sol. (d) Let f ( x ,y ) = x 2 + y 2 + 2gx + 2 fy + c
Y
⇒
b
C2
O
(–1, 0)
C1
a
1– a a
(1, 0)
2
2

1
1 
( (1 − 2a ))2 + a −   = a + 


2
2  

2
or
1
1


1 − 2a + a −  = a + 



2
2
1 − 2a + a 2 +
or
1

Qb = 2
Ex. 7 A variable circle C has the equation
x + y 2 − 2(t 2 − 3t + 1) x − 2(t 2 + 2t )y + t = 0, where t is a
parameter. If the power of point (a , b ) w.r.t. the circle C is
constant, then the ordered pair (a , b ) is
l
2
1
1
− a = a2 + + a
4
4
1
a=
4
(c) 8
Its centre C 1:(0,0) and radius r1 = 3
Its centre C 2 :( 4,3) and radius r 2 = (25 − n 2 )
⇒
…(i)
⇒
(25 − n 2 ) > 2
⇒
25 − n 2 > 4
or
n 2 < 21
or
− 21 < n < 21
2
…(ii)
− 21 < n < 21
But n ∈ I . So, n = −4, − 3, − 2, − 1,0, 1, 2, 3, 4
Hence, number of possible values of n is 9.
l Ex. 6 Suppose f ( x , y ) = 0 is the equation of a circle such
that f ( x ,1) = 0 has equal roots (each equal to 2) and
f (1, x ) = 0 also has equal roots (each equal to zero). The
equation of circle is
(c) x 2 + y 2 + 4 x − 3 = 0 (d) x 2 + y 2 − 4 x + 3 = 0
∴
a 2 + b 2 − 2(t 2 − 3t + 1 )a − 2 (t 2 + 2t )b + t = constant
⇒
−2(a + b )t 2 + (6a − 4b + 1)t + (a 2 + b 2 − 2a ) = constant
Ex. 8 If the radii of the circles ( x − 1) 2 + (y − 2 ) 2 = 1 and
( x − 7 ) 2 + (y − 10 ) 2 = 4 are increasing uniformly w.r.t. time
as 0.3 unit/s and 0.4 unit/s respectively, then they will touch
each other at t equals to
l
From Eqs. (i) and (ii), we get
(a) x 2 + y 2 + 4 x + 3 = 0 (b) x 2 + y 2 + 4y + 3 = 0
1
 1
(d)  − , − 
 10 10
 1 1
Hence, required ordered pair is  − , 
 10 10
3 + (25 − n ) > ( 4 + 3 )
2
 1 1
(c)  − , 
 10 10
Q Power of circle is constant, then
a + b = 0 and 6a − 4b + 1 = 0
or
b = −a, then 6a + 4a + 1 = 0
1
1
∴
a = − ,b =
10
10
S 2 : x 2 + y 2 − 8x − 6y + n 2 = 0
2
 1 1
(b)  , 
 10 10
given power of circle = constant
(d) 9
Here,
25 − n 2 > 0 ⇒ −5 < n < 5
For exactly two common tangents,
r1 + r 2 > C 1C 2
1
1
(a)  ,− 
 10 10
Sol. (c) QC : x 2 + y 2 − 2(t 2 − 3t + 1)x − 2(t 2 + 2t )y + t = 0
Sol. (d) Given circles are S1: x 2 + y 2 − 9 = 0
and
…(ii)
2
Ex. 5 There are two circles whose equations are
x 2 + y 2 = 9 and x 2 + y 2 − 8 x − 6y + n 2 = 0, n ∈I . If the two
circles having exactly two common tangents, then the number
of possible values of n is
(b) 7
(given)
x 2 + y 2 − 4x + 3 = 0
l
(a) 2
(given)
…(i)
2
then,
f = 0, 2g + c = −1
From Eqs. (i) and (ii), we get
g = −2, f = 0,c = 3
Thus, equation of circle is
Now, C 1C 2 = a + b
or
f (1, x ) = 1 + x + 2g + 2 fx + c ≡ ( x − 0)
2
Also,
X
b
Y¢
⇒
g = −2, 2 f + c = 3
then,
1– b
b
X¢
f ( x ,1) = x 2 + 1 + 2gx + 2 f + c ≡ ( x − 2)2
(a) 45 s
(c) 11 s
(b) 90 s
(d) 135 s
Sol. (b) Given circles are S1 :( x − 1)2 + (y − 2)2 = 1
Its centre C 1 :(1,2) and radius r1 = 1
and
S 2 :( x − 7 )2 + (y − 10)2 = 4
Its centre C 2 :(7,10) and radius r 2 = 2
Q
C 1C 2 = 10 > r1 + r 2
Hence, the two circles are separated.
The radii of the two circles at time t are (1 + 03
. t ) and
(2 + 0.4 t )
Chap 4 Circle
For the two circles touch each other, then
C 1C 2 = |(1 + 03
. t ) ± (2 + 0.4 t )|
⇒
10 = |3 + 07
. t | or 10 = | −1 − 01
. t|
⇒
07
. t + 3 = ±10 or −1 − 01
. t = ±10
⇒
t = 10 or t = 90
or
l
(a) 4y + 3 x + 22 = 0
(c) 4y + 2x + 20 = 0
(b) 3y + 4 x + 20 = 0
(d) y + x + 6 = 0
Sol. (a) Any tangent of x 2 + y 2 = 4 is y = mx ± 2 (1 + m 2 ).
i.e.
Ex. 10 If a circle having centre at (α, β) radius r
completely lies with in two lines x + y = 2 and x + y = −2,
then, min.(| α + β + 2|,| α + β − 2| ) is
(a) greater than 2 r
(b) less than 2 r
Y
x=–2
(–2, –4)
4y + 3x + 22 = 0
l
If it passes through ( −2, − 4 ), then −4 = −2m ± 2 (1 + m 2 )
X¢
( m − 2) 2 = 1 + m 2
or
m = ∞, m = 3 / 4
Hence, the slope of the reflected ray is 3/4.
Thus, the equation of the incident ray is
3
y + 4 = − ( x + 2)
4
[Qt > 0]
Ex. 9 A light ray gets reflected from x = −2. If the
reflected ray touches the circle x 2 + y 2 = 4 and the point of
incident is ( −2, − 4 ), then the equation of the incident ray is
(c) greater than 2r
(d) less than 2r
Sol. (a) Minimum distance of the centre from line > radius of
X
O
 | α + β + 2| | α + β − 2| 
circle i.e. min. 
,
>r
2
2 

or min.{| α + β + 2|,| α + β − 2| } > 2r
Y¢
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
Ex. 11 If point P ( x , y ) is called a lattice point, if x , y ∈I .
Then, the total number of lattice points in the interior of the
circle x 2 + y 2 = a 2 , a ≠ 0 cannot be
l
(a) 202
(b) 203
311
(c) 204
(d) 205
Sol. (a, b, c) Given circle is x 2 + y 2 = a 2
…(i)
Clearly (0, 0) will belong the interior of circle Eq. (i). Also,
other points interior to circle Eq. (i) will have the coordinates
of the form
( ± λ , 0 ),( 0, ± λ ), where λ2 < a 2
Ex. 12 Let x , y be real variable satisfying
x + y 2 + 8 x − 10y − 40 = 0. Let
l
2
a = max. { ( x + 2 ) 2 + (y − 3 ) 2 } and
b = min. { ( x + 2 ) 2 + (y − 3 ) 2 }, then
(a) a + b = 18
(b) a − b = 4 2
(c) a + b = 4 2
(d) a ⋅ b = 73
Sol. (a, b, d) Given circle is
x 2 + y 2 + 8x − 10y − 40 = 0
The centre and radius of the circle are ( −4,5 ) and 9,
respectively.
Distance of the centre ( −4,5 ) from ( −2, 3 ) is
(4 + 4) = 2 2.
and ( ± λ , ± µ ) and ( ± µ, ± λ ), where λ2 + µ 2 < a 2 and λ ,µ ∈ I
∴Number of lattice points in the interior of the circle will be of
the form 1 + 4r + 8t, where r , t = 0, 1,2, …
∴Number of such points must be of the form 4n + 1, where
n = 0, 1, 2, …
Therefore,
a =2 2 +9
and
b = −2 2 + 9
∴
a + b = 18, a − b = 4 2, ab = 73
312
Textbook of Coordinate Geometry
Ex. 13 The equation of the tangents drawn from the
origin to the circle x 2 + y 2 − 2rx − 2hy + h 2 = 0, are
l
Sol. (b, c) Given circle is
( x − 4 )2 + (y − 8)2 = 20
(a) x = 0
(b) y = 0
(c) (h 2 − r 2 )x − 2rhy = 0
or
−2 ⋅ x + 0 ⋅ y − 4( x − 2 ) − 8(y + 0 ) + 60 = 0
3 x + 4y − 34 = 0
Solving Eqs. (i) and (ii), we get
or
Sol. (a, c) The given equation is ( x − r ) + (y − h ) = r
2
2
2
2
Y
r
D
C ( r, h )
p/2–2a
X
O

π
y = x tan  − 2α  = x cot 2α

2
x(1 − tan 2 α )
2 tan α
 r
x 1 − 2 
 h 
y =
r
2 
 h
5 x 2 − 28 x − 12 = 0
or
( x − 6 )(5 x + 2 ) = 0
x = 6, −
2
5
 2 44
Therefore, the points are (6, 4 ) and  − ,  .
 5 5
E
a
or
or
r
=
…(ii)
 34 − 3 x 
 34 − 3 x 
x2 + 
 + 60 = 0
 − 8 x − 16 
 4 
 4 
tangents are x = 0
and
…(i)
Equation of chord of contact from ( −2, 0 ) is
(d) (h 2 − r 2 )x + 2rhy = 0
a
x 2 + y 2 − 8 x − 16y + 60 = 0
2
Ex. 15 The equations of four circles are
( x ± a ) 2 + (y ± a ) 2 = a 2 . The radius of a circle touching all
the four circles is
l
(a) ( 2 − 1)a
(b) 2 2 a
(c) ( 2 + 1)a
(d) ( 2 + 2 )a
Sol. (a, c) Radius of inner circle = OR − a
r

Qin ∆ODC , tanα = 

h
= (a 2 + a 2 ) − a
= a( 2 − 1 )
Radius of outer circle = OR + RQ
or (h 2 − r 2 ) x − 2rhy = 0
= a 2 + a = a( 2 + 1 )
Ex. 14 Point M moved on the circle
( x − 4 ) 2 + (y − 8 ) 2 = 20. Then it broke away from it and
moving along a tangent to the circle cut the X-axis at point
( −2, 0 ). The coordinates of the point on the circle at which the
moving point broke away is
Y
l
 42 36 
(a)  , 
5 5
 2 44 
(b)  − , 
 5 5
(c) (6, 4 )
(d) (2, 4)
Q
(a, a)
R
Pa
X′
O
Y′
X
JEE Type Solved Examples :
Paragraph Based Questions
n
for fixed point −3 x + y + 4 = 0, x − y − 3 = 0
1
5
x = ,y = −
∴
2
2
 1 −5 
∴ Fixed point is  , 
2 2 
This section contains 2 solved paragraphs based upon
each of the paragraph 3 multiple choice questions have
to be answered. Each of these questions has four choices
(a), (b), (c) and (d) out of which ONLY ONE is correct.
Paragraph I
18. (c) Let S ≡ x 2 + y 2 − 6x + 4 = 0.
(Q. Nos. 16 to 18)
Consider the relation 4l 2 − 5m 2 + 6l + 1 = 0, where l , m ∈R.
∴
16. The line lx + my + 1 = 0 touches a fixed circle whose
equation is
= 4 + 9 − 12 − 4
=5 > 0
Therefore, point (2, − 3 ) lies outside the circle from which two
tangents can be drawn.
(a) x 2 + y 2 − 4 x − 5 = 0 (b) x 2 + y 2 + 6 x + 6 = 0
(c) x 2 + y 2 − 6 x + 4 = 0 (d) x 2 + y 2 + 4 x − 4 = 0
17. Tangents PA and PB are drawn to the above fixed circle
from the point P on the line x + y − 1 = 0. Then, the
chord of contact AB passes through the fixed point
 1 5
 1 4
(a)  ,−  (b)  , 
 2 2
3 3 
 1 3
(c)  − , 
 2 2
 1 5
(d)  , 
 2 2
(b) 1
(c) 2
If α- chord of a circle be that chord which subtends an angle
α at the centre of the circle.
16. (c) Let the equation of the circle be
…(i)
The line lx + my + 1 = 0 touch circle Eq. (i), then
| −lg − mf + 1|
= (g 2 + f 2 − c )
(l 2 + m 2 )
π
4
(c)
(c) x − y − 3 = 0
(d) x − y − 2 3 = 0
1
2
(b) 1
Sol.
…(iii)
19. (c) From figure
Y
(0, 1)
y=
x+
1
a
X¢
O
(1, 0)
17. (a) Let any point on the line x + y − 1 = 0 is
P ( λ ,1 − λ ), λ ∈ R.
Then, equation of AB is
⇒
λx + (1 − λ )y − 3( x + λ ) + 4 = 0
( −3 x + y + 4 ) + λ ( x − y − 3 ) = 0
(d) 2
(c) 2
Then, we get g = −3, f = 0, c = 4
Substituting these values in Eq. (i), the equation of the circle is
x + y − 6x + 4 = 0
3π
4
(b) x − y + 3 = 0
−1
f 2 − c g 2 − c −2 gf
g 2f
=
=
= =
=
4
−5
0
3
0
1
2
(d)
(a) x − y + 6 = 0
(a)
…(ii)
Comparing Eqs. (ii) and (iii), we get
2
π
2
2π
— chord of x 2 + y 2 + 2 x + 4y + 1 = 0
3
from the centre is
But the given condition is
4l 2 − 5m 2 + 6l + 1 = 0
(b)
21. Distance of
(lg + mf − 1 ) 2 = (l 2 + m 2 )( g 2 + f 2 − c )
or ( f 2 − c )l 2 + ( g 2 − c )m 2 − 2 gflm + 2 gl +2 fm − 1 = 0
π
6
π
20. If slope of a -chord of x 2 + y 2 = 4 is 1, then its
3
equation is
Sol.
⇒
(Q. Nos. 19 to 21)
(a)
(d) 1 or 2
x 2 + y 2 + 2 gx + 2 fy + c = 0
Paragraph II
19. If x + y =1 is α-chord of x 2 + y 2 = 1, then α is equal to
18. The number of tangents which can be drawn from
the point ( 2, − 3 ) are
(a) 0
S1 = (2 ) 2 + ( −3 ) 2 − 6(2 ) + 4
Y¢
π
α=
2
X
314
Textbook of Coordinate Geometry
λ=± 6
⇒
20. (a) Q Slope of chord is 1.
Let the equation of chord be x − y + λ = 0.
 π
OM = 2 cos  = 3
Q
 6
Hence, equation of chords are
x − y ± 6 = 0.
21. (b) From figure,
B
B
M
p/6
A
2
2
2
p/6
O (0, 0)
| 0 − 0 + λ|
= 3
2
∴
M
p/3
p/3
O
2
(–1, –2)
A
 π
OM = 2 cos  = 1
 3
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
From Eqs. (i) and (ii), we get
This section contains 2 examples. The answer to each
example is a single digit integer, ranging from 0 to 9
(both inclusive).
l Ex. 22 A circle with centre in the first quadrant is tangent
to y = x +10, y = x − 6 and the Y-axis. Let ( p, q ) be the centre
of the circle. If the value of ( p + q ) = a + b a , when a , b ∈Q ,
then the value of | a − b| is
CP = CR
Sol. (6) Q
| p − q + 10|
=p
2
⇒
p − q + 10 = p 2
or
and
…(i)
CP = CQ
p − q + 10
 p − q − 6
= −
 or p − q = −2

2
2 
Y
P
p
R
X¢
y=x+
…(ii)
10
C (p, q)
X
O
6
Q
Y¢
y=x–
p = 4 2 and q = 4 2 + 2
Now,
p+q =2+8 2 =a +b 2
∴
a = 2, b = 8
(given)
Hence, | a − b| = |2 − 8| = 6
Ex. 23 If the circles x 2 + y 2 + (3 + sin θ ) x + 2 cos φy = 0
and x 2 + y 2 + ( 2 cos φ ) x + 2 λy = 0 touch each other, then
the maximum value of λ is
l
Sol. (1) Since, both the circles are passing through the origin
(0, 0), the equation of tangent at (0,0) of first circle will be
same as that of the tangent at (0, 0) of second circle.
Equation of tangent at (0, 0) of first circle is
(3 + sin θ ) x + (2 cos φ )y = 0
Equaton of tangent at (0, 0) of second circle is
…(i)
(2 cos φ ) x + 2 λy = 0
Therefore, Eqs. (i) and (ii) must be identical, then
3 + sin θ 2 cos φ
=
2 cos φ
2λ
…(ii)
2 cos2 φ
(3 + sin θ )
or
λ=
or
λ max = 1
(when sinθ = −1 and cos φ = 1)
315
Chap 4 Circle
JEE Type Solved Examples :
Matching Type Questions
n
(C)QC1 and C 2 intersect, the common chord is
This section contains 2 examples. Examples 24 and 25
have four statements (A, B, C and D) given in Column I
and four statements (p, q, r and s) in Column II. Any
given statement in Column I can have correct matching
with one or more statement(s) given in Column II.
2(b − a )( x + y ) = b 2 − a 2
given common chord is longest, then passes through (a, a )
Ex. 24. Consider the circles C 1 of radius a and C 2 of
radius b, b > a both lying the first quadrant and touching the
coordinate axes.
l
Column I
(p) λ + µ is a prime
number
(B) C 1 and C 2 cut orthogonally and
b
= λ + µ, λ ∈prime number and
a
µ ∈whole number, then
(q) λ + µ is a composite
number
(C) C 1 and C 2 intersect so that the
common chord is longest and
b
= λ + µ, λ ∈prime number and
a
µ ∈whole number, then
(r) 2λ + µ is a perfect
number
(D) C 2 passes through the centre of C 1
b
and = λ + µ, λ ∈prime number
a
and µ ∈whole number, then
(s) | λ − µ | is a prime
number
or
(b − 3a )(b − a ) = 0
Q
∴
b −a ≠ 0
b − 3a = 0
[b > a ]
b
= 3 ⇒ λ = 3, µ = 0
a
or
(D)QC 2 passes through (a, a ), then a 2 + a 2 − 2ab − 2ab + b 2 = 0
or b 2 − 4ab + 2a 2 = 0
2
b
b
  − 4  + 2 = 0
a
a
or
l
or
b 4 ± (16 − 8 )
=
=2+ 2
a
2
⇒
λ = 2, µ = 2
Ex. 25. Match the following
Column I
Column II
2
(p)
(A) The circles x + y + 2x + c = 0
(c > 0) and x 2 + y2 + 2 y + c = 0 touch
each other, then the value of 2c is
1
(B) The circles x 2 + y2 + 2x + 3 y + c = 0 (q)
(c > 0) and x 2 + y2 − x + 2 y + c = 0
intersect orthogonally, then the
value of 2c is
2
(C) The circle x 2 + y2 = 9 is contains the (r)
circle x 2 + y2 − 2x + 1 − c2 = 0 (c > 0),
then 2c can be
2
2
(D) The circle x + y = 9 is contains in (s)
c2
the circle x 2 + y2 − 2x + 1 − = 0
4
(c > 0), then (c − 6) can be
3
2
Sol. (A)→ (p, s); (B) → (p); (C) → (p, r, s); (D) → (q, r)
C1 : x 2 + y 2 − 2ax − 2ay + a 2 = 0
Centre : (a, a ) and radius : a
C 2 : x 2 + y 2 − 2bx − 2by + b 2 = 0
and
2(b − a )(2a ) = b 2 − a 2
Column II
(A) C 1 and C 2 touch each other and
b
= λ + µ, λ ∈ prime number and
a
µ ∈whole number, then
Q
⇒
Centre : (b, b ) and radius : b
(A)QC1 and C 2 touch each other, then
b
2(b − a ) = b + a ⇒ = ( 2 + 1 ) 2 = 3 + 8
a
4
⇒
λ = 3, µ = 8
(B)QC1 and C 2 intersect orthogonally, then
Sol. (A) → (p); (B) → (q); (C) → (p, q, r); (D) → (r, s)
2(b − a ) 2 = b 2 + a 2
S1 :( x + 1 ) 2 + y 2 = ( (1 − c ) ) 2
a 2 + b 2 − 4ab = 0
Centre C1 :( −1, 0 ), radius r1 : (1 − c )
⇒
2
or
b
b
  − 4  + 1 = 0
a
a
∴
b 4 ± (16 − 4 )
=
=2+ 3
a
2
⇒
λ = 2, µ = 3]
(A) The circles
and
S 2 : x 2 + (y + 1 ) 2 = ( (1 − c ) ) 2
Centre C 2 :( 0, − 1 ), radius : r2 = (1 − c )
Now,
C1C 2 = 2 and r1 = r2
∴The circles will touch externally only and C1C 2 = r1 + r2
⇒
2 = 2 (1 − c ) or 2c = 1
316
Textbook of Coordinate Geometry
2
  13  
3

(B) The circles S1:( x + 1 ) 2 + y +  =   − c 



2
 4
2
Centre C 2 :(1, 0 ), radius r2 :c
Now, S 2 will be contained in S1, then
3

 13 
Centre C1 :  −1, −  , radius r1 :  − c


4

2
and
2
 5 
1

S 2 :  x −  + (y + 1 ) 2 =   − c 


2
 4 
S 2 :( x − 1 ) 2 + y 2 = c 2
and
C1C 2 < r1 − r2
or
1 < 3 − c or c < 2 ⇒2c < 4
(D) The circles
2
S1 : x 2 + y 2 = 9
1

5 
Centre C 2 :  , − 1 , radius r2 :  − c
2

4 
Centre C1 :( 0, 0 ), radius r1 :3 and
c
S 2 :( x − 1 ) 2 + y 2 =  
 2
For intersect orthogonally
(C1C 2 ) 2 = r12 + r22
2
⇒
Centre C 2 :(1, 0 ), radius r2 :
2
13
5
 3
 1
  +   = −c + −c
 2
 2
4
4
c
2
Now, S1 will be contained in S 2,
r2 − r1 > C1C 2
c
− 3 > 1 or c > 8
2
then,
2c = 2
or
2
⇒
(C) The circles
S1 : x 2 + y 2 = 3 2
∴
Centre C1 :( 0, 0 ), radius r1 :3
(c − 6 ) > 2
JEE Type Solved Examples :
Statement I and II Type Questions
n
r=
Ex. 26 C 1 is a circle of radius 2 touching X-axis and
Y -axis. C 2 is another circle of radius greater than 2 and
touching the axes as well as the circle C 1 .
Statement I Radius of Circle C 2 = 2 ( 2 + 1)( 2 + 2)
l
Statement II Centres of both circles always lie on the line
y = x.
Sol. (c) C 1 :( x − 2) + (y − 2) = 2
2
2
2
C 2 :( x − r ) 2 + (y − r ) 2 = r 2
According to question,
(r − 2 ) 2 + (r − 2 ) 2 = r + 2
(r − 2 ) 2 + (r − 2 ) 2 = (r + 2 ) 2
(r > 2 )
12 ± (144 − 16 )
2
=6± 4 2
Each of these examples also has four alternative choices only
one of which is the correct answer. You have to select the
correct choice as given below :
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Y
r 2 − 12r + 4 = 0
Directions (Ex. Nos. 26 and 27) are Assertion-Reason
Type examples. Each of these examples contains two
statements :
Statement I (Assertion) and Statement II (Reason)
(r, r)
∴ r =6+ 4 2
= 2( 2 + 1 )
[Qr > 2 ]
2
(2, 2)
X¢
= 2( 2 + 1 )(2 + 2 )
C2
r
2
C1
O
X
Y¢
∴Statement I is true and Statement II is always not true
(where circles in II or IV quadrants)
Ex. 27 From the point P( 2 , 6 ) tangents PA and PB are
drawn to the circle x 2 + y 2 = 4
l
Statement I Area of the quadrilateral OAPB (O being
origin) is 4.
Statement II Tangents PA and PB are perpendicular to each
other and therefore quadrilateral OAPB is a square.
Sol. (a) Clearly, P( 2, 6 ) lies on x 2 + y 2 = 8, which is the
director circle of x 2 + y 2 = 4.
Therefore, tangents PA and PB are perpendicular to each other.
So, OAPB is a square.
Hence, area of OAPB = ( S1 ) 2 = S1
= ( 2 )2 + ( 6 )2 − 4 = 4
∴Both statements are true and statement II is correct
explanation of statement I.
Chap 04 Circle
317
Subjective Type Examples
n
In this section, there are 16 subjective solved examples.
and let
∴ Equation of chord AB is T = S1
a
b
hx + 0 − ( x + h ) − (y + 0) = h 2 + 0 − ah − 0
2
4
Since, its passes through (a, b / 2) we have
Ex. 28 Find the equation of a circle having the lines
2
x + 2 xy + 3 x + 6y = 0 as its normals and having size just
sufficient to contain the circle
l
x ( x − 4 ) + y (y − 3 ) = 0.
Sol. Given pair of normals is
ah −
x + 2xy + 3x + 6y = 0
2
or
( x + 2y ) ( x + 3) = 0
∴ Normals are x + 2y = 0 and x + 3 = 0 the point of
intersection of normals x + 2y = 0 and x + 3 = 0 is the
centre of required circle, we get centre C 1 ≡ ( − 3, 3 /2) and
other circle is
x 2 + y 2 − 4 x − 3y = 0
…(i)
9
5
=
4
2
its centre C 2 ≡ (2, 3 / 2) and radius r = 4 +
∴
⇒ radius of the required circle = | C 1 − C 2 | + r
2
5 15
=5+ =
2 2
Hence, equation of required circle is
2
x 2 + y 2 + 6x − 3y − 54 = 0
Ex. 29 Let a circle be given by
( a ≠ 0, b ≠ 0)
2 x ( x − a ) + y (2y − b ) = 0
Find the condition on a and b if two chords, each bisected by
the X-axis, can be drawn to the circle from (a, b/2).
or
λ2 +
2x ( x − a ) + y (2y − b ) = 0
x + y − ax − by / 2 = 0
2
Let AB be the chord which is bisected by X -axis at a point
M. Let its coordinates be M (h , 0)
b2
b2
−aλ +
=0
4
4
λ2 − a λ +
or
5
 3 3
= ( − 3 − 2) 2 +  −  +
 2 2
2
 15
( x + 3) 2 + ( y − 3 / 2) 2 =  
 2
a 2 > 2b 2 .
Q Mid-point of ordinates of A and B is origin.
∴ B ( λ , b / 2) lies on Eq. (i)
Since, the required circle just contains the given circle(i),
the given circle should touch the required circle internally
from inside.
2
2
 a2 b2 
 − 3a 
 − 4 ⋅1 ⋅  +  > 0

 2 
8
2
Aliter : Given circle is
2x ( x − a ) + y (2y − b ) = 0
by
…(i)
or
=0
x 2 + y 2 − ax −
2
Let chords bisected at M (h , 0) but given chords can be
 b
drawn A a ,  then chord cut the circle at B ( λ , − b / 2)
 2
x ( x − 4 ) + y ( y − 3) = 0
Sol. The given circle is
B 2 − 4 AC > 0
⇒
C2
l
h2 −
∴
C1
or
a
b2
(a + h ) −
= h 2 − ah
2
8
3 ah a 2 b 2
+
+
=0
2
2
8
Now, there are two chords bisected by the X-axis, so there
must be two distinct real roots of h.
⇒
⇒
or
S ≡ x 2 + y 2 − ax − by / 2 = 0
b2
=0
2
Q
λ is real
∴
B 2 − 4 AC > 0 or a 2 − 4 ⋅
b2
> 0 or a 2 > 2b 2
2
Ex. 30 Let C 1 and C 2 be two circles with C 2 lying inside
C 1 . A circle C lying inside C 1 touches C 1 internally and C 2
externally. Identify the locus of the centre of C.
l
Sol. Let the given circles C 1and C 2 have centres O1 and O 2
with radii r1 and r 2 , respectively. Let centre of circle C is
at O radius is r.
Q
OO 2 = r + r 2
OO1 = r1 − r
⇒
OO1 + OO 2 = r1 + r 2
which is greater than O1 O 2 as O1O 2 < r1 + r 2 .
∴ Locus of O is an ellipse with foci O1 and O 2 .
318
Textbook of Coordinate Geometry
∴
O2
r
r O r
O1
A
Aliter :
Let O1 ≡ (0, 0), O 2 ≡ (a, b ) and O ≡ (h , k )
∴
C 1 : x 2 + y 2 = r12
Substituting the values of a and b in Eq. (ii), we get
 1
1
( x 2 + y 2 ) 2  2 + 2  = 4r 2
x
y 
C : ( x − h )2 + (y − k )2 = r 2
⇒
OO 2 = r + r 2
⇒
(h − a ) + (k − b )2 = r + r 2
2
…(i)
OO1 = r1 − r
⇒
(h 2 + k 2 ) = r1 − r
…(ii)
On adding Eqs. (i) and (ii) we get
or
( x 2 + y 2 ) 2 ( x −2 + y −2 ) = 4r 2
which is the required locus.
Aliter :
Q AB is the diameter of circle. If ∠OAB = α, then
OA = 2r cos α, OB = 2r sin α
(h − a )2 + (k − b )2 + (h 2 + k 2 ) = r1 + r 2
B
∴ Locus of O is ( x − a ) + (y − b ) + ( x + y ) = r1 + r 2
2
2
2
2
(x 2
which represents an ellipse with foci are at (a,b ) and (0, 0).
l Ex. 31 A circle of constant radius r passes through the
origin O, and cuts the axes at A and B. Show that the locus
of the foot of the perpendicular from O to AB is
( x 2 + y 2 ) 2 ( x −2 + y −2 ) = 4r 2
M
X´
)
X
A (a, 0)
O
O
y
A
x
y
+
=1
2r cos α 2r sin α
⇒
(
+y2
)
Equation of AB is
Y
(0, b) B
x
α
Sol. Let the coordinates of A and B are (a, 0) and (0, b).
C a, b
2 2
…(iii)
x2 + y2
x2 + y2
and b =
x
y
a=
C 2 : ( x − a )2 + (y − b )2 = r 22
and
…(ii)
Equation of OM which is ⊥ to AB is
ax − by = λ
It passes through (0, 0)
∴
0= λ
∴ Equation of OM is
ax − by = 0
On solving Eq. (i) and Eq. (iii), we get
r2
r2
a 2 + b 2 = 4r 2
x
y
+
= 2r
cos α sin α
…(i)
and equation of OM is y = x tan (90°− α)
y
cot α =
⇒
x
Y´
Y
x y
…(i)
+ =1
∴ Equation of AB is
a b
Centre of circle lie on line AB, since AB is diameter of the
circle (Q ∠ AOB = π / 2)
a b
∴ Coordinate of centre C is C ≡  , 
 2 2
Since, the radius of circle = r
∴
r = AC = CB = OC
2
a


= 0 −  + 0 −


2
(0, b) B
X´
90°– α
α
O
Y´
∴
sin α =
and
cos α =
2
b
a2 + b2
 =
2
4
M
x
(x + y 2 )
2
y
(x + y 2 )
2
X
A (a, 0)
Chap 04 Circle
Then, from Eq. (i),
y
x
(x 2 + y 2 ) +
x
y
a b
Let centre be (h , k ) ≡  ,  then a = 2h and b = 2k .
 2 2
( x + y ) = 2r
2
2
Substituting the values of a and b in Eq. (i), then
(x + y ) (x + y )
= 2r
xy
2
⇒
2
On squaring, we have ( x 2 + y 2 )2
2
2
(x 2 + y 2 )
2
x y
2
2
2
1 
 1
+
− 1 = −2  2 +

 4h
2h 2k
4k 2 
⇒
= 4r 2
( x 2 + y 2 ) 2 ( x −2 + y −2 ) = 4r 2
⇒
Ex. 32 The circle x 2 + y 2 − 4 x − 4y + 4 = 0 is inscribed
in a triangle which has two of its sides along the coordinate
axes. The locus of the circumcentre of the triangle is
x + y − xy + k ( x 2 + y 2 ) 1/ 2 = 0. Find k.
Sol. The given circle is x 2 + y 2 − 4 x − 4y + 4 = 0. This can
be re-written as ( x − 2)2 + (y − 2)2 = 4 which has centre
C (2, 2) and radius 2.
Let the equation of third side is
x y
+ =1
a b
∴ Locus of M (h , k ) is
x + y − xy + ( x 2 + y 2 ) = 0
Hence, the required value of k is 1.
Ex. 33 P is a variable on the line y = 4. Tangents are
drawn to the circle x 2 + y 2 = 4 from P to touch it at A and
B. The parallelogram PAQB is completed . Find the equation
of the locus of Q.
l
Sol. Let P (h , 4 ) be a variable point. Given circle is
(equation of AB)
x2 + y2 = 4
…(i)
Draw tangents from P (h , 4 ) and complete parallelogram
PAQB.
(0, b)
Equation of the diagonal AB which is chord of contact of
…(ii)
x 2 + y 2 = 4 is hx + 4y = 4
M
2
1 1
1
 1
+ − 1 = −  2 + 2
h
h k
k 
h + k − hk + (h 2 + k 2 ) = 0
or
l
B
319
2
Y
C (2, 2)
2
(0, 4)
P (h, 4)
O
A
(a, 0)
Length of perpendicular from (2, 2) on AB = radius = CM
2 + 2 − 1
a b 
=2
∴
1
1
 2 + 2
a
b 
or

2 2
 + − 1

a b
=2
−
1
1
 2 + 2
a
b 
2 2
1
1
+ − 1 = −2  2 + 2
a
a b
b 
…( i )
π
2
Hence, AB is the diameter of the circle passing through
a b
∆OAB, mid-point of AB is the centre of the circle i.e.  ,  .
 2 2
Since,
∠ AOB =
X¢
O
Q (a, b)
Since, origin and (2, 2) lie on the same side of AB
∴
(x1, y1) A
X
B (x2, y2)
Y¢
Let coordinates of A and B are ( x 1, y1 ) and ( x 2 , y 2 ),
respectively.
Since, A ( x 1, y1 ) and B ( x 2 , y 2 ) lies on Eq. (ii)
∴
hx 1 + 4y1 = 4 and hx 2 + 4y 2 = 4
∴
h ( x 1 + x 2 ) + 4 (y 1 + y 2 ) = 8
Since, PAQB is parallelogram
Mid-point of AB = Mid-point of PQ
∴
x1 + x 2 α + h
⇒
=
2
2
y1 + y 2 β + 4
and
=
2
2
…(iii)
…(iv)
320
Textbook of Coordinate Geometry
Eliminating x from Eqs. (i) and (ii), then
=
2
 4 − 4y 
2
 +y = 4

 h 
⇒
⇒
∴
16 + 16y 2 − 32y + h 2y 2 = 4h 2
(16 + h 2 ) y 2 − 32y + 16 − 4h 2 = 0
y1 + y 2 =
32
16 + h 2
From Eqs. (iii) and (v), we get
8h
x1 + x 2 =
16 + h 2
…(v)
∴
λ=
and
µ=
or
…(vi)
ab +
⇒
32
(b 2 − ac ) (c 2 + a 2 )
(16 + h ) (β + 4 ) = 32
…(vii)
⇒
bc (a 2 − bc ) (b 2 + c 2 )
(c − ab ) (a + b )
2
2
2
+
8h
ca (b 2 − ac ) (c 2 + a 2 )
(c 2 − ab ) (a 2 + b 2 )
=0
(c 2 − ab ) (a 2 + b 2 ) ab + (a 2 − bc )(b 2 + c 2 )bc
abc 2 (a 2 + b 2 ) + a 2bc (b 2 + c 2 ) + b 2ca (c 2 + a 2 )
+ c 2a 2 (c 2 + a 2 )
⇒
…(viii)
abc {c (a 2 + b 2 ) + a (b 2 + c 2 ) + b (c 2 + a 2 )}
= a 2b 2 (a 2 + b 2 ) + b 2c 2 (b 2 + c 2 )
Dividing Eq. (viii) by Eq. (vii), then
4α
α +h h
or h =
=
β
β+4 4
+ c 2a 2 (c 2 + a 2 )
⇒
abc {(a + b ) (b + c ) (c + a ) − 2abc }
= a 2b 2 (a 2 + b 2 ) + b 2c 2 (b 2 + c 2 )
Substituting the value of h in Eq. (vii) then
2

16α
16 + 2  (β + 4 ) = 32
β 

⇒
…(v)
= a 2b 2 (a 2 + b 2 ) + b 2c 2 (b 2 + c 2 )
16 + h 2
(16 + h 2 ) (α + h ) = 8h
…(iv)
(c 2 − ab ) (a 2 + b 2 )
+ (b 2 − ca ) (c 2 + a 2 ) ca = 0
From Eqs. (iv) and (vi)
or
(c 2 − ab ) (a 2 + b 2 )
16 + h 2
2
α +h =
(a 2 − bc ) (b 2 + c 2 )
and given, Eq. (i) passes through the origin then
ab + bcλ + ca µ = 0
From Eqs. (iv) and (v), we get
From Eqs. (iv) and (vi)
β+4=
µ
(b 2 − ac ) (c 2 + a 2 )
+ c 2a 2 (c 2 + a 2 )
⇒
( α 2 + β 2 ) ( β + 4 ) = 2β 2
abc (a + b ) (b + c ) (c + a )
= 2a 2b 2c 2 + a 2b 2 (a 2 + b 2 ) + b 2c 2 (b 2 + c 2 )
Hence, locus of Q (α, β ) is ( x 2 + y 2 ) (y + 4 ) = 2y 2
+ c 2a 2 (c 2 + a 2 )
⇒
Ex. 34 Show that the circumcircle of the triangle formed
by the lines ax + by + c = 0; bx + cy + a = 0 and
cx + ay + b = 0 passes through the origin if
(b 2 + c 2 ) (c 2 + a 2 ) (a 2 + b 2 ) = abc (b + c ) (c + a ) (a + b ) .
l
abc (a + b ) (b + c ) (c + a )
= (a 2 + b 2 ) (b 2 + c 2 ) (c 2 + a 2 )
Hence, (a 2 + b 2 ) (b 2 + c 2 ) (c 2 + a 2 )
= abc (a + b ) (b + c ) (c + a )
Sol. Equation of conic is
(bx + cy + a ) (cx + ay + b ) + λ (cx + ay + b )(ax + by + c )
+ µ (ax + by + c ) (bx + cy + a ) = 0 …(i)
where, λ and µ are constants.
Eq. (i) represents a circle if the coefficient of x 2 and y 2 are
equal and the coefficient of xy is zero such that
bc + λca + µab = ca + λab + µbc
or
…(ii)
(a − b ) c + λ (b − c ) a + µ (c − a ) b = 0
and
(c 2 + ab ) + λ (a 2 + bc ) + µ (b 2 + ac ) = 0 …(iii)
on solving Eq. (ii) and Eq. (iii) by cross multiplication rule,
we get
1
λ
= 2
2
2
2
(c − ab ) (a + b ) (a − bc ) (b 2 + c 2 )
Ex. 35 If four points P , Q , R, S in the plane be taken and
the square of the length of the tangents from P to the circle
on QR as diameter be denoted by {P , QR }, show that
l
{ P , RS } − { P , QS } + {Q, PR } − {Q, RS } = 0
Sol. Let P ≡ ( x 1, y1 ), Q ≡ ( x 2 , y 2 ), R ≡ ( x 3 , y 3 ) and S ≡ ( x 4 , y 4 ).
Equation of circle with RS as diameter is
( x − x 3 ) ( x − x 4 ) + (y − y 3 ) (y − y 4 ) = 0
∴
{ P , RS } = ( x 1 − x 3 )( x 1 − x 4 ) + (y1 − y 3 )(y1 − y 4 )
Now, equation of circle with QS as diameter is
( x − x 2 ) ( x − x 4 ) + (y − y 2 ) (y − y 4 ) = 0
∴
{ P , QS } = ( x 1 − x 2 )( x 1 − x 4 ) + (y1 − y 2 )(y1 − y 4 )
Chap 04 Circle
Equations of common tangents to circle (i) and circle C are
1
x = − 1, y = ±
( x + 2), {T1 and T 2 }
3
Equation of circle with PR as diameter is
( x − x 1 ) ( x − x 3 ) + (y − y 1 ) (y − y 3 ) = 0
∴
{Q , PR } = ( x 2 − x 1 )( x 2 − x 3 ) + (y 2 − y1 )(y 2 − y 3 )
Equation of circle with RS as diameter is
( x − x 3 ) ( x − x 4 ) + (y − y 3 ) (y − y 4 ) = 0
∴
{Q , RS } = ( x 2 − x 3 )( x 2 − x 4 ) + (y 2 − y 3 )(y 2 − y 4 )
Hence, { P , RS } − { P , QS } + {Q , PR } − {Q , RS } = 0
and equations of common tangents to circle (ii) and circle C
are
1
x = 1, y = ±
( x + 2) ({T1 and T 2 }
3
To find the remaining two transverse common tangents to
Eqs. (i) and (ii). If I divides C 1 and C 2 in the ratio
r1 :`r 2 = 1 / 3 : 3 = 1 : 9.
Therefore coordinates of I are ( −4 / 5, 0).
Equation of any line through I is y − 0 = m ( x + 4 / 5). If it
will touch Eq. (ii)
| m ( 4 + 4 / 5) − 0 |
then
=3
(1 + m 2 )
Ex. 36 Let T1 , T 2 be two tangents drawn from ( −2 , 0 ) on
the circle C : x 2 + y 2 = 1. Determine the circles touching C
and having T1 , T 2 as their pair of tangents. Further, find the
equations of all possible common tangents to these circles
when taken two at a time.
l
Sol. In figure OS = 1, OP = 2
∴
sin ∠SPO =
1
= sin 30°
2
Y
A1
P (– 2, 0)
30°
30° C1
S
2
B1
I1
O
Q
A2
T1
R
B2
C2
X
 1
C 1 : ( x + 4 / 3) 2 + y 2 =  
 3
∴
 24 
2
2
  m = 9 (1 + m )
5
⇒
64m 2 = 25 + 25m 2
⇒
39m 2 = 25 ⇒ m = ±
Ex. 37 Find the equation of the circle of minimum radius
which contains the three circles
x 2 − y 2 − 4y − 5 = 0
l
x 2 + y 2 + 12 x + 4y + 31 = 0
x 2 + y 2 + 6 x + 12y + 36 = 0
and
…(i)
circles are as given below :
C 1 ≡ (0, 2) ; r1 = 3
C 2 ≡ ( −6, − 2) ; r 2 = 3
and
C 3 ≡ ( −3, − 6) ; r 3 = 3
C1
3 2
a
3
= 4
=
a
3a
2
−a
2
a /2
6
⇒ a=
tan 30° =
3
3
Radius =
h, k
C(
P
3 6
=3
.
2
3
⇒ coordinates of C 2 are (4, 0)
∴ Equation of C 2 : ( x − 4 )2 + y 2 = 32
5
39
Sol. The coordinates of the centres and radii of three given
2
The other circle C 2 touches the equilateral triangle PB1B 2
externally.
∆
its radius is given by =
, where B1B 2 = a
s −a
but
⇒
Therefore, equations of transverse common tangents are
5
y=±
( x + 4 / 5)
39
T2
∴
∠SPO = 30°
Q
PA1 = PA 2 ⇒ ∠PA1A 2 = ∠PA 2 A1
⇒
∆ PA1A 2 is an equilateral triangle.
Therefore, centre C 1 is centroid of ∆ PA1A 2 ,C 1 divides PQ in
the ratio 2 : 1.
1
 4 
C 1 ≡  − , 0 and its radius = C 1Q =
∴
 3 
3
⇒
321
…(ii)
C3
)
C2
Let C ≡ (h , k ) be the centre of the circle passing through the
centres C 1 (0, 2), C 2 ( −6, − 2) and C 3 ( −3, − 6).
Then,
CC 1 = CC 2 = CC 3
⇒
(CC 1 )2 = (CC 2 )2 = (CC 3 )2
322
Textbook of Coordinate Geometry
⇒
( h − 0) 2 + ( k − 2) 2 = ( h + 6) 2 + ( k + 2) 2
= ( h + 3) 2 + ( k + 6) 2
⇒
⇒
− 4k + 4 = 12h + 4k + 40 = 6h + 12k + 45
12h + 8k + 36 = 0
3h + 2k + 9 = 0
6h − 8k − 5 = 0
or
and
⇒
9 (1 − sin 2 θ ) = 4 (1 + sin θ )2
⇒
9 (1 − sin θ ) = 4 (1 + sin θ )
5
12
and cosθ =
sinθ =
13
13
Eq. (ii) OP will be maximum, if P becomes the point
extended part of OC cuts the circle. Let this point be Q
Now, CP = CC 3 + C 3 P = CC 3 + 3
2
then maximum value of OP = OQ = OC + CQ = ( 13 + 2)
31
23



5
(949 ) + 3
= − 3 +  + − 6 +  +3 = 






18
12
36
Let
∠ COX = α
then, Q ≡ (OQ cos α, OQ sin α )
Hence, equation of required circle is
2
5
31
23




949 
 x +  + y +  = 3 +




18
12
36
2
≡ ((2 + 13 ) cos α, (2 + 13 ) sin α )
Now, in ∆COL,
cosα =
∴
sinα =
Remark
If radii of three given circles are distinct say r1 < r2 < r3 then the
radius of the required circle will be equal to (CC1 or CC2 or CC3)
+ r3 (QCC1 = CC2 = CC3)
Ex. 38 Find the point P on the circle
x 2 + y 2 − 4 x − 6y + 9 = 0 such that
(i) ∠ POX is minimum,
(ii) OP is maximum, when O is the origin and OX is the
X-axis.
Sol. Given circle is
OL NC
2
=
=
OC OC
13
3
13
Ex. 39 The circle x 2 + y 2 − 4 x − 8y + 16 = 0 rolls up the
tangent to it at ( 2 + 3 , 3 ) by 2 units, assuming the X-axis as
horizontal, find the equation of the circle in the new
position.
l
Sol. Given circle is
x 2 + y 2 − 4 x − 6y + 9 = 0
( x − 2) 2 + ( y − 3) 2 = 22
... (iii)
4
6 

Now, from Eq. (iii), Q ≡ 2 +
,3 +


13
13 
l
or
(Q sin θ ≠ − 1)
5
12

 36 15
From Eq. (ii), P ≡ 3 × , 3 ×  i.e. P ≡  , 


 13 13
13
13
31
23
On solving Eqs. (i) and (ii), we get h = − , k = −
18
12
2
3 cos θ = 2 + 2 sin θ
∴
...(i)
...(ii)
2
⇒
.... (i)
Its centre is C ≡ (2, 3) and radius r = 2
Eq. (i) Let OP and ON be the two tangents from O to the
circle Eq. (i), then OP = ON = 3
x 2 + y 2 − 4 x − 8y + 16 = 0
Let
…(i)
P ≡ ( 2 + 3, 3)
B
Y
2
Q
Q
C
N
A 2
(2,4)
P (2 + 3, 3)
60°
θ
α
X´
O
H
θ
P
θ
L MK
X
Equation of tangent to the circle Eq. (i) at P (2 + 3, 3) is
Y´
then ∠ POX is minimum when OP is tangent to the circle
Eq. (i) at P
Let ∠ POX = θ
∴
P ≡ (OP cos θ, OP sin θ )
i.e.
...(ii)
P ≡ (3 cos θ, 3 sin θ )
From figure, OM = OL + LM = NC + HP = NC + CP sinθ
⇒
OP cos θ = NC + CP sin θ
(2 + 3 ) x + 3y − 2 ( x + 2 + 3 ) − 4 (y + 3) + 16 = 0
or
3 x −y −2 3 =0
…(ii)
Let A and B be the centres of the circles in old and new
positions, then
B ≡ (2 + 2 cos 60° , 4 + 2 sin 60° )
(Q AB makes an angle 60° with X-axis)
or
B ≡ ( 3, 4 + 3 )
Chap 04 Circle
and
⇒
radius = 22 + 4 2 − 16 = 2
+
( x − 3) 2 + ( y − 4 − 3 ) 2 = 22
x 2 + y 2 − 6x − 2 ( 4 + 3 ) y + 24 + 8 3 = 0
Ex. 40 Find the intervals of the values of ‘a’ for which the
line y + x = 0 bisects two chords drawn from a point
 1 + 2a 1 − 2a 
,
 to the circle

2 
 2
l
⇒
 1 + 2a 1 − 2a 
Sol. The point A 
,
 lies on the given circle as
2 
 2
its coordinate satisfy the equation of the circle. Let AB
and AC are two chords drawn from A. Let M and N are
the mid-points of AB and AC.
Let coordinate of M be (h , − h ) and coordinate of B is (α, β ),
then
16h 2 − 12 2 ah + 2 + 4a 2 = 0
or
8h 2 − 6 2 ah + 1 + 2a 2 = 0
Hence, for two real and different values of h, we must have
N
(h, – h)
B
h=
and
∴
−h =
β+
α = 2h −
1 + 2a
2
2
2
or
72a 2 − 32 (1 + 2a 2 ) > 0
⇒
8a 2 − 32 > 0
⇒
a2 − 4 > 0
4 xh − 4yh − (1 + 2a ) ( x + h ) − (1 − 2a ) (y − h )
= 8h 2 − 2 (1 + 2a ) h + 2 (1 − 2a ) h
⇒
x [ 4h − (1 + 2a )] − y [ 4h + (1 − 2a )] − h (1 + 2a )
+ h (1 − 2a ) = 8h 2 − 2 (1 + 2a ) h + 2 (1 − 2a ) h
1 − 2a
2
2
or
8h 2 − (1 + 2a ) h + (1 − 2a ) h − x [ 4h − (1 + 2a )]
+ y [ 4h + (1 − 2a )] = 0
1 − 2a
2
 1 + 2a 1 − 2a 
It passes through A 
,
 , then
2 
 2
β = − 2h −
2
⇒
–
( −6 2a )2 − 4 ⋅ 8 (1 + 2a 2 ) > 0
⇒
1 − 2a
2
Since, B (α, β ) lies on the given circle, we have
and
+
–2
= 2h 2 + 2h 2 − (1 + 2a ) h + (1 − 2a ) h
y+ x =0
α+
+
( a + 2) ( a − 2) > 0
Hence, the required value of a (from wavy curve)
a ∈ ( −∞, − 2) ∪ (2, ∞ )
Aliter : Equation of chord AB whose mid-point is (h , − h ) is
T = S1
 x + h
y − h
2xh − 2yh − (1 + 2a ) 
 − (1 − 2a ) 

 2 
 2 
C
M
16h 2 − 12 2 ah + (1 + 2a )2 + (1 − 2a )2 = 0
⇒
2 x 2 + 2y 2 − (1 + 2 a ) x − (1 − 2a )y = 0.
A
16h 2 − 4h (1 + 2a ) + 4h (1 − 2a )
(1 + 2a )2 (1 − 2a )2
+
− 2h (1 + 2a )
2
2
(1 + 2a )2
(1 − 2a )2
+
+ 2h (1 − 2a ) +
=0
2
2
∴ Equation of the required circle is
or
323


1 − 2a 
1 + 2a 
2 2h −
 + 2 −2h −
2 
2



 1 + 2a 
8h 2 − 2 2ah − 
 [ 4h − (1 + 2a )]
 2 
2

1 + 2a 
− (1 + 2a ) 2h −
2 


1 − 2a 
=0
− (1 − 2a ) −2h −
2 

 1 − 2a 
+
 [ 4h + (1 − 2a )] = 0
 2 
or
or
(1 + 2a )2
2
(1 − 2a )2
+ 2h (1 − 2a ) +
=0
2
8h 2 − 6 2ah + 1 + 2a 2 = 0
8h 2 − 2 2ah − 2h (1 + 2a ) +
324
Textbook of Coordinate Geometry
Hence, for two real and different values of h, we must have
+
⇒
2 cos 2α
.cos α = 1
sin 2α
⇒
2(1 − 2sin 2 α) cos α
=1
2sin α cos α
+
–
–2
2
( −6 2a )2 − 4 ⋅ 8 ⋅ (1 + 2a 2 ) > 0
or
a2 − 4 > 0
∴
( a + 2) ( a − 2) > 0
∴
a ∈ ( −∞ , − 2) ∪ (2, ∞ )
⇒
⇒
⇒
⇒
2sin α + sin α −1 = 0
(2sin α −1) (sin α +1) = 0
sinα ≠ −1
1
∴
sinα =
2
∴
α = 30°
Tangent at P( −2, − 2) is
−2x − 2y − ( x − 2) − 2(y − 2) − 20 = 0
⇒
3x + 4y + 14 = 0
Slope of PM = −3 / 4
Q
∠PMQ = 90°− α = 90°−30° = 60°
4m + 3
m +3/ 4
, 3=
tan 60° =
∴
4 − 3m
1 − 3m / 4
Ex. 41 A ball moving around the circle
x 2 + y 2 − 2 x − 4y − 20 = 0 in anti-clockwise direction leaves
it tangentially at the point P ( − 2 , − 2 ). After getting reflected
from a straight line, it passes through the centre of the circle.
Find the equation of the straight line if its perpendicular
distance from P is 5/2. You can assume that the angle of
incidence is equal to the angle of reflection.
l
Sol. Radius of the circle = CP = 9 + 16 = 5
Let the equation is of surface is y = mx + c
5
given
PQ =
2
5
−2m + 2 + c
=±
∴
2
2
(1 + m )
∴
…(i)
P
(–2, –2)
5/2
Q
we get
rm
c =
 −39 + 2 3 
 4 3 − 3
y=

x +
 4 +3 3 
 4 + 3 3
α
–α α
M
4 3 −3
4 +3 3
11 + 2 3
−39 + 2 3
or
4 +3 3
4 +3 3
c being intercept on Y-axis made by surface is clearly–ve.
Hence, the required line is
al
N
90°
m=
5 2(1 − m ) + c
From Eq. (i) ± =
2
1 + m2
C (1, 2)
No
1 − 2sin 2 α = sin α
2
⇒
Surface
Tangent at P strikes it at the point M and after reflection
passes through the centre C (1, 2).
Let MN be the normal at M.
∠ PMN = ∠NMC = α
PC
In ∆ PCM,
tan2α =
PM
5
⇒
tan2α =
PM
…(ii)
⇒
PM = 5 cot 2α
and in ∆ PQM
5/2
sin(90°−α ) =
PM
5
…(iii)
PM =
∴
2 cosα
5
From Eqs. (ii) and (iii), 5 cot 2α =
2 cos α
⇒
2 cot 2α cos α = 1
( 4 3 − 3) x − ( 4 + 3 3 )y − (39 − 2 3 ) = 0.
Ex. 42 Find the limiting points of the circles
( x 2 + y 2 + 2 gx + c ) + λ ( x 2 + y 2 + 2 fy + d ) = 0 and show
that the square of the distance between them is
(c − d ) 2 − 4 f 2 g 2 + 4cf 2 + 4dg 2
l
f 2 +g2
Sol. The given circles are
( x 2 + y 2 + 2gx + c ) + λ ( x 2 + y 2 + 2 fy + d ) = 0
⇒
x2 + y2 +
2g
2 fλ
( c + λd )
=0
x+
y+
1+ λ
1+ λ
1+ λ
 −g − f λ 
Centre of the circle 
,

1 + λ 1 + λ 
Equating the radius of this circle to zero, we get
g2
(1 + λ )
2
+
f 2 λ2
(1 + λ )2
−
( c + λd )
=0
(1 + λ )
Chap 04 Circle
⇒
g 2 + f 2 λ 2 − ( c + λd ) ( 1 + λ ) = 0
⇒
( f − d ) λ − (c + d )λ + g − c = 0
2
2
Ex. 43 One vertex of a triangle of given species is fixed
and another moves along circumference of a fixed circle.
Prove that the locus of the remaining vertex is a circle and
find its radius.
l
2
Let the roots be λ 1 and λ 2
then
∴
λ1 + λ 2 =
(c + d )
(f
− d)
2
f
Sol. Let OPQ be a triangle of given species. Then the angles
α,β, γ will be fixed.
−d
( λ 1 − λ 2 ) = ( λ 1 + λ 2 )2 − 4 λ 1λ 2
(c + d )
=
(f
λ1 =
and λ 2 =
2
2
− d )2
−
Y
4(g − c )
− d)
2
(f
(c + d ) − 4 f g + 4cf
2
2
r
+ 4dg
2
…(i)
(f −d)
(c + d ) + (c − d )2 − 4 f 2 g 2 + 4cf
2
+ 4dg 2
2( f 2 − d )
(c + d ) − {(c − d ) − 4 f g + 4cf + 4dg }
2
2
2
2
2( f 2 − d )
…(iii)
Substituting the values of λ 1 and λ 2 from Eqs. (ii) and (iii)
square of the distance between limiting points
=
(g + f ) (λ1 − λ 2 )
2
2
⇒
cos θ1 =
∴
r1 =
(f
2
g2 + f 2

 2
 f −d 
+ 4dg 2 }
− d )2
2
[(c − d )2 − 4 f 2 g 2 + 4cf
(g 2 + f 2 )
2
…(i)
…(ii)
r sin γ
sin β
…(iii)
Substituting the values of θ1 and r1 from Eqs. (ii )and (iii)
in Eq. (i)
2b
{(c − d )2 − 4 f 2 g 2 + 4cf
r12 + b 2 − a 2
2r1b
Q
θ = α + θ1 , ∴ θ1 = θ − α
using sine rule in ∆OPQ
r
r
= 1
sin β sin γ
[1 + ( λ 1 + λ 2 ) + λ 1λ 2 ]2
(g 2 + f 2 )
=
2
X
Let the polar coordinates of Q be (r1,θ1 ), we have to find the
locus of P (r ,θ ). In ∠OCQ
 −g
−f λ2
and 
,

1 + λ 2 1 + λ 2 
2
a
G(b, 0)
b
O
Y´
Hence, limiting points are
2
X´
r1
γ
β Q (r1, θ1)
…(ii)
2
 −g
f λ2 
g   − f λ1
+
+
=

 +
1 + λ1 1 + λ 2  1 + λ1 1 + λ 2 
α
θ1
2
2
 −g − f λ1 
,


1 + λ1 1 + λ1
P (r, θ)
2
2
=
∴
2
g2 − c
, λ 1λ 2 =
⇒
r sin γ
r 2 sin 2 γ
+ b2 − a2
cos(θ − α ) =
sin β
sin 2 β
a 2 sin 2 β
sin 2 γ
=r2 +
b 2 sin 2 β
sin 2 γ
− 2b
r sin β
cos(θ − α)
sin γ
This is an equation of circle in polar form with radius
2
325
+ 4dg 2 ]
sin β
.
sin γ
#L
Circle Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. The sum of the square of the length of the chord
intercepted by the line x + y = n, n ∈ N on the circle
x 2 + y 2 = 4 is
(a) 11
(c) 33
(b) 22
(d) None of these
2. Tangents are drawn to the circle x 2 + y 2 = 50 from a
point ‘P’ lying on the X -axis. These tangents meet the
Y -axis at points ‘P1 ’ and ‘P2 ’. Possible coordinates of ‘P’
so that area of triangle PP1 P2 is minimum, is
(b) (10 2, 0) (c) ( −10 2, 0 ) (d) (10 3, 0 )
(a) (10, 0 )
3. Equation of chord AB of circle x 2 + y 2 = 2 passing
through P(2, 2) such that
PB
= 3, is given by
PA
(a) x = 3y
(c) y − 2 = 3( x − 2 )
(b) x = y
(d) y − 3 = 3( x − 1 )
4. If r1 and r 2 are the radii of smallest and largest circles
which passes through (5, 6) and touches the circle
( x − 2) 2 + y 2 = 4, then r1r 2 is
4
(a)
41
41
(b)
4
5
(c)
41
41
(d)
5
5. Equation of circle S ( x , y ) = 0, (S (2, 3) = 16) which touches
the line 3x + 4y − 7 = 0 at (1, 1) is given by
(a) x 2 + y 2 + x + 2y − 5 = 0
(c) x + y + 4 x − 6y + 13 = 0
2
(d) x 2 + y 2 − 4 x + 6y − 7 = 0
6. If P(2, 8) is an interior point of a circle
x 2 + y 2 − 2x + 4y − λ = 0 which neither touches nor
intersects the axes, then set for λ is
(a) ( − ∞, − 1 )
(c) (96, ∞ )
if a and b both are rational numbers) on the
circumference of a circle having centre (π, e) is
(a) atmost one
(c) exactly two
(b) atleast two
(d) infinite
9. Three sides of a triangle have the equations
L r ≡ y − mr x − c r = 0; r = 1, 2, 3. Then
λL 2 L 3 + µL 3 L 1 + vL 1 L 2 = 0, where λ ≠ 0, µ ≠ 0,v ≠ 0 is
the equation of circumcircle of triangle, if
(a) λ (m2 + m3 ) + µ(m3 + m1 ) + v(m1 + m2 ) = 0
(b) λ (m2m3 − 1 ) + µ(m3m1 − 1 ) + v(m1m2 − 1 ) = 0
(c) Both (a) and (b)
(d) None of the above
10. f ( x , y ) ≡ x 2 + y 2 + 2ax + 2by + c = 0 represent a circle. If
f ( x , 0) = 0 has equal roots, each being 2 and f (0, y ) = 0
has 2 and 3 as its roots, then the centre of the circle is
 5
(a) 2, 
 2
(b) Data are not consistent
5

(c)  −2, − 

2
(d) Data are inconsistent
11. If (1 + αx )n = 1 + 8x + 24 x 2 + ... and a line through P(α, n )
cuts the circle x 2 + y 2 = 4 in A and B, then PA. PB is
equal to
(a) 4
(b) 8
(c) 16
(d) 32
12. A region in the xy-plane is bounded by the curve
y = (25 − x 2 ) and the line y = 0. If the point (a, a + 1) lies
in the interior of the region, then
(b) x 2 + y 2 + 2 x + 2y − 7 = 0
2
8. The number of rational point(s) (a point (a, b) is rational,
(b) ( − ∞, − 4 )
(d) φ
7. The difference between the radii of the largest and
smallest circles which have their centre on the
circumference of the circle x 2 + y 2 + 2x + 4y − 4 = 0 and
pass through the point (a, b ) lying outside the given
circle is
(a) 6
(b) (a + 1 ) 2 + (b + 2 ) 2
(c) 3
(d) (a + 1 ) 2 + (b + 2 ) 2 − 3
(a) a ∈ ( −4,3 )
(c) a ∈ ( −1, 3 )
(b) a ∈ ( − ∞, − 1 ) ∪ (3, ∞ )
(d) None of these
13. S( x , y ) = 0 represents a circle. The equation S( x , 2) = 0
gives two identical solutions x = 1 and the equation
S(1, y ) = 0 gives two distinct solutions y = 0, 2, then the
equation of the circle is
(a) x 2 + y 2 + 2 x − 2y + 1 = 0 (b) x 2 + y 2 − 2 x + 2y + 1 = 0
(c) x 2 + y 2 − 2 x − 2y − 1 = 0 (d) x 2 + y 2 − 2 x − 2y + 1 = 0
π
be a fixed angle. If P = (cos θ,sin θ) and
2
Q = (cos(α − θ), sin (α − θ)), then Q is obtained from P by
14. Let 0 < α <
(a) clockwise rotation around origin through an angle α
(b) anti-clockwise rotation around origin through an angle α
(c) reflection in the line through origin with slope tanα
α
(d) reflection in the line through origin with slope tan  
 2
Chap 04 Circle
15. The number of points ( x , y ) having integral coordinates
satisfying the condition x + y < 25 is
2
(a) 69
(b) 80
2
(c) 81
(d) 77
16. The point ([P + 1],[P ]), (where [. ] denotes the greatest
integer function) lying inside the region bounded by the
circle x 2 + y 2 − 2x − 15 = 0 and x 2 + y 2 − 2x − 7 = 0, then
(a) P ∈ [ −1, 0 ) ∪ [ 0,1 ) ∪ [1, 2 ) (b) P ∈ [ −1,2 ) − { 0,1 }
(c) P ∈ ( −1, 2 )
(d) None of these
x 2 + y 2 − 8x + 7 = 0. The point P starts moving under the
conditions that its path encloses greatest possible area
and it is at a fixed distance from any arbitrarily chosen
fixed point in its region. The locus of P is
(a) 4 x 2 + 4y 2 − 12 x + 1 = 0 (b) 4 x 2 + 4y 2 + 12 x − 1 = 0
(d) x 2 + y 2 − 3 x + 2 = 0
18. The set of values of ‘c’ so that the equations y = | x | + c
and x 2 + y 2 − 8 | x |−9 = 0 have no solution is
(a) ( − ∞, − 3 ) ∪ (3, ∞ )
(c) ( − ∞, − 5 2 ) ∪ (5 2, ∞ )
24. One of the diameter of the circle circumscribing the
rectangle ABCD is 4y = x + 7. If A and B are the points
( −3, 4 ) and (5, 4) respectively, then the area of the
rectangle is
(a) 16 sq units
(c) 32 sq units
(b) 24 sq units
(d) None of these
25. A, B, C and D are the points of intersection with the
coordinate axes of the lines ax + by = ab and
bx + ay = ab, then
17. A point P lies inside the circles x 2 + y 2 − 4 = 0 and
(c) x 2 + y 2 − 3 x − 2 = 0
(a) A, B, C, D are concyclic
(b) A, B, C, D form a parallelogram
(c) A, B, C, D form a rhombus
(d) None of the above
26. α, β and γ are parametric angles of three points P, Q and
R respectively, on the circle x 2 + y 2 = 1 and A is the
point ( −1, 0). If the lengths of the chords AP, AQ and AR
β
 α
γ
are in GP, then cos   , cos   and cos   are in
 2
 2
 2
(a) AP
(c) HP
(b) ( −3, 3 )
(d) (5 2 − 4, ∞ )
(b) GP
(d) None of these
27. The equation of the circle passing through (2, 0) and
(0, 4) and having the minimum radius is
19. If a line segment AM = a moves in the plane XOY
remaining parallel to OX so that the left end point A
slides along the circle x 2 + y 2 = a 2 , the locus of M is
(a) x 2 + y 2 = 20
(b) x 2 + y 2 − 2 x − 4y = 0
(a) x 2 + y 2 = 4a 2
(b) x 2 + y 2 = 2ax
(c) ( x 2 + y 2 − 4 ) + λ ( x 2 + y 2 − 16 ) = 0
(c) x 2 + y 2 = 2ay
(d) x 2 + y 2 − 2ax − 2ay = 0
(d) None of the above
28. A circle of radius unity is centred at the origin. Two
20. The four points of intersection of the lines
(2x − y + 1)( x − 2y + 3) = 0 with the axes lie on a circle
whose centre is at the point
 7 5
(a)  − , 
 4 4
 3 5
(b)  , 
 4 4
 5
(d)  0, 
 4
 9 5
(c)  , 
 4 4
21. The number of integral values of λ for which
x 2 + y 2 + λx + (1 − λ )y + 5 = 0 is the equation of a circle
whose radius cannot exceed 5, is
(a) 14
(b) 18
(c) 16
(d) None of these
22. Let φ( x , y ) = 0 be the equation of a circle. If φ (0, λ ) = 0
has equal roots λ = 2, 2 and φ ( λ , 0) = 0 has roots λ =
4
, 5,
5
then the centre of the circle is
 29
(a) 2, 
 10
 29 
(b)  , 2
 10 
 29
(c)  −2, 
 10
(d) None of these
23. The locus of the point of intersection of the tangents to
the circle x = r cos θ, y = r sin θ at points whose
π
parametric angles differ by is
3
(a) x + y = 4(2 − 3 )r
2
2
(c) x 2 + y 2 = (2 − 3 )r 2
2
(b) 3( x + y ) = 1
2
2
(d) 3( x 2 + y 2 ) = 4r 2
327
particles start moving at the same time from the point
(1, 0) and move around the circle in opposite direction.
One of the particle moves anticlockwise with constant
speed v and the other moves clockwise with constant
speed 3v. After leaving (1, 0), the two particles meet first
at a point P and continue until they meet next at point
Q. The coordinates of the point Q are
(a) (1, 0)
(b) (0, 1)
(c) (−1, 0)
(d) (0, −1)
29. The circle x + y = 4 cuts the line joining the points
2
2
A(1, 0) and B(3, 4 ) in two points P and Q. Let
BP
= α and
PA
BQ
= β, then α and β are roots of the quadratic equation
QA
(a) x 2 + 2 x + 7 = 0
(c) 2 x 2 + 3 x − 27 = 0
(b) 3 x 2 + 2 x − 21 = 0
(d) None of these
30. The locus of the mid-points of the chords of the circle
x 2 + y 2 + 4 x − 6y − 12 = 0 which subtend an angle of
radians at its circumference is
(a) ( x + 2 ) 2 + (y − 3 ) 2 = 6.25 (b) ( x − 2 ) 2 + (y + 3 ) 2 = 6.25
(c) ( x + 2 ) 2 + (y − 3 ) 2 = 18.75 (d) ( x + 2 ) 2 + (y + 3 ) 2 = 18.75
π
3
328
#L
Textbook of Coordinate Geometry
Circle Exercise 2 :
More than One Correct Option Type Questions
This section contains 15 multiple choice questions. Each
question has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
31. If OA and OB are two perpendicular chords of the circle
r = a cos θ + b sin θ passing through origin, then the locus
of the mid-point of AB is
a
2
b
(d) y =
2
(a) x 2 + y 2 = a + b
(b) x =
(c) x 2 − y 2 = a 2 − b 2
32. If A and B are two points on the circle
x 2 + y 2 − 4 x + 6y − 3 = 0 which are farthest and nearest
respectively, from the point (7, 2), then
(a) A ≡ (2 − 2 2, − 3 − 2 2 )
37. An equation of a circle touching the axes of coordinates
and the line x cos α + y sin α = 2 can be
2
(cosα + sin α + 1 )
2
2
2
2
(b) x + y − 2 gx − 2 gy + g = 0, where g =
(cosα + sin α − 1 )
2
(c) x 2 + y 2 − 2 gx + 2 gy + g 2 = 0, where g =
(cosα − sin α + 1 )
2
(d) x 2 + y 2 − 2 gx + 2 gy + g 2 = 0, where g =
(cosα − sin α − 1 )
(a) x 2 + y 2 − 2 gx − 2 gy + g 2 = 0, where g =
38. If α is the angle subtended at P( x 1 , y 1 ) by the circle
S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0, then
(a) cot α =
(b) A ≡ (2 + 2 2, − 3 + 2 2 )
(c) B ≡ (2 + 2 2, − 3 + 2 2 )
(c) tan α =
(d) B ≡ (2 − 2 2, − 3 − 2 2 )
33. If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 cuts each of the
circles x + y − 4 = 0, x + y − 6x − 8y + 10 = 0 and
2
2
2
2
x 2 + y 2 + 2x − 4y − 2 = 0 at the extremities of a diameter,
then
(a) c = −4
(c) g 2 + f 2 − c = 17
(b) g + f = c − 1
(d) gf = 6
S1
(g + f − c )
2
2
between the pair of tangents from point ( λ , 0) to the
 π 2π 
circle x 2 + y 2 = 4 lies in interval  ,  is
2 3 
S1
α
=
2
2
(g + f 2 − c )
 (g 2 + f 2 − c ) 
2 (g 2 + f 2 − c )

(d) α = 2 tan −1 


S1
S1


39. The equation of the circle which touches the axis of
x y
+ = 1 and whose centre lies
3 4
in the first quadrant is x 2 + y 2 − 2λx − 2λy + λ 2 = 0, then
λ is equal to
coordinates and the line
(a) 1
(c) 3
34. The possible value of λ ( λ > 0) such that the angle
(b) cot
(b) 2
(d) 6
40. If P is a point on the circle x 2 + y 2 = 9, Q is a point on
the line 7 x + y + 3 = 0, and the line x − y + 1 = 0, is the
perpendicular bisector of PQ, then the coordinates of P
are
 4

(a)  , 2 2
 3

(b) ( 0, 2 )
(a) (3, 0)
 72 21
(b)  , − 
 25 25
(c) (1, 2 )
 4 4
(d)  , 
 3 3
(c) (0, 3)
 72 21
(d)  − , 
 25 25
35. If a chord of the circle x 2 + y 2 − 4 x − 2y − c = 0 is
 1 1
 8 8
trisected at the points  ,  and  ,  , then
 3 3
 3 3
(a) c = 10
(c) c = 15
(b) c = 20
(d) c 2 − 40c + 400 = 0
36. From the point A(0, 3) on x + 4 x + (y − 3) = 0, a chord
2
2
AB is drawn and extended to a point M, such that
AM = 2AB. An equation of the locus of M is
(a) x 2 + 6 x + (y − 2 ) 2 = 0
(b) x 2 + 8 x + (y − 3 ) 2 = 0
(c) x 2 + y 2 + 8 x − 6y + 9 = 0
(d) x 2 + y 2 + 6 x − 4y + 4 = 0

41. If a circle passes through the point 3,

7
 and touches
2
x + y = 1 and x − y = 1, then the centre of the circle is
(a) ( 4, 0 )
(b) (4, 2)
(c) (6, 0)
(d) (7, 9)
42. The equation of a circle C 1 is x 2 + y 2 = 4. The locus of
the intersection of orthogonal tangents to the circle is
the curve C 2 and the locus of the intersection of
perpendicular tangents to the curve C 2 is the curve C 3 .
Then,
(a) C 3 is a circle
(b) the area enclosed by the curve C 3 is 8π
(c) C 2 and C 3 are circles with the same centre
(d) None of the above
Chap 04 Circle
43. The equation of a tangent to the circle x 2 + y 2 = 25
passing through ( −2, 11) is
(a) 4 x + 3y = 25
(c) 24 x − 7y + 125 = 0
(b) 3 x + 4y = 38
(d) 7 x + 24y = 230
(a) L is the radical axis of C1 and C 2
(b) L is the common tangent of C1 and C 2
(c) L is the common chord of C1 and C 2
(d) L is perpendicular to the line joining centres of C1 and C 2
45. A square is inscribed in the circle
44. Consider the circles
x 2 + y 2 − 10x − 6y + 30 = 0. One side of the square is
parallel to y = x + 3, then one vertex of the square is
C1 ≡ x 2 + y 2 − 2 x − 4y − 4 = 0 and
C 2 ≡ x 2 + y 2 + 2 x + 4y + 4 = 0
(a) (3, 3)
(c) (6, 3 − 3 )
and the line L ≡ x + 2y + 2 = 0, then
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329
(b) (7, 3 )
(d) (6, 3 + 3 )
Circle Exercise 3 :
Paragraph Based Questions
This section contains 7 paragraphs based upon each of the
paragraph 3 multiple choice questions have to be answered.
Each of these questions has four choices (a), (b), (c) and (d)
out of which ONLY ONE is correct.
Paragraph I
(Q. Nos. 46 to 48)
46. Length of the chord AB is
(b) 10
(c) 2 5
(d) 5 2
47. The angle subtended by the chord AB is the minor arc of
S is
(a)
π
4
(b)
2π
3
(c)
3π
4
(d)
5π
6
48. Acute angle between the line L and the circle S is
(a)
π
6
(b)
π
4
(c)
π
3
(d)
π
2
Paragraph II
P is a variable point on the line L = 0. Tangents are drawn to
the circle x 2 + y 2 = 4 from P to touch it at Q and R. The
parallelogram PQSR is completed.
49. If L ≡ 2x + y − 6 = 0, then the locus of the circumcenter of
∆PQR is
value of λ is
(a) 2
(b) 3
(c) 5
 46 68 
(c)  − , 
 25 25
 68 51
(d)  − , 
 25 25
(Q. Nos. 52 to 54)
Equation of the circumcircle of a triangle formed by the lines
L1 = 0, L2 = 0 and L3 = 0 can be written as
L1 L2 + λL2 L3 + µL3 L1 = 0, where λ and µ are such that
coefficient of x 2 = coefficient of y 2 and coefficient of xy = 0.
52. L 1 L 2 2 + λL 2 L23 + µL 3 L21 = 0 represents
(a) a curve passing through point of intersection of L1 = 0,
L2 = 0 and L3 = 0
(b) a circle is coefficient of x 2 = coefficient of y 2 and
coefficient of xy = 0
(c) a parabola
(d) pair of straight lines
be two other distinct parallel lines which are not parallel
to L 1 = 0. The equation of a circle passing through the
vertices of the parallelogram formed must be of the form
(a) λL1L4 + µL2L3 = 0
(b) λL1L3 + µL2L4 = 0
(c) λL1L2 + µL3L4 = 0
(d) λL12L3 + µL22L4 = 0
54. If L 1 L 2 + λL 2 L 3 + µL 3 L 1 = 0 is such that µ = 0 and λ is
(b) 2 x + y = 3
(d) x + 2y = 3
50. If P ≡ (6, 8), then area of ∆QRS is
 51 68
(b)  − , − 
 25 25
53. L 1 = 0, L 2 = 0 be the distinct parallel lines, L 3 = 0, L 4 = 0
(Q. Nos. 49 to 51)
(a) 2 x − y = 4
(c) x − 2y = 4
 46 63
(a)  − , 
 25 25
Paragraph III
Consider the circle S : x 2 + y 2 − 4x − 1 = 0 and the line
L : y = 3x − 1. If the line L cuts the circle at A and B.
(a) 5
51. If P ≡ (3, 4 ), then the coordinates of S are
non-zero, then it represents
192
λ sq units. The
25
(d) 6
(a) a parabola
(b) a pair of straight lines
(c) a circle
(d) an ellipse
330
Textbook of Coordinate Geometry
Paragraph IV
Paragraph VI
(Q. Nos. 55 to 57)
Given two circles intersecting orthogonally having the length of
24
unit. The radius of one of the circles is 3 units.
common chord
5
(Q. Nos. 61 to 63)
55. If radius of other circle is λ units, then λ is
(a) 2
(b) 4
(c) 5
(d) 6
Two variable chords AB and BC of a circle x 2 + y 2 = a 2 are
such that AB = BC = a, M and N are the mid-points of AB
and BC respectively such that line joining MN intersect the
circle at P and Q, where P is closer to AB and O is the centre
of the circle.
61. ∠OAB is
56. If angle between direct common tangents is 2θ, then
(a) 15°
(c) 45°
sin 2θ is
(a)
4
5
(b)
4 6
25
(c)
12
25
(d)
62. Angle between tangents at A and C is
24
25
(a) 60°
(c) 120°
57. If length of direct common tangent is λ units, then λ is
2
(a) 12
(b) 24
(c) 36
Paragraph V
(Q. Nos. 58 to 60)
Consider the two circles C1 : x 2 + y 2 = a 2 and
C 2 : x 2 + y 2 = b 2 ( a > b ). Let A be a fixed point on the circle
C1 , say A ( a, 0) and B be a variable point on the circle C 2 . The
line BA meets the circle C 2 again at C. ‘O’ being the origin.
58. If (OA ) 2 + (OB ) 2 + ( BC ) 2 = λ, then λ ∈
(b) [ 4b 2, 4b 2 + a 2 ]
(c) [ 4a 2, 4b 2 ]
(d) [5b 2 − 3a 2, 5b 2 + 3a 2 ]
2
a
a2

(b)  x −  + y 2 =

2
4
2
b
b2

(d)  x −  + y 2 =

2
4
b
a2

(c)  x −  + y 2 =

2
4
60. If ( BC ) 2 is maximum, then the locus of the mid-point of
AB is
#L
(c) x 2 + y 2 = (a − b ) 2
(d) None of these
(c) x 2 + y 2 = 4a 2
(d) x 2 + y 2 = 8a 2
t 1 , t 2 , t 3 are lengths of tangents drawn from a point ( h, k ) to
the circles x 2 + y 2 = 4, x 2 + y 2 − 4x = 0 and x 2 + y 2 − 4 y = 0
respectively further, t 14 = t 22 t 32 + 16. Locus of the point ( h, k )
consist of a straight line L1 and a circle C1 passing through
origin. A circle C 2 , which is equal to circle C1 is drawn
touching the line L1 and the circle C1 externally.
2
(b) x 2 + y 2 = (a + b ) 2
(b) x 2 + y 2 = 2a 2
(Q. Nos. 64 to 66)
2
(a) x 2 + y 2 = b 2
(a) x 2 + y 2 = a 2
Paragraph VII
59. The locus of the mid-point of AB is
a
b2

(a)  x −  + y 2 =

2
4
(b) 90°
(d) 150°
63. Locus of point of intersection of tangents at A and C is
(d) 48
(a) [5b 2 − 3a 2, 5b 2 + a 2 ]
(b) 30°
(d) 60°
64. Equation of L 1 is
(a) x + y = 0
(c) 2 x + y = 0
(b) x − y = 0
(d) x + 2y = 0
65. Equation of C 1 is
(a) x 2 + y 2 − x − y = 0
(b) x 2 + y 2 − 2 x + y = 0
(c) x 2 + y 2 − x + 2y = 0
(d) x 2 + y 2 − 2 x − 2y = 0
66. The distance between the centres of C 1 and C 2 is
(a) 2
(c) 2 2
(b) 2
(d) 4
Circle Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 10 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both
inclusive).
67. The point (1, 4) lies inside the circle x 2 + y 2 − 6x − 10y + λ = 0. If the circle neither touches nor cuts the axes, then the
difference between the maximum and the minimum possible values of λ is
68. Consider the family of circles x 2 + y 2 − 2x − 2λy − 8 = 0 passing through two fixed points A and B. Then the distance
between the points A and B is
Chap 04 Circle
69. If C 1 : x 2 + y 2 = (3 + 2 2 ) 2 be a circle and PA and PB are
pair of tangents on C 1 , where P is any point on the
director circle of C 1 , then the radius of the smallest circle
which touches C 1 externally and also the two tangents
PA and PB, is
70. If a circle S( x , y ) = 0 touches the point (2, 3) of the line
1
λ
x + y = 5 and S(1, 2) = 0, then radius of such circle is
units, then the value of λ 2 is.
71. If real numbers x and y satisfy ( x + 5) 2 + (y − 12) 2 = 196,
then the maximum value
1
2
2 3
of ( x + y )
is
72. If the equation of circle circumscribing the quadrilateral
formed by the lines in order are
73. A circle x 2 + y 2 + 4 x − 2 2y + c = 0 is the director circle
of the circle C 1 and C 1 is the director circle of circle C 2
and so on. If the sum of radii of all these circles is 2 and
if c = λ 2, then the value of λ is
74. If the area bounded by the circles x 2 + y 2 = r 2 , r = 1, 2
and the rays given by 2x 2 − 3xy − 2y 2 = 0 , y > 0 is
units, then the value of λ is
λπ
sq
4
75. The length of a common internal tangent of two circles
is 5 and that of a common external tangent is 13. If the
product of the radii of two circles is λ, then the value of
λ is
4
76. Consider a circles S with centre at the origin and radius 4.
2x + 3y = 2, 3x − 2y = 3, x + 2y = 3 and 2x − y = 1 is given
by x 2 + y 2 + λx + µy + v = 0. Then the value of
| λ + 2 µ + ν| is
#L
331
Four circles A, B, C and D each with radius unity and
centres ( −3, 0), ( −1, 0), (1, 0) and (3, 0) respectively are drawn.
A chord PQ of the circle S touches the circle B and passes
through the centre of the circle C. If the length of this
λ
chord can be expressed as λ , then the value of is
9
Circle Exercise 5 :
Matching Type Questions
n
This section contains 4 questions. Questions 77 and 78
have four statements (A, B, C and D) given in Column I
and four statements (p, q, r and s) in Column II, and
questions 79 and 80 have three statements (A, B and C)
given in Column I and five statements (p, q, r, s and t) in
Column II. Any given statement in Column I can have
correct matching with one or more statement(s) given in
Column II.
77. Consider the circles S 1 : x 2 + y 2 − 4 x − 6y + 12 = 0 and
S 2 :( x − 5) 2 + (y − 6) 2 = r 2 > 1
Column I
Column II
(A) S1 and S 2 touch internally, then (r − 1) is
divisible by
(p)
3
(B) S1 and S 2 touch externally, then r 2 + 2r + 3
is divisible by
(q)
4
(C) S1 and S 2 intersect orthogonally, then r 2 − 1
is divisible by
(r)
5
(D) S1 and S 2 intersect so that the common chord
is longest, then r 2 + 5 is divisible by
(s)
6
2
78. Match the following
Column I
Column II
(A) If ax + by − 5 = 0 is the equation of the chord
of the circle ( x − 3 ) 2 + (y − 4 ) 2 = 4, which
passes through (2 , 3 ) and at the greatest
distance from the centre of the circle, then
(p)
a + b =1
(B) Let O be the origin and P be a variable point
on the circle x 2 + y 2 + 2 x + 2y = 0. If the
locus of mid-point of OP is x 2 + y 2 + 2ax
+ 2 by = 0, then
(q)
a + b =2
(C) If (a, b) be coordinates of the centre of the (r)
smallest circle which cuts the circle
and
x 2 + y 2 − 2 x −4y − 4 = 0
orthogonally,
x 2 + y 2 − 10 x +12y + 52 = 0
then
a2 + b2 = 2
(D) If a and b are the slope of tangents which are
drawn to the circle x 2 + y 2 − 6 3 x −
6y + 27 = 0 from the origin, then
a2 + b2 = 3
(s)
332
Textbook of Coordinate Geometry
79. Match the following
80. Match the following
Column I
Column II
(A) If the shortest and largest distance from the
point (10 , 7 ) to the circle x 2 + y 2 − 4 x − 2y
−20 = 0 are L and M respectively, then
(p) M + L = 10
(A) If the straight lines y = a1 x + b and
y = a 2 x + b (a1 ≠ a 2 ) and b ∈ R meet the
coordinate axes in concyclic points, then
(B) If the shortest and largest distance from the
point (3, −6) to the circle x 2 + y 2 − 16 x
−12y − 125 = 0 are L and M respectively,
then
(q) M + L = 20
(B) If the chord of contact of the tangents drawn (q)
to x 2 + y 2 = b 2 and b ∈ R from any point on
(C) If the shortest and largest distance from the
point (6, −6) to the circle x 2 + y 2 − 4 x + 6y
−12 = 0 are L and M respectively, then
(r)
M + L = 30
(s)
M − L = 10
(s)
a1a 2 = 1
M − L = 26
(t)
a1a 2 = b 2
(p)
a12 + a 22 = 4
a1 + a 2 = 3
x 2 + y 2 = a12 , touches the circle x 2 + y 2 = a 22
(a1 ≠ a 2 ) , then
(t)
#L
Column I
Column I
(C) If the circle x 2 + y 2 + 2a1 x + b = 0 and
(r)
a1a 2 = b
x 2 + y 2 + 2a 2 x + b = 0 (a1 ≠ a 2 ) and b ∈ R
cuts orthogonally, then
Circle Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos. 81 to 88) are Assertion-Reason type
questions. Each of these questions contains two
statements:
Statement I (Assertion) and Statement II (Reason) Each
of these questions also has four alternative choices, only
one of which is the correct answer. You have to select the
correct choice as given below :
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II
is not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
81. Statement I Only one tangent can be drawn from the
point (1, 3) to the circle x + y = 1
2
Statement II Solving
2
|3 − m|
(1 + m 2 )
= 1, we get only one
real value of m
82. Statement I Tangents cannot be drawn from the point
(1, λ ) to the circle x 2 + y 2 + 2x − 4y = 0
Statement II (1 + 1) 2 + ( λ + 2) 2 < 1 2 + 2 2
83. Statement I Number of circles passing through (1, 4),
(2, 3), (−1, 6) is one
Statement II Every triangle has one circumcircle
84. Statement I Two tangents are drawn from a point on
the circle x 2 + y 2 = 50 to the circle x 2 + y 2 = 25, then
π
angle between tangents is
3
Statement II x 2 + y 2 = 50 is the director circle of
x 2 + y 2 = 25.
85. Statement I Circles x 2 + y 2 = 4 and x 2 + y 2 − 6x + 5 = 0
intersect each other at two distinct points
Statement II Circles with centres C 1 , C 2 and radii r1 , r 2
intersect at two distinct points if | C 1C 2 | < r1 + r 2
86. Statement I The line 3x − 4y = 7 is a diameter of the
circle x 2 + y 2 − 2x + 2y − 47 = 0
Statement II Normal of a circle always pass through
centre of circle
87. Statement I A ray of light incident at the point (−3, −1)
gets reflected from the tangent at (0, − 1) to the circle
x 2 + y 2 = 1. If the reflected ray touches the circle, then
equation of the reflected ray is 4y − 3x = 5
Statement I The angle of incidence = angle of reflection
i.e. ∠i = ∠r
88. Statement I The chord of contact of the circle
x 2 + y 2 = 1 w.r.t. the points (2, 3), (3, 5) and (1, 1) are
concurrent.
Statement II Points (1, 1), (2, 3) and (3, 5) are collinear.
Chap 04 Circle
333
Circle Exercise 7 :
Subjective Type Questions
n
In this section, there are 16 subjective questions.
89. Find the equation of the circle passing through (1, 0) and
(0, 1) and having the smallest possible radius.
90. Find the equation of the circle which touches the circle
x 2 + y 2 − 6x + 6y + 17 = 0 externally and to which the
lines x 2 − 3xy − 3x + 9y = 0 are normals.
91. A line meets the coordinate axes at A and B. A circle is
circumscribed about the triangle OAB. If the distance of
the points A and B from the tangent at O, the origin, to
the circle are m and n respectively, find the equation of
the circle.
92. Find the equation of a circle which passes through the
point (2 , 0) and whose centre is the limit of the point of
intersection of the lines 3x + 5y = 1 and
(2 + c ) x + 5c 2 y = 1 as c → 1.
93. Tangents are drawn from P (6, 8) to the circle x 2 + y 2 = r 2 .
Find the radius of the circle such that the area of the ∆
formed by tangents and chord of contact is maximum.
94. 2x − y + 4 = 0 is a diameter of the circle which
circumscribed a rectangle ABCD. If the coordinates of A
and B are A ( 4, 6) and B (1, 9 ) , find the area of rectangle
ABCD.
95. Find the radius of smaller circle which touches the straight
line 3x − y = 6 at (1, − 3) and also touches the line y = x .
96. If the circle C 1 , x 2 + y 2 = 16 intersects another circle C 2
of radius 5 in such a manner that the common chord is of
maximum length and has a slope equal to (3/4), find the
coordinates of centre C 2 .
97. Let 2x 2 + y 2 − 3xy = 0 be the equation of a pair of
tangents drawn from the origin O to a circle of radius 3
with centre in the first quadrant. If A is one of the points
of contact, find the length of OA.
98. The circle x 2 + y 2 = 1 cuts the X -axis at P and Q.
another circle with centre at Q and variable radius
intersects the first circle at R above the X -axis and the
line segment PQ at S. Find the maximum area of the
∆QSR .
99. If the two lines a1 x + b1y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0
cut the coordinate axes in concyclic points, prove that
a1a 2 = b1b 2 and find the equation of the circle.
100. The centre of the circle S = 0 lie on the line
2x − 2y + 9 = 0 and S = 0 cuts orthogonally the circle
x 2 + y 2 = 4. Show that circle S = 0 passes through two
fixed points and find their coordinates.
101. Find the condition on a , b , c such that two chords of
the circle
x 2 + y 2 − 2ax − 2by + a 2 + b 2 − c 2 = 0
passing through the point (a , b + c ) are bisected by the
line y = x .
102. Two straight lines rotate about two fixed points. If
they start from their position of coincidence such that
one rotates at the rate double that of the other. Prove
that the locus of their point of intersection is a circle.
103. The base AB of a triangle is fixed and its vertex C moves
such that sin A = k sin B (k ≠ 1) . Show that the locus of
C is a circle whose centre lies on the line AB and whose
ak
, a being the length of the base
radius is equal to
(1 − k 2 )
AB.
104. Consider a curve ax 2 + 2hxy + by 2 = 1 and a point P
not on the curve. A line drawn from the point P
intersects the curve at points Q and R. If the product
PQ ⋅ PR is independent of the slope of the line, then
show that the curve is a circle.
334
#L
Textbook of Coordinate Geometry
Circle Exercise 8 :
Questions Asked in Previous 13 Year’s Exams
n
This section contains questions asked in IIT-JEE, AIEEE,
JEE Main & JEE Advanced from year 2005 to 2017.
105. A circle is given by x + (y − 1) = 1, another circle C
2
2
touches it externally and also the X -axis, then the locus
of its centre is
[IIT-JEE 2005, 3M]
(a) {( x, y ): x = 4y } ∪ {( x, y ):y ≤ 0 }
2
(b) {( x, y ): x 2 + (y − 1 ) 2 = 4 } ∪ {( x, y ):y ≤ 0 }
(c) {( x,y ): x 2 = y } ∪ {( 0,y ):y ≤ 0 }
111. A line L′ through A is drawn parallel to BD. Point S
moves such that its distances from the line BD and the
vertex A are equal. If locus of S cuts L ′ at T 2 and T 3 and
AC at T 1 , then area of ∆T 1T 2 T 3 is
[IIT-JEE 2006, 5+5+5 M]
1
(a) sq units
2
(c) 1 sq units
2
(b) sq units
3
(d) 2 sq units
112. If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two
(d) {( x,y ): x 2 = 4y } ∪ {( 0, y ):y ≤ 0 }
106. If the circles x 2 + y 2 + 2ax + cy + a = 0 and
diameters of a circle of area 49π square units, the
equation of the circle is
[AIEEE 2006, 6M]
x 2 + y 2 − 3ax + dy − 1 = 0 intersect in two distinct points P
and Q, then the line 5x + by − a = 0 passes through P and
[AIEEE 2005, 6M]
Q for
(a) x 2 + y 2 + 2 x − 2y − 47 = 0
(a) exactly one value of a
(b) no value of a
(c) infinitely many values of a (d) exactly two values of a
(d) x 2 + y 2 − 2 x + 2y − 47 = 0
107. A circle touches the X -axis and also touches the circle
with centre at (0, 3) and radius 2. The locus of the centre
of the circle is
[AIEEE 2005, 3M]
(a) an ellipse
(c) a hyperbola
(b) a circle
(d) a parabola
108. If a circle passes through the point (a, b ) and cuts the
circle x 2 + y 2 = p 2 orthogonally, then the equation of
the locus of its centre is
[AIEEE 2005, 3M]
(a) x 2 + y 2 − 3ax − 4by + (a 2 + b 2 − p 2 ) = 0
(b) 2ax + 2by − (a 2 − b 2 + p 2 ) = 0
(c) x 2 + y 2 − 2ax − 3by +(a 2 − b 2 − p 2 ) = 0
(d) 2ax + 2by − (a 2 + b 2 + p 2 ) = 0
109. If P is any point of C 1 and Q is another point on C 2 , then
(b) 1.25
is equal to
(c) 1
(d) 0.5
110. If a circle is such that it touches the line L and the circle
C 1 externally, such that both the circles are on the same
side of the line, then the locus of centre of the circle is
(a) ellipse
(c) parabola
The equation of the locus of the mid-points of the
2π
at its
chords of the circle C that subtend an angle of
3
centre is
[AIEEE 2006, 6M]
3
2
27
2
2
(c) x + y =
4
(a) x 2 + y 2 =
(b) x 2 + y 2 = 1
(d) x 2 + y 2 =
9
4
Statement I The tangents are mutually perpendicular.
because
ABCD is a square of side length 2 units. C1 is the circle
touching all the sides of the square ABCD and C 2 is the
circumcircle of square ABCD. L is a fixed line in the same
plane and R is a fixed point.
(a) 0.75
113. Let C be the circle with centre (0, 0) and radius 3 units.
x 2 + y 2 = 169.
(Q. Nos. 109 to 111)
QA 2 + QB 2 + QC 2 + QD 2
(c) x 2 + y 2 − 2 x + 2y − 62 = 0
114. Tangents are drawn from the point (17, 7) to the circle
Paragraph
PA 2 + PB 2 + PC 2 + PD 2
(b) x 2 + y 2 + 2 x − 2y − 62 = 0
(b) hyperbola
(d) pair of straight line
Statement II The locus of the points from which
mutually perpendicular tangents can be drawn to the
given circle is x 2 + y 2 = 338.
[IIT-JEE 2007, 3M]
(a) Statement I is True, statement II is True; statement II is a
correct explanation for statement I
(b) Statement I is True, statement II is True; statement II is
not a correct explanation for statement I
(c) Statement I is True, statement II is False
(d) Statement I is False, statement II is True
115. Consider a family of circles which are passing through
the point ( −1, 1) and are tangent to X -axis. If (h, k ) are
the coordinate of the centre of the circles, then the set
of values of k is given by the interval [AIEEE 2007, 3M]
1
1
(a) − ≤ k ≤
2
2
1
(c) 0 ≤ k ≤
2
1
2
1
(d) k ≥
2
(b) k ≤
Chap 04 Circle
Paragraph
122. The centres of two circles C 1 and C 2 each of unit
(Q. Nos. 116 to 118)
A circle C of radius 1 is inscribed in an equilateral triangle PQR.
The points of contact of C with the sides PQ, QR, RP are D, E, F,
respectively. The line PQ is given by the equation 3x + y − 6 = 0
 3 3 3
and the point D is 
,  . Further, it is given that the origin
 2 2
and the centre of C are on the same side of the line PQ.
116. The equation of circle C is
2
1

(a) ( x − 2 3 ) + (y − 1 ) = 1 (b) ( x − 2 3 ) + y +  = 1

2
2
2
2
(c) ( x − 3 ) 2 + (y + 1 ) 2 = 1 (d) ( x − 3 ) 2 + (y − 1 ) 2 = 1
117. Points E and F are given by
 3 3
(a)  ,  ,( 3, 0 )
 2 2
 3 1
(b)  ,  ,( 3, 0 )
 2 2
 3 3  3 1
(c)  ,  ,  , 
 2 2  2 2
 3 3   3 1
(d)  ,  ,  , 
 2 2   2 2
118. Equations of the sides QR, RP are
[IIT-JEE 2008, (4 + 4 + 4) M]
where, p is a real number, and C : x 2 + y 2 + 6 x − 10y + 30 = 0
Statement I If line L 1 is a chord of circle C, then line L 2 is
not always a diameter of circle C and
Statement II If line L 1 is a diameter of circle C, then line
[IIT-JEE 2008, 3M]
L 2 is not a chord of circle C.
(a) Statement I is True, statement II is True; statement II is a
correct explanation for statement I
(b) Statement I is True, statement II is True; statement II is not a
correct explanation for statement I
(c) Statement I is True, statement II is False
(d) Statement I is False, statement II is True
120. The point diametrically opposite to the point P(1, 0) on the
(b) ( −3, 4 )
(c) ( −3, − 4 )
123. If P and Q are the points of intersection of the circles
x 2 + y 2 + 3x + 7y + 2p − 5 = 0 and
x 2 + y 2 + 2x + 2y − p 2 = 0 then there is a circle passing
through P, Q and (1, 1) for :
[AIEEE 2009, 4M]
(a) all except one value of p
(b) all except two values of p
(c) exactly one value of p
(d) all values of p
3x − 4y = m at two distinct points if
119. Consider L 1 : 2x + 3y + p − 3 = 0 ; L 2 : 2x + 3y + p + 3 = 0
(a) (3, − 4 )
radius are at a distance of 6 units from each other. Let
P be the mid point of the line segment joining the
centres of C 1 and C 2 and C be a circle touching circles
C 1 and C 2 externally. If a common tangent to C 1 and C
passing through P is also a common tangent to C 2 and
C, then the radius of the circle C is [IIT-JEE 2009, 4M]
124. The circle x 2 + y 2d = 4 x + 8y + 5 intersects the line
2
2
1
(a) y =
x + 1, y = −
x − 1 (b) y =
x, y = 0
3
3
3
3
3
(c) y =
x + 1, y = −
x − 1 (d) y = 3 x, y = 0
2
2
circle x 2 + y 2 + 2x + 4y − 3 = 0 is
335
[AIEEE 2008, 3M]
(d) (3, 4)
121. Tangents drawn from the point P(1, 8) to the circle
x 2 + y 2 − 6x − 4y − 11 = 0 touch the circle at the points A
and B. The equation of the circumcircle of the triangle PAB
is
[IIT-JEE 2009, 3M]
(a) −35 < m < 15
(c) 35 < m < 85
[AIEEE 2010, 4M]
(b) 15 < m < 65
(d) −85 < m < −35
125. The circle passing through the point ( −1, 0) and
touching the Y -axis at (0, 2) also passes through the
point.
[IIT-JEE 2011, 3M]
 3 
(a)  − , 0
 2 
 5 
(b)  − , 2
 2 
 3 5
(c)  − , 
 2 2
(d) (−4, 0)
126. The straight line 2x − 3y = 1 divides the circular region
x 2 + y 2 ≤ 6 into two parts.
 3   5 3   1 1   1 1  
If S =  2,  ,  ,  ,  , −  ,  ,   then the number
 4   2 4   4 4   8 4  
of point(s) in S lying inside the smaller part is
[IIT-JEE 2011, 4M]
127. The two circles x + y = ax and x + y 2 = c 2 (c > 0)
2
2
2
touch each other if
(a) | a| = c
(c) | a| = 2c
[AIEEE 2011, 4M]
(b) a = 2c
(d) 2| a | = c
128. The locus of the mid-point of the chord of contact of
tangents drawn from points lying on the straight line
4 x − 5y = 20 to the circle x 2 + y 2 = 9 is
[IIT-JEE 2012, 3M]
(a) 20 ( x 2 + y 2 ) − 36 x + 45y = 0
(b) 20 ( x 2 + y 2 ) + 36 x − 45y = 0
(a) x 2 + y 2 + 4 x − 6y + 19 = 0 (b) x 2 + y 2 − 4 x − 10y + 19 = 0
(c) 36 ( x 2 + y 2 ) − 20 x + 45y = 0
(c) x 2 + y 2 − 2 x + 6y − 29 = 0 (d) x 2 + y 2 − 6 x − 4y + 19 = 0
(d) 36 ( x 2 + y 2 ) + 20 x − 45y = 0
336
Textbook of Coordinate Geometry
[JEE Advanced 2014, 3M]
Paragraph
(Q. Nos. 129 and 130)
A tangent PT is drawn to the circle x 2 + y 2 = 4 at the point
P( 3 , 1) . A straight line L, perpendicular to PT is a tangent to
the circle ( x − 3) 2 + y 2 = 1.
129. A possible equation of L is
(a) x − 3y = 1
(b) x + 3y = 1
(c) x − 3y = −1
(d) x + 3y = 5
130. A common tangent of the two circles is
[IIT-JEE 2012, (3 + 3 ) M]
(a) x = 4
(c) x + 3y = 4
(b) y = 2
(d) x + 2 2y = 6
131. The length of the diameter of the circle which touches
the X -axis at the point (1, 0) and passes through the
point (2, 3) is
[AIEEE 2012, 4M]
10
3
6
(c)
5
(a)
3
5
5
(d)
3
(b)
132. The circle passing through (1, − 2) and touching the axis
of x at (3, 0) also passes through the point
[JEE Main 2013, 4M]
(a) ( −5, 2 )
(b) (2, − 5 )
(c) (5, − 2 )
(d) ( −2,5 )
(a) radius of S is 8
(b) radius of S is 7
(c) centre of S is ( −7,1 )
(d) centre of S is ( −8, 1 )
136. Locus of the image of the point (2, 3) in the line
(2x − 3y + 4 ) +k ( x − 2y + 3) = 0, k ∈ R, is a
[JEE Main 2015, 4M]
(a) circle of radius 2
(b) circle of radius 3
(c) straight line parallel to X -axis
(d) straight line parallel to Y -axis
137. The number of common tangents to the circles
x 2 + y 2 − 4 x −6x − 12 = 0 and x 2 + y 2 + 6x + 18y + 26 = 0,
is
[JEE Main 2015, 4M]
(a) 3
(c) 1
(b) 4
(d) 2
138. The centres of those circles which touch the circle,
x 2 + y 2 − 8x − 8y − 4 = 0, externally and also touch the
[JEE Main 2016, 4M]
X -axis, lie on
(a) a hyperbola
(b) a parabola
(c) a circle
(d) an ellipse which is not a circle
139. If one of the diameters of the circle, given by the
equation, x 2 + y 2 − 4 x + 6y − 12 = 0, is a chord of a circle
S, whose centre is at ( −3, 2) , then the radius of S is
[JEE Main 2016, 4M]
133. Circle(s) touching X -axis at a distance 3 from the origin
and having an intercept of length 2 7 on Y -axis is (are)
(a) x 2 + y 2 − 6 x + 8y + 9 = 0
[JEE Advanced 2013, 3M]
(b) x 2 + y 2 − 6 x + 7y + 9 = 0
(c) x 2 + y 2 − 6 x − 8y + 9 = 0
(d) x 2 + y 2 − 6 x − 7y + 9 = 0
134. Let C be the circle with centre at (1, 1) and radius = 1. If T
is the circle centred at (0, y ), passing through origin and
touching the circle C externally, then the radius of T is
equal to
[JEE Main 2014, 4M]
(a)
(c)
1
2
(b)
3
2
(d)
1
4
3
2
(b) 10
(d) 5 3
140. Let RS be the diameter of the circle x 2 + y 2 = 1, where S
is the point (1, 0). Let P be a variable point (other than R
and S) on the circle and tangents to the circle at S and P
meet at the point Q. The normal to the circle at P
intersects a line drawn through Q parallel to RS at point
E. Then the locus of E passes through the point(s)
[JEE Advanced 2016, 4M]
1 1 
(a)  , 
3 3
1 1 
(c)  ,− 
3
3
 1 1
(b)  , 
 4 2
 1 1
(d)  , − 
 4 2
141. For how many values of p, the circle
135. A circle S passes through the point (0, 1) and is
orthogonal to the circles ( x − 1) + y = 16 and
2
x 2 + y 2 = 1. Then
(a) 5
(c) 5 2
2
x 2 + y 2 + 2x + 4y − p = 0 and the co-ordinate axes have
exactly three common points? [JEE Advanced 2017, 3M]
Chap 04 Circle
Answers
Exercise for Session 7
Exercise for Session 1
1. (d)
6. (c)
2. (b)
7. (d)
11. (b)
12. (a)
3. (b)
8. (a)
4. (b)
9. (c)
5. (a)
10. (c)
− 2 4
13. 
, ; 2
 5 5
1. (d)
6. (a)
12. (a)
15. x2 + y2 − 2x − 4 y − 4 = 0
14. x + y − 6x − 6 y + 9 = 0
2
2
5. (c)
10. (a)
11. (d)
− 16 − 31

16. 
,

 21
63 
18. x + y − 5 = 0
Chapter Exercises
Exercise for Session 2
1. (c)
6. (a)
2. (b)
7. (c)
11. (c)
12. (a)
14. (− 2, − 7)
3. (b)
8. (b)
4. (c)
9. (d)
5. (a)
10. (c)
13. x2 + y2 − y − 16 = 0 ;  0,
1
;
 2
15. x2 + y2 − 2x − 3 y − 18 = 0
65
2
16. (x2 + y2 − 4x − 2 y + 4) = 0
17. x2 + y2 − 6x + 2 y − 15 = 0
Exercise for Session 3
2. (a)
7. (a)
12. (c)
15. λ ∈ (− 1, 4)
3. (a)
4. (d)
5. (d)
8. (a)
9. (d)
10. (b)
13. x2 + y2 ± 10x − 6 y + 9 = 0
16. x2 + y2 − 4x − 6 y = 0
Exercise for Session 4
1. (c)
2. (c)
3. (c)
4. (d)
5. (b)
6. (b)
7. (b)
8. (a)
9. (d)
10. (b)
11. (d)
12. (a)
14. (i) 3x − 4 y + 20 = 0 and 3x − 4 y − 10 = 0 (ii) 4x + 3 y + 5 = 0
and 4x + 3 y − 25 = 0,
15. centre of the circle (0, 1, ± r 2 ), where r is radius
16. 15, − 35
2. (b)
7. (b)
12. (d)
16. 3x + 2 y − 13 = 0
2. (a)
3. (b)
4. (b)
5. (a)
6. (d)
7. (a)
8. (a)
9. (c)
10. (b)
11. (c)
12. (c)
13. (d)
14. (d)
15. (a)
16. (d)
17. (d)
18. (d)
19. (b)
20. (a)
21. (c)
22. (b)
23. (d)
24. (c)
25. (a)
26. (b)
27. (b)
28. (c)
29. (b)
30. (a)
31. (b,d) 32. (b,d)
33. (a,b,c,d)
34. (a,d) 35. (b,d)
36. (b,c) 37. (a,b,c,d)
38. (b,d) 39. (a,d) 40. (a,d)
41. (a,c) 42. (a,c)
43. (a,c)
44. (a,c,d) 45. (a,b) 46. (b)
47. (c)
48. (b)
49. (b)
50. (d)
51. (b)
52. (a)
53. (c)
54. (b)
55. (b)
56. (b)
57. (b)
58. (a)
59. (a)
60. (d)
61. (d)
62. (a)
63. (c)
64. (a)
65. (d)
66. (c)
67. (4)
68. (6)
69. (1)
70. (4)
71. (9)
72. (3)
73. (4)
74. (3)
75. (9)
76. (7)
77. (A) → (p, s); (B) → (q, r); (C) → (q); (D) → (p, q, s)
78. (A) → (q, r) ; (B) → (p); (C) → (p) (D) → (s)
79. (A) → (q, s); (B) → (r,t); (C) → (p,s)
80. (A) → (p, q, s); (B) → (p, q, s, t); (C) → (p, q, r, s)
81. (d)
82. (a)
83. (d)
84. (d)
85. (c)
86. (b)
87. (b)
88. (a)
89. (x2 + y2 − x − y) = 0
90. x2 + y2 − 6x − 2 y + 1 = 0
91. x2 + y2 ±
m (m + n ) x ±
93. (5)
5. (d)
10. (a)
97. 3 (3 +
10 )
 − 1,
 2
100. (− 4, 4) or 
101. 4a 2 + 4b 2 − c2 −
Exercise for Session 6
5. (c)
11. (b)
14. 2 2
6. (b)
12. (b)
16. Direct common tangents are 3x + 4 y = 57, 7x − 24 y = 233,
Transverse common tangents are 4x − 3 y = 26, 24x + 7 y = 156
105. (d)
111. (c)
117. (a)
123. (a)
129. (a)
135. (b,c)
141. (2)
 9 − 12  or  − 9 , 12 



5 5 
 5 5
4 3
sq units
98.
9
(a1 c2 + a2 c1 ) x + (b1 c2 + b2 c1 ) y = 0
1

2
8ab + 4bc − 4ca < 0
96.  ,
99. a1 a2 (x2 + y2 ) +
 5
 
 16 
1. (a)
2. (c)
3. (a)
4. (d)
7. (c)
8. (b)
9. (b)
10. (d)
13. 4x2 + 4 y2 + 30x − 13 y − 25 = 0
n (m + n ) y = 0
94. 18 sq units
95. 1.5 units
3. (c)
4. (b)
8. (c)
9. (b)
14. 8 sq units
17.
1. (b)
92. 25x2 + 25 y2 − 20x + 2 y − 60 = 0
Exercise for Session 5
1. (c)
6. (d)
11. (c)
4. (b)
9. (b)
2
18. (x + 1)2 + ( y − 3)2 = 4; (− 1, 3) ; 2
1. (d)
6. (c)
11. (a)
3. (a)
8. (d)
17. 4x2 + 4 y2 + 6x + 10 y − 1 = 0
16. x + y − 2x − 8 y + 15 = 0
2
2. (c)
7. (b)
106. (b)
112. (d)
118. (d)
124. (a)
130. (d)
136. (a)
107. (d)
113. (d)
119. (c)
125. (d)
131. (a)
137. (a)
108. (d)
114. (a)
120. (c)
126. (2)
132. (c)
138. (b)
109. (a)
115. (d)
121. (b)
127. (a)
133. (a,c)
139. (d)
110. (b)
116. (d)
122. (8)
128. (a)
134. (b)
140. (a,c)
337
Solutions
(5, 6)
2
(2, 0)
1.

n2 
AB 2 = 4 AM 2 = 4  4 −  = 2(8 − n 2 )
2

Hence, required sum = 2(8 − 1 2 + 8 − 2 2 ) = 22
B
i.e.
M
(0, 0) O
∴
A
2
3 5
 3 5 − 2  3 5 + 2
r1r2 = 

.
 2   2 
41
4
5. Any circle which touches 3x + 4y − 7 = 0 at (1, 1) will be of the
form
S ( x,y ) ≡ ( x − 1 ) 2 + (y − 1 ) 2 + λ (3 x + 4y − 7 ) = 0
=
2. Tangent at (5 2 cosθ,5 2 sin θ ) is
x cosθ + y sin θ = 5 2
∴
OP = 5 2 secθ, OP, = 5 2 cosecθ
Since,
S(2, 3 ) = 16 ⇒1 + 4 + λ(11 ) = 16
∴
λ =1
So, required circle will be x 2 + y 2 + x + 2y − 5 = 0
∴ Area (∆PP1P2 ) = 2 × area of ∆OPP1
1
= 2 × × 5 2 secθ × 5 2 cosec θ
2
100
,
=
sin 2θ
Area ( ∆PP1P2 ) min = 100
π
θ=
⇒ OP = 10
⇒
4
⇒
P ≡ (10, 0 ), ( −10, 0 )
3. Let S ≡ x 2 + y 2 − 2 = 0, P ≡ (2,2)
∴
6. Let S ≡ x 2 + y 2 − 2x + 4y − λ = 0
for interval point P(2,8 ),
4 + 64 − 4 + 32 − λ < 0
⇒
λ > 96
and x-intercept = 2 (1 + λ ) and y-intercept = 2 ( 4 + λ )
S1 = 2 2 + 2 2 − 2 = 6 > 0
⇒ P lies outside the circle
PA. PB = ( PT ) 2 = S1 = 6
PB
given
=3
PA
…(i)
…(ii)
given
2 (1 + λ ) < 0 ⇒ λ < −1
…(ii)
and
2 ( 4 + λ ) < 0 ⇒ λ < −4
…(iii)
from Eqs. (i), (ii) and (iii), we get
λ ∈φ
7. Let S ≡ x 2 + y 2 + 2x + 4y − 4 = 0
Its centre C ≡ ( −1, − 2 ) and radius r = 3. The points on the circle
which are nearest and farthest to the point P (a, b ) are Q and R
respectively. Thus, the circle centred at Q having radius PQ will
be the smallest circle while the circle centred at R having
radius PR will be the largest required circle.
P (a, b)
B
T
Q
(0, 0)
A
(2, 2) P
From Eqs. (i) and (ii), we get
PA = 2, PB = 3 2
⇒
AB = PB − PA = 2 2 = Diameter of circle
Hence, chord AB passes through the centre ( 0, 0 ), y = x
4. Let S ≡ ( x − 2) 2 + y 2 − 4 = 0
Its centre C ≡ (2, 0 ) and radius r = 2
Distance between C ≡ (2, 0 ) and (5, 6) is (2 − 5 ) 2 + ( 0 − 6 ) 2
…(i)
C
R
Hence, the difference between their radii is
PR − PQ = QR = 2 × 3 = 6
8. Radius = (a − π ) 2 + (b − e ) 2
= irrational = k
∴ Circle ( x − π ) 2 + (y − e ) 2 = k 2
Chap 04 Circle
9. Given λL2L3 + µL3L1 + vL1L2 = 0
⇒
⇒ λ(y − m2x − c 2 )(y − m3x − c 3 ) + µ(y − m3x − c 3 )(y − m1x − c1 )
+ v (y − m1x − c1 )(y − m2x − c 2 ) = 0
2
for circle coefficient of x = coefficient of y 2 and coefficient of
xy = 0,
then, λ (m2m3 − 1 ) + µ(m3m1 − 1 ) + v (m1m2 − 1 ) = 0
and
(m2 + m3 ) λ + (m3 + m1 )µ + (m1 + m2 )v = 0
10. Q f ( x,y ) ≡ x 2 + y 2 + 2ax + 2by +c = 0
⇒
f ( x, 0 ) = 0 ⇒ x 2 + 2ax + c = ( x − 2 ) 2
∴
and
a = −2, c = 4
f ( 0, y ) = 0 ⇒ y 2 + 2by + c = (y − 2 )(y − 3 )
∴
2b = −5, c = 6
5
or
b = − ,c =6
2
Clearly, that data are not consistent.
11. Given, (1 + αx )n = 1 + 8x + 24x 2 + ...
n(n − 1 )
⇒
1 + n(αx ) +
(αx ) 2 + ...
1.2
= 1 + 8 x + 24 x 2 + ...
339
a 2 + (a + 1 ) 2 < 5
⇒
∴
2a 2 + 2a − 24 < 0
a 2 + a − 12 < 0
(a + 4 )(a − 3 ) < 0
⇒ −4 < a < 3 and for I and II quadrant
a +1 > 0
∴
a > −1
Hence, −1 < a < 3
13. Let S ( x,y ) ≡ x 2 + y 2 + 2gx + 2 fy + c = 0
⇒
⇒
S ( x, 2 ) = 0
x 2 + 4 + 2 gx + 4 f + c = ( x − 1 ) 2
∴
and
…(i)
g = −1, 3 + 4 f + c = 0
S (1, y ) = 0 ⇒1 + y 2 + 2 g + 2 fy + c = (y − 0 )(y − 2 )
∴
f = −1, 1 + 2 g + c = 0
From Eqs. (i) and (ii), we get
g = −1, f = −1, c = 1
∴ Equation of required circle is
x 2 + y 2 − 2 x − 2y + 1 = 0
…(ii)
14. See the diagram. Since, there is no condition on θ, Q can be
placed either at Q1 or Q2 for a particular position of P. So
option (a) and (b) cannot be definitely true.
B
C
Q2(a–q)
A
P(q)
Q1(a–q)
q
P (2, 4)
O
Equating the coefficients of x and x 2, we get
nα (nα − α )
nα = 8,
= 24
1.2
8 (8 − α )
or
= 24 ⇒ 8 − α = 6
2
∴
α = 2 and n = 4
x −2 y − 4
Equation of line is
=
= r , then point
cosθ sin θ
(2 + r cosθ, 4 + r sin θ ) lies on the circle x 2 + y 2 = 4,
then,
(2 + r cosθ ) 2 + ( 4 + r sin θ ) 2 = 4
or
r 2 + 4r (cosθ + 2 sin θ ) + 16 = 0
16
PA ⋅ PB = r1r2 =
= 16
1
PA ⋅ PB = ( PC ) 2 = 2 2 + 4 2 − 4 = 16
∴
Aliter :
12. For interior point OP < 5
C
5
B
(–5, 0)
(a, a+1)
P
5
O (0, 0)
y=0
A
(5, 0)
P¢(– q)
Consider a line through origin y = mx. If Q and P are reflection
of each other with line mirror y = mx
(Slope of PQ) ×m = −1
 sin θ − sin(α − θ ) 
or
m
 = −1
 cosθ − cos(α − θ )

 2θ − α  
α

 2 cos  .sin 
2
2 
m
= −1
 α − 2θ  
α


 2 sin   .sin 

2
2 
or
α

m  − cot  = −1

2
or
α
m = tan  
 2
15. Since, x 2 + y 2 < 25 and x and y are integers, the possible values
of x and y ∈ ( 0, ± 1, ± 2, ± 3, ± 4 ). Thus, x and y can be chosen in
9 × 9 = 81 ways. However, we have to exclude cases (3, 4), ( 4, 3 )
and ( 4, 4 ) i.e. 3 × 4 = 12 cases. Hence, the number of permissible
values
= 81 − 12 = 69
340
Textbook of Coordinate Geometry
16. Q The point ([ P + 1],[ P ]) lies inside the circle
From Eqs. (i) and (ii), we get
(x − a )2 + y 2 = a 2
x 2 + y 2 − 2 x − 15 = 0, then
[ P + 1 ]2 + [ P ]2 − 2[ P + 1 ] − 15 < 0
⇒
x 2 + y 2 − 2ax = 0
Y
(1, 0)
O
A
q
X′
⇒
([ P ] + 1 ) 2 + [ P ]2 − 2([ P ] + 1 ) − 15 < 0
⇒
2[ P ]2 − 16 < 0 ⇒[ P ]2 < 8
([ P ] + 1 ) + [ P ] − 2([ P + 1 ]) − 7 > 0
⇒
([ P ] + 1 ) 2 + [ P ]2 − 2([ P ] + 1 ) − 7 > 0
[ P ]2 > 4
for circle coefficient of xy = 0
i.e.
−5 + λ = 0,
∴
λ =5
2
∴Circle is 2 x + 2y 2 + 7 x − 5y + 3 = 0
7
5
3
⇒
x2 + y 2 + x − y + = 0
2
2
2
 7 5
∴Centre is  − , 
 4 4
…(ii)
From Eqs. (i) and (ii), we get
4 < [ P ]2 < 8 which is impossible
∴For no value of ‘P’ the point will be within the region.
17. The circles are x 2 + y 2 = 2 2 and ( x − 4) 2 + (y − 0) 2 = 3 2
21.
Y

 λ2 (1 − λ ) 2
− 5 ≤ 5
 +
4

4
⇒
For the point P to enclose greatest area, the arbitrarily chosen
1
3 
point should be  , 0 and P should move in a circle of radius
2 
2
2
⇒
2
18. Since, y = | x| + c and x + y 2 − 8 | x | − 9 = 0 both are symmetrical
2
about Y -axis for x > 0, y = x + c.
Equation of tangent to circle x 2 + y 2 − 8 x − 9 = 0
Parallel to y = x + c is y = ( x − 4 ) + 5 (1 + 1 )
⇒
y = x + (5 2 − 4 )
for no solution c > 5 2 − 4,
c ∈ (5 2 − 4, ∞ )
∴
⇒
and
A ≡ (a cosθ, a sin θ )
M ≡ (a + a cosθ, a sin θ )
x = a + a cosθ
( x − a ) = a cosθ
y = a sinθ
φ( 0, λ ) = 0 + λ2 + 0 + 2 fλ + c = 0
have equal roots,
2f
c
Then, 2 + 2 = −
and 2.2 =
1
1
∴
f = −2 and c = 4
and φ( λ , 0 ) ≡ λ2 + 0 + 2 gλ + 0 + c = 0
∴
λ2 + 2 gλ + c = 0
Here, c = 4
∴
λ2 + 2 gλ + 4 = 0
have roots 4/5, 5
4
+ 5 = −2 g
∴
5
29
⇒
g=−
10
19. Let ∠AOL = θ
∴
∴
2 λ2 − 2 λ − 119 ≤ 0
1 − 239
1 + 239
∴
≤λ≤
2
2
⇒
−7.2 ≤ λ ≤ 8.2
∴ λ = −7, − 6, − 5,...,7, 8
22. Let φ( x,y ) ≡ x 2 + y 2 + 2gx + 2 fy + c = 0
∴
2
x + y − 3 x + 2 = 0.
2
λ2 + (1 − λ ) 2 − 20 ≤ 100
⇒
X
O
3

 1
The locus of P is  x −  + (y − 0 ) 2 =  

 2
2
x 2 + y 2 = 2ax
20. Equation of circle is (2x − y + 1)( x − 2y + 3) + λxy = 0
∴
∴
X
Y′
⇒
2
2[ P ]2 − 8 > 0
L
O
…(i)
Q Circles are concentric
∴point ([ P + 1 ],[ P ]) out side the circle
x 2 + y 2 − 2x − 7 = 0
⇒
M
a
√8
2
a
…(i)
∴
 29 
Centre ≡ ( − g, − f ) =  , 2
 10 
(approx.)
Chap 04 Circle
23. Circle is x 2 + y 2 = r 2 cos2 θ + r 2 sin 2 θ
26. Coordinates of P, Q, R are (cosα ,sin α ), (cosβ,sin β) and
(cos γ,sin γ ) respectively.
and
A ≡ ( −1, 0 )
x2 + y 2 = r 2
Equation of tangent at θ is
x cosθ + y sin θ = r
π
π
π



and at θ +  is x cosθ +  + y sin θ +  = r



3
3
3
…(i)
∴
⇒
x cosθ + y sin θ − x 3 sin θ + y 3 cosθ = 2r
⇒
r − 3( x sin θ − y cosθ ) = 2r
or
x sin θ − y cosθ = −
α
AP = ((1 + cosα ) 2 + sin 2 α ) = 2 cos 
 2
 β
AQ = ((1 + cosβ ) 2 + sin 2 β ) = 2 cos 
 2

1

1
3
3
⇒ x  cosθ −
sin θ + y  sin θ +
cosθ = r

2

2
2
2
 γ
AR = ((1 + cos γ ) 2 + sin 2 γ ) = 2 cos  
 2
r
3
Squaring and adding Eqs. (i) and (ii), then we get
4r 2
x2 + y 2 =
⇒ 3( x 2 + y 2 ) = 4r 2
3
24. Let MN be the diameter of the circle whose equation is
4y = x + 7
…(ii)
Q AP , AQ, AR are in GP, then
α
 β
 γ
cos   , cos   , cos   are also in GP.
 2
 2
 2
27. Let equation of circle be
x 2 + y 2 + 2 gx + 2 fy + c = 0
…(i)
M
C
It pass through (2, 0) and (0, 4), then 4 + 0 + 4 g + 0 + c = 0
(c + 4 )
and 0 + 16 + 0 + 8 f + c = 0
g=−
⇒
4
(c + 16 )
⇒
f =−
8
Q
Radius r = ( g 2 + f 2 − c )
(c + 4 ) 2 (c + 16 ) 2

= 
+
−c
64
 16

B
O
D
L
 4 {c 2 + 8c + 16 } + {c 2 + 32c + 256 } − 64c 
= 

64


A
N
and coordinates of A and B are ( −3, 4 ) and (5, 4 ) respectively.
Equation of ⊥ bisector of AB is [ L ≡ (1, 4 )]
1
y − 4 = − (x − 1)
(Qslope of AB = 0 )
0
...(ii)
∴
x =1
Solving Eqs. (i) and (ii), we get the coordinates of the centre of
the circle as (1, 2)
∴
OL = (1 − 1 ) + ( 4 − 2 ) = 2
2
341
2
 5c 2 + 320
= 

 64 
For minimum radius c = 0
∴
g = −1, f = −2
Required circle is x 2 + y 2 − 2 x − 4y = 0
28. The particle which moves clockwise is moving three times as
fast as the particle moving anticlockwise (Q speed in clockwise
3v and in anticlockwise v).
P
∴
BC = 2OL = 4 unit
AB = 8 unit
∴Area of rectangle ABCD = 4 × 8
= 32 sq units.
25. QOC ⋅ OA = ab = OB ⋅ OD
∴A, B, C, D are concyclic.
v
Q
(1, 0)
3v
Y
(0, b) D
B
(0, a)
O
ax+
b y=
ab
C
(a, 0)
A
(b, 0 )
bx+ay=ab
X
 3
This mean the clockwise particle travels   th of the way
 4
 1
around the circle, the anticlockwise particle will travel   th
 4
of the way around the circle. So, the second particle will meet
at P( 0,1 ).
Using the same logic, they will meet at Q( −1, 0 ), when they
meet the second time.
342
Textbook of Coordinate Geometry
BP
=α
PA
BP : PA = α :1
2
29. Q
or
⇒
3 + α 4 
2
2
,
∴Coordinates of P is 
 , P lie on x + y = 4
 1 + α α + 1
⇒
⇒
(α + 3 ) 2 + 16 = 4(α + 1 ) 2
3α 2 + 2α − 21 = 0
…(i)
O
A
(1, 0)
27
=0
4
( x + 2 ) 2 + (y − 3 ) 2 = 6.25
x 2 + y 2 + 4 x − 6y +
⇒
B (3, 4)
(–2, 0)
25
= x12 + y12 + 4 x1 − 6y1 + 13
4
4 x12 + 4y12 + 16 x1 − 24y1 + 27 = 0
∴ Locus of mid-point is
Y
(0, 2)
 5
2
2
  = ( x1 + 2 ) + (y1 − 3 )
 2
31. Q r = a cosθ + b sin θ
P
⇒
r 2 = a(r cosθ ) + b(r sin θ )
X
(2, 0)
or
x 2 + y 2 = ax + by
or
x + y − ax − by = 0
2
(Qx = r cosθ, y = r sin θ )
2
…(i)
Circle pass through O (origin)
given
∠AOB = 90 °
Q
O
and
BQ β
=
QA 1
90°
BQ
−1 = β −1
QA
BQ :QA = β :1 or
B
AB (β − 1 )
=
QA
1
AB :QA = (β − 1 ):1
 β − 3 −4 
∴Coordinates of Q is 
,

 β −1 β − 1
a b
∴AB is Diameter of circle Eq. (i), Centre is  , 
 2 2
Q lie on x 2 + y 2 = 4
∴
⇒
A
C a, b
2 2
(β − 3 ) 2 + 16 = 4 (β − 1 ) 2
3β 2 − 21β − 21 = 0
Hence, α is a root of 3 x 2 + 2 x − 21 = 0
and β is a root of 3 x 2 − 2 x − 21 = 0
π
30. Q ∠ACB =
3
2π
∴ ∠AOB =
3
(given)
∴Locus of mid point AB (Q mid-point of AB is C)
a
b
x = ,y =
2
2
−3 − 2
32. Slope of PC =
=1
2 −7
If tanθ = 1
∴
θ = 45 °
x −7 y −2
Equation of PA is
=
=r
1
1
2
2
B
4
B
5
(–2, 3)
p/3
O
p/3
p/3
5
P
(7, 2)
A
∴
π
3
Let mid-point of chord AB is ( x1, y1 )
∴ In ∆ AOM,
∠AOM = ∠BOM =
C
45°
r
r 

,2 +
7 +
 lie on circle,

2
2
2
2
r 
r 
r 
r 




then, 7 +
 −3 = 0
 + 2 +
 − 47 +
 +6  2 +




2
2
2
2
( x1 + 2 ) + (y1 − 3 )
π OM
=
=
3 OA
5
2
cos
4
M(x1, y1)
C
⇒
A
2
∴
r2 = −5 2 ± 4
∴
r = −5 2 ± 4
Chap 04 Circle
36. Let M(α ,β)

−5 2 ± 4
−5 2 ± 4 
,2 +
∴ Points 7 +


2
2 
Q
∴
⇒ (2 ± 2 2, − 3 ± 2 2 )
AM = 2 AB ⇒ AB + BM = 2 AB
AB = BM
Taking + ve sign for A, −ve sign for B.
33. Let S ≡ x 2 + y 2 + 2gx + 2 fy + c = 0
M(a, b)
S1 ≡ x 2 + y 2 − 4 = 0
B
S 2 ≡ x + y − 6 x − 8y + 10 = 0
2
2
(–2, 3)
S 3 ≡ x 2 + y 2 + 2 x − 4y − 2 = 0
∴Common chords are
…(i)
S − S1 ≡ 2 gx + 2 fy + c + 4 = 0
…(ii)
S − S 2 ≡ (2 g + 6 ) x + (2 f + 8 )y + c − 10 = 0
…(iii)
S − S 3 ≡ (2 g − 2 ) x + (2 f + 4 )y + c + 2 = 0
For cutting the extremities of diameter, chords Eqs. (i), (ii) and
(iii) pass through the centres of S1, S 2 and S 3 respectively, then
∴
c + 4 = 0, (2 g + 6 ) 3 + (2 f + 8 ) 4 + c − 10 = 0
and (2 g − 2 )( −1 ) + (2 f + 4 )(2 ) + c + 2 = 0
after solving c = −4, g = −2, f = −3
A (0, 3)
∴B is the mid point of AM.
 α 3 + β
∴Coordinates of B are  ,
 which lie on
2 2 
x 2 + 4 x + (y − 3 ) 2 = 0
2
α2
α 3 + β 
+ 4× + 
− 3 = 0

4
2  2
34.
α 2 + 8α + (β − 3 ) 2 = 0
A
∴Requires locus is
x 2 + 8 x + (y − 3 ) 2 = 0
2
O
q/2
q/2
l
P (l, 0)
2
B
Q
⇒
⇒
or
π
2π
π θ π
<θ <
⇒
< <
2
3
4 2 3
1
3
 θ
< sin   <
 2
2
2
1
2
3
< <
2 λ
2
4
2 2>λ>
3
2
λ
>
2
3
or
2>
or
4
< λ <2 2
3
x 2 + y 2 + 8 x − 6y + 9 = 0
x
y
37. x cosα + y sin α = 2 or
+
=1
2 secα 2 cosec α
or
2secα = + ve in IV
and 2 cosec α = −ve in IV
value of g in I quadrant
 g cosα + g sin α − 2
g=



1
xc
N
M
(8/3, 8/3)
(1/3, 1/3)
16 1
16 1
− = 5 and β = − = 5
3 3
3 3
(α ,β ) ≡ (5, 5 ) lie on the circle
∴
52 + 52 − 4 × 5 − 2 × 5 − c = 0
c = 20
2
or
c − 40c + 400 = 0
α=
ys
ina
=2
2
(g, –g)
1
Q
∴
a+
(0, 0)
P(a, b)
1
os
(g , g )
35. It is clear that N is the mid-point of M and P
1
O
xc
os
a+
in
ys
a=
2
cosα + sin α ± 1
and value g of in IV quadrant
g cosα − g sin α − 2
g=
1
∴
g=
⇒
± g = g(cosα − sin α ) − 2
2
g=
cosα − sin α ± 1
Equation of circles in I quadrant
( x − g ) 2 + (y − g ) 2 = g 2
And in IV quadrant is ( x − g ) 2 + (y − g ) 2 = g 2
343
344
Textbook of Coordinate Geometry
38. ∴
S1
α
cot   =
2
 2
(g + f 2 − c )
∴
(g + f − c )
α
tan   =
 2
S1
∴
2
41. Circle possible in I region and centre of circle on X -axis.
If centre is (h, 0 ), then
| h − 0 − 1|
7
= (h − 3 ) 3 +
2
2
∴
h = 6,4
2
 (g 2 + f 2 − c ) 

α = 2 tan −1 


S1


Y
0
y=
x– (3, √7/2)
(0, 1)
B
I
S1
a/2
P
a/2
(x1 , y )
1
S2
C
(–g,–f)
(1, 0)
(g2+f2–c)
X
(h, 0)
x+y=1
A
(0, –1)
39. For condition of tangency
∴
| 4 λ + 3 λ − 12|
= λ ⇒ 7 λ − 12 = ±5 λ
5
λ = 1, 6
Then, centres are (6, 0) and (4, 0)
42. QC 2 is the director circle of C1
∴Equation of C 2 is
x 2 + y 2 = 2(2 ) 2 = 8
Y
B
(0, 4)
(1, 1)
Again C 3 is the director circle of C 2. Hence, the equation of C 3
is
x 2 + y 2 = 2(8 ) = 16
l
43. The equation of tangent in terms of slope of x 2 + y 2 = 25 is
P
O
A
(3, 0)
y = mx ± 5 (1 + m 2 )
X
Given Eq. (i), pass through ( −2,11 ), then
11 = −2m ± 5 (1 + m 2 )
40. Let coordinates of P ≡ (3 cosα ,3 sin α )
Let x-coordinate of Q is x1, then
y-coordinate of Q is −7 x1 − 3
∴
Q ≡ ( x1, − 7 x1 − 3 )
Q x − y + 1 = 0 is the perpendicular bisector of PQ, then
mid-point of PQ lie on x − y + 1 = 0
3 cosα + x1 3 sin α − 7 x1 − 3
−
+1 = 0
⇒
2
2
⇒
8 x1 + 3 cosα − 3 sin α + 5 = 0
⇒
24 x1 + 9 cosα − 9 sin α + 15 = 0
and slope of ( x − y + 1 = 0 ) × slope of PQ = −1
3 sin α + 7 x1 + 3
⇒
1×
= −1
3 cosα − x1
⇒
3 sin α + 7 x1 + 3 = −3 cosα + x1
⇒
6 x1 + 3 sin α + 3 cosα + 3 = 0
⇒
24 x1 + 12 sin α + 12 cosα + 12 = 0
Subtracting Eqs. (i) and (ii), we obtain
−3 cosα − 21 sin α + 3 = 0
⇒
(1 − cosα ) = 7 sin α
⇒
(1 − cosα ) 2 = 49(1 − cos2 α )
⇒
⇒
(1 − cosα ) 2 = 49(1 + cosα )(1 − cosα )
(1 − cosα )(1 − cosα − 49 − 49 cosα ) = 0
24
∴
cosα = 1 and cosα = −
25
 72 21
∴Coordinates of P are (3, 0) and  − ,  .
 25 25
…(i)
…(i)
squaring both sides, then we get
21m 2 − 44m − 96 = 0
⇒
(7m − 24 )(3m + 4 ) = 0
4 24
m=− ,
∴
3 7
There from Eq. (i) we get required tangents are
24 x − 7y ± 125 = 0 and 4 x + 3y = ±25
Hence, tangents are 24 x − 7y + 125 = 0 and 4 x + 3y = 25
44. C1 ≡ x 2 + y 2 − 2x − 4y − 4 = 0
and
…(ii)
C 2 ≡ x + y + 2 x + 4y + 4 = 0
2
2
…(i)
…(ii)
∴Radical axis is C1 − C 2 = 0
⇒
−4 x − 8y − 8 = 0
or x + 2y + 2 = 0 which is L = 0
(a) Option is correct.
Centre and radius of C1 = 0 are (1, 2) and 3.
Q Length of ⊥ from (1, 2) on L = 0
|1 + 4 + 2|
7
is
=
≠ radius
1+ 4
5
∴(b) Option is wrong.
L is also the common chord of C1 and C 2.
∴(c) Option is correct.
Q Centres of C1 = 0 and C 2 = 0 are (1, 2) and ( −1, − 2 )
∴Slope of Line joining centres of circles C1 = 0 and C 2 = 0 is
−2 − 2 4
(say)
= = 2 = m1
−1 − 1 2
Chap 04 Circle
1
(say)
slope of L = 0 is − = m2
2
∴
m1m2 = −1
Hence, L is perpendicular to the line joining centres of C1 and C 2.
∴ (d) Option is correct.
45. Let slope of OA is m,
m −1
Then,
= tan 45 °
1+m
and
From Eqs. (i) and (ii), we get x 2 + (3 x − 1 ) 2 − 4 x − 1 = 0
or
10 x 2 − 10 x = 0
∴
x = 0 and x = 1
From Eq. (ii), y = −1 and y = 2
∴
A ≡ ( 0, − 1 ) and B ≡ (1,2 )
46. AB = ( 0 − 1) 2 + ( −1 − 2) 2 = 10
47. Let ∠ACB = 2θ
∴
C
⇒
B
2
π 3π
=
4
4
48. Equation of tangent at A( 0, − 1) to the circle S is
λ
45°
0. x + ( −1 ). y − 2( x + 0 ) − 1 = 0
or
2x + y + 1 = 0
∴ Slope (m1 ) = −2
and slope of line L is m2 = 3
A
m −1
m −1
= ±1 or
= −1
m+1
1+m
⇒
m − 1 = −m − 1
∴
m=0
∴Equation of OA is y = 3
Solving y = 3 and x 2 + y 2 − 10 x − 6y + 30 = 0
⇒
⇒
If θ be the angle between L and S, then tan θ =
∴
x 2 + 9 − 10 x − 18 + 30 = 0
⇒
x − 10 x + 21 = 0
⇒
( x − 7 )( x − 3 ) = 0
or
x = 3, 7
∴Two vertices are (3, 3) and (7, 3) and other diagonal is ⊥ to
y = 3 and through centre (5, 3) is x = 5.
Now, solving x = 5 and x 2 + y 2 − 10 x − 6y + 30 = 0
25 + y 2 − 50 − 6y + 30 = 0
⇒
y 2 − 6y + 5 = 0
Q
PQ = PR,
parallelogram PQRS is a rhombus.
P
Q
O
…(i)
…(ii)
Q
q
B
S
∴Mid-point of QR = mid point of PS and QR ⊥ PS
Therefore, S in the mirror image of P w.r.t. QR.
49. Let P( λ,6 − 2λ ) be any point on L = 0
Q Circumcircle of ∆PQR always pass through O
∴OP is diameter of circle.
λ

Then, centre is  ,3 − λ  So,
2

λ
and y = 3 − λ
2
or 2 x + y = 3 is the locus of the circumcenter of ∆PQR.
x=
P
L=0
2
R
L :y = 3 x − 1
A
π
4
2
⇒
(y − 1 ) (y − 5 ) = 0 or y = 1, 5
∴Other two vertices are (5, 1) and (5, 5).
Sol. (Q. Nos. 46 to 48)
S : x 2 + y 2 − 4x − 1 = 0
C (2, 0)
q Ö5
q
Ö5
M
θ=
−2 − 3
=1
1 + ( −2 )(3 )
Sol. (Q. Nos. 49 to 51)
2
⇒
5
2
1
2
π
θ=
4
Required angle = ∠APB = π − θ = π −
2
y=
x+
m
5
=
10
CM
cosθ =
=
5
CM =
∴
O (5, 3)
D
345
346
Textbook of Coordinate Geometry
50. Area of ∆QRS = Area of ∆PQR =
RL3
R + L2
…(i)
2
R = 2, L = S1 = (6 2 + 8 2 − 4 ) = 96
Here,
⇒
d =
⇒
From Eq. (i),
2 × 96 × 96 192 6 192 λ
=
=
4 + 96
25
25
∴
λ =6
51. Let S ≡ (α,β)
Area of ∆QRS =
(given)
∴
(given)
2
r =4
55. λ = r = 4
56. Q(C1C 2 ) 2 = r 2 + 3 2 = 16 + 9 = 25
∴
QS is the mirror image of P w.r.t. QR.
Eq. (i) of QR is 3 x + 4y = 4, then
α − 3 β − 4 −2 (3 .3 + 4 . 4 − 4 )
=
=
3
4
32 + 42
42
=−
25
51
68
α = − and β = −
∴
25
25
 51 68
Hence,
S ≡ − ,− 
 25 25 
1  24
=  
2 5 
9+r
r
4
=
2
(9 − r ) 5
3r
C1C 2 = 5
BM 4 − 3 1
sinθ =
=
=
AB C1C 2 5
In ∆ABM,
2
1
4 6
 1
sin 2θ = 2 sin θ cosθ = 2 × × 1 −   =
 5
5
25
∴
57. Length of direct common tangent = (C1C 2 ) 2 − (r − 3) 2
= (5 ) 2 − (1 ) 2
= 24 = λ
Sol. (Q. Nos. 52 to 54)
52. This is third degree equation which satisfy the point of
(given)
∴
λ2 = 24
Sol. (Q. Nos. 58 to 60)
Y
intersection of L1 = 0, L2 = 0 and L3 = 0.
53. λL1L2 + µL3L4 = 0 will always pass through the vertices of the
C
parallelogram for all λ, µ ∈ R ~ { 0 }
L2=0
D
B
C
X¢
L4=0
B
L1=0
54. Since, µ = 0
Y¢
So,
L1L2 + λL2L3 = 0
⇒
L2( L1 + λL3 ) = 0
This is the equation of pair of straight lines.
Sol. (Q. Nos. 55 to 57)
Let 2d = length of common chord and ∠PC1C 2 = φ, then
d
sin φ =
3
M
A
and
X
A(a, 0)
O
L3=0
A
Q
C1
C2
q
q P
f
r
Let equation of line AB be
x −a y − 0
=
=r
cosθ sin θ
Coordinates of any point on the line AB is (a + r cosθ,r sin θ ),
then
B ≡ (a + r1 cosθ,r1 sin θ )
and
C ≡ (a + r2 cosθ,r2 sin θ )
58. λ = (OA ) 2 + (OB ) 2 + ( BC ) 2
= a 2 + b 2 + (r1 − r2 ) 2 cos2 θ+ (r1 − r2 ) 2 sin 2 θ
= (a 2 + b 2 ) + (r1 − r2 ) 2
B
= (a 2 + b 2 ) + 4a 2 cos2 θ − 4(a 2 − b 2 )
(90°–f)
C1
d
d
sin(90°− φ ) = or cos φ =
r
r
d2 d2
1=
+
9 r2
C2
(Q B and C lie on x 2 + y 2 = b 2
∴
(a + r cosθ ) 2 + (r sin θ ) 2 = b 2
or
r 2 + 2ar cosθ + a 2 − b 2 = 0
∴
r1 − r2 = 4a 2 cos2 θ − 4(a 2 − b 2 )
⇒
λ = 5b 2 − 3a 2 + 4a 2 cos2 θ
Q
0 ≤ cos2 θ ≤ 1
∴
λ ∈ [5b 2 − 3a 2,5b 2 + a 2 ]
Chap 04 Circle
59. Let (h,k ) be the mid-point of AB and let (α ,β) be the coordinates
of B, then
65. Locus of (h,k ) is
66. Q Circle C 2 is equal to circle C1
∴Radius of circle C 2 = radius of circle C1 = 2
(2h − a ) 2 + (2k ) 2 = b 2
∴Distance between the centres of C1 and C 2
= 2+ 2
(Q circles C1 and C 2 touch externally)
=2 2
2
a
b2

2
h −  + k =


2
4
or
Hence, locus of (h, k ) is
67. The given circle is
2
a
b2

2
x −  + y =

2
4
S :( x − 3 ) 2 + (y − 5 ) 2 = ( (34 − λ ) ) 2
60. The locus of mid-point of AB, when BC is maximum is a fixed
point M on X -axis.
Sol. (Q. Nos. 61 to 63)
61. From the figure OA = OB = AB = a
A
P
M
a
N
Q
C
5 > (34 − λ ) or λ > 9
…(iii)
or ( x 2 + y 2 − 2 x − 8 ) − 2 λy = 0
which is of the form S + λL = 0
All the circles pass through the point of intersection of the
circle x 2 + y 2 − 2 x − 8 = 0 and y = 0
∴
∠OAB = 60 °
62. Let T be the point of intersection of tangents.
Q
∠AOC = 120 °
∴
∠ATC = 180 °−120 ° = 60 °
63. Locus of point of intersection of tangents at A and C is
a2
x2 + y 2 =
α
sin 2  
 2
=
…(ii)
a2
= 4a 2
sin (30 ° )
Solving, we get
x 2 − 2x − 8 = 0
or
( x − 4 )( x + 2 ) = 0
∴
A ≡ ( −2, 0 )
and
B ≡ ( 4, 0 )
Hence, | AB| = 6
(Here, α = 60 °)
2
69. Q AQ = BQ = 3 + 2 2
and PQ = radius of director circle of C1
Hence, x 2 + y 2 = 4a 2
t12
t 22
t 32
t14
given
= h + k − 4,
2
A
2
r
R
= h 2 + k 2 − 4h,
2
= t 22t 32 + 16
2
2
2
Q
2
B
2
or (h 2 + k 2 ) 2 − 8(h 2 + k 2 ) + 16 = (h 2 + k 2 ) 2 − 4(h 2 + k 2 )(h + k )
+16hk + 16
or
(h 2 + k 2 )(h + k ) − 2(h 2 + k 2 ) − 4hk = 0
or
(h + k )(h + k ) − 2(h + k ) = 0
or
(h + k )(h 2 + k 2 − 2(h + k )) = 0
or
and
B¢
= h 2 + k 2 − 4k
or (h + k − 4 ) = (h + k − 4h )(h + k − 4k ) + 16
2
P
A¢
Sol. (Q. Nos. 64 to 66)
Here,
…(i)
From Eqs. (i), (ii) and (iii), we get
25 < λ < 29
Hence, difference = 29 − 25 = 4
68. x 2 + y 2 − 2x − 2λy − 8 = 0
B
60°
a
Since, point P(1, 4 ) lies inside the circle
∴
S1 < 0
⇒ 1 + 16 − 6 − 40 + λ < 0
or
λ < 29
Also, circle neither touches nor cuts the axes, then
3 > (34 − λ ) or λ > 25
and
T
60°
O
[From Eq. (ii)]
C1 : x 2 + y 2 − 2 x − 2y = 0
Since, (α , β ) lies on x 2 + y 2 = b 2
⇒
[From Eq. (i)]
L1 : x + y = 0
a +α
0+β
= h and
=k
2
2
α = 2h − a and β = 2k
⇒
64. Locus of (h,k ) is
347
2
2
Let ‘r’ be required radius
∴
2
h+k = 0
h 2 + k 2 − 2h − 2k = 0
= 2(3 + 2 2 ) = 3 2 + 4
…(i)
…(ii)
⇒
3 2 + 4 =3+2 2 +r +r 2
r=
( 2 + 1)
=1
( 2 + 1)
 PR A′R 
=
Q

 PQ AQ 
348
Textbook of Coordinate Geometry
70. Since, A. x + y − 5 = 0 is a tangent to the circle at P(2,3), so the
centre of the circle lies on x − y + 1 = 0. Now Q(1,2 ) lies on the
circle.
Equation of right bisector of PQ is x + y − 4 = 0
∴Centre of circle is point of intersection of x − y + 1 = 0 and
 3 5
x + y − 4 = 0 which is  , 
 2 2
3 5
+ −5
1
1
2 2
∴ Radius
=
=
λ
2
2
∴
λ =2
Hence,
λ2 = 4
a
=2
1
2
a = 2 − 2 = 6 −c
4 + 2 − 4 2 = 6 −c
c=4 2
Then,
1−
or
[From Eq. (i)]
or
∴
Hence, λ = 4
74. The rays are given as 2x 2 − 3xy − 2y 2 = 0 i.e.
(2 x + y )( x − 2y ) = 0
x
or y = −2 x which are perpendicular.
⇒ y =
2
∴Required area
71. Since, given circle is ( x + 5) 2 + (y − 12) 2 = (14) 2
2
Y
2
 y − 12
 x + 5
 =1
 +

 14 
 14 
x +5
y − 12
Let,
= cosθ and
= sinθ
14
14
or
x = 14 cosθ − 5 and y = 14 sinθ + 12
∴
x 2 + y 2 = (14 cosθ − 5 ) 2 + (14 sin θ + 12 ) 2
y=–2x
or
y=
O
X¢
X
= 196 + 25 + 144 + 28 (12 sin θ − 5 cosθ )
= 365 + 28 (12 sin θ − 5 cosθ )
Maximum value of ( x 2 + y 2 ) = 365 + 28 × 13 = 729
(Q −13 ≤ 12 sin θ − 5 cosθ ≤ 13)
1
or maximum value of ( x 2 + y 2 ) 3 = 9
72. Let the given lines are represented by L1, L2, L3
and L4 , then
L1 ≡ 2 x + 3y − 2 = 0
L2 ≡ 3 x − 2y − 3 = 0
L3 ≡ x + 2y − 3 = 0
L4 ≡ 2 x − y − 1 = 0
Equation of second degree conic circumscribing a quadrilateral
whose sides are L1 = 0, L2 = 0, L3 = 0 and L4 = 0 is
L1L3 + kL2L4 = 0
or (2 x + 3y − 2 )( x + 2y − 3 ) + k (3 x − 2y − 3 )(2 x − y − 1 ) = 0 …(i)
For circle, coefficient of x 2 = coefficient of y 2
or
2 + 6k = 6 + 2k
∴
k =1
From, Eq. (i), required circle is
8 x 2 + 8y 2 − 17 x − 8y + 9 = 0
 9
 17
or
x + y −   x −y +   = 0
 8
8
17
9
Here,
λ = − , µ = −1, ν =
8
8
then,
| λ + 2µ + ν| = 3
73. Let radius of given circle is a
…(i)
∴
a = ( 4 + 2 − c ) = (6 − c )
a
a
Now, radius of circle C1 =
and radius of circle C 2 =
2
( 2 )2
and so on.
a
a
(gjven)
Also,
a+
+
+ ... ∞ = 2
2 ( 2 )2
2
2
x
2
Y¢
1
= ( π (2 ) 2 − π (1 ) 2 )
4
3 π λπ
(given)
=
=
4
4
∴
λ =3
75. Let radii of circles be r1 and r2 and distance between centres is
d, then
and
d 2 − (r1 + r2 ) 2 = 5
…(i)
d 2 − (r1 − r2 ) 2 = 13
…(ii)
From Eqs. (i) and (ii), we get
(r1 + r2 ) 2 − (r1 − r2 ) 2 = (13 ) 2 − (5 ) 2
⇒
4r1r2 = 144 or λ = 36
λ
∴
=9
4
76. Let ax + by = 1
(Qr1r2 = λ)
…(i)
be a chord PQ.
Y
P
M
X¢
X
O
(–1, 0)
(1, 0)
Y¢
Q
Chap 04 Circle
which pass through (1, 0), then a + 0 = 1
∴
a =1
and touch circle B with centre ( −1, 0 ) and radius 1, then
| −a + 0 − 1|
=1
a2 + b2
or
(B) Let P be the point (h, k ), then h 2 + k 2 + 2h + 2k = 0 …(i)
h k
mid-point of OP is  , 
 2 2
h
k
x = ,y =
2
2
or
h = 2 x, k = 2y
From Eq. (i), 4 x 2 + 4y 2 + 4 x + 4y = 0
Let
a 2 + b 2 = (a + 1 ) 2
or
b 2 = 2a + 1 = 3
b= 3
∴
From Eq. (i), Equation of chord is x + 3 y = 4
1
OM = and OP = 4
∴
2
∴
or
Hence,
PQ = 2 PM = 2 (OP ) 2 − (OM ) 2
λ = 63
λ
=7
9
77. Q S1 :( x − 2) 2 + (y − 3) 2 = 1
Centre C1 :(2,3 ) and radius r1 :1 and S 2 :( x − 5 ) 2 + (y − 6 ) 2 = r 2
Centre C 2 :(5,6 ) and radius r2 :r
C1C 2 = 3 2
⇒
(A)QS1 and S 2 touch internally, then
C1C 2 = r2 − r1
or
3 2 = r −1
or
S1 = 0 2 + 0 2 − 0 − 0 + 27 = 27
T = 0. x + 0. y − 3 3( x + 0 ) − 3(y + 0 ) + 27
(r − 1 ) 2 = 18
= −3 3 x − 3y + 27
(B)QS1 and S 2 touch externally, then
C1C 2 = r2 + r1
or
3 2 =r +1
or
Eq. of the pair of tangents is given by
SS1 = T 2
⇒ 27( x 2 + y 2 − 6 3 x − 6y + 27 ) = ( −3 3 x − 3y + 27 ) 2
r 2 + 2r + 3 = (r + 1 ) 2 + 2 = 18 + 2 = 20
or 3 ( x 2 + y 2 − 6 3 x − 6y + 27 ) = ( 3 x + y − 9 ) 2
(C)QS1 and S 2 intersect orthogonally, then
(C1C 2 ) 2 = r12 + r22
⇒
18 = r 2 + 1
or
r 2 − 1 = 16
(D)QS1 and S 2 intersect, the common chord is S1 − S 2 = 0
i.e.
6 x + 6y + r 2 − 49 = 0
or
18y (y − 3 x ) = 0
a = 0, b = 3
then,
a2 + b2 = 3
79. (A)Q P ≡ (10,7)
and
r = 19
∴
r 2 + 5 = 24
78. (A) Since (2, 3) lies on ax + by − 5 = 0
From Eq. (i), a = b = 1
∴
a + b = 2 and a 2 + b 2 = 2
18y 2 − 18 3 xy = 0
∴
2
∴
2a + 3b − 5 = 0
Since, line is at greatest distance from centre
 4 − 3  a 
⇒
  −  = −1 i.e. a = b

 3 − 2  b 
or
the tangents are y = 0, y = 3 x
Given, common chord is longest, then passes through (2, 3)
⇒
12 + 18 + r 2 − 49 = 0
or
x2 + y 2 + x + y = 0
On comparing, we get
2a = 1, 2b = 1
1
or
a =b =
2
∴
a +b =1
(C) Centre of circles are C1 :(1, 2 ) and C 2 :(5, − 6 )
−6 − 2
Equation of C1C 2 is y − 2 =
(x − 1)
5 −1
or
2x + y − 4 = 0
Equation of radical axis is 8 x − 16y − 56 = 0
or
x − 2y − 7 = 0
Point of intersection of Eqs. (i) and (ii) are (3, −2)
∴
a = 3, b = −2
or
a +b =1
(D) Let S ≡ x 2 + y 2 − 6 3 x − 6y + 27 = 0
1

= 2 16 −  = 63 = λ

4
∴
349
…(i)
S = x 2 + y 2 − 4 x − 2y − 20
∴
S1 = 100 + 49 − 40 − 14 − 20 > 0
∴P outside the circle, radius r = 4 + 1 + 20 = 5
centre C ≡ (2,1 )
∴Shortest distance L = CP − r
= (10 − 2 ) 2 + (7 − 1 ) 2 − 5
= (64 + 36 ) − 5 = 10 − 5 = 5
And largest distance M = CP + r
= 10 + 5 = 15
M + L = 20, M − L = 10
...(i)
…(ii)
350
Textbook of Coordinate Geometry
(B)Q
and
P ≡ (3, − 6 )
S ≡ x 2 + y 2 − 16 x − 12y − 125
∴
S1 = 9 + 36 − 48 + 72 − 125 < 0
∴ P inside the circle radius r = (64 + 36 + 125 ) = 15
centre C ≡ (8,6 )
∴ Shortest distance L = r − CP
= 15 − (8 − 3 ) 2 + (6 + 6 ) 2
= 15 − 13 = 2
And largest distance M = r + CP
= 15 + 13 = 28
M + L = 30,
M − L = 26
(C) P ≡ (6, − 6 )
and
S ≡ x 2 + y 2 − 4 x + 6y − 12
∴
S1 = (6 ) 2 + ( −6 ) 2 − 24 − 36 − 12 = 0
P on the circle
∴ radius r = ( 4 + 9 + 12 ) = 5
∴ Shortest distance L = 0
and largest distance M = 2r = 10
M + L = 10, M − L = 10
80. (A) Equation of conic
(y − a1x − b )(y − a 2x − b ) + λxy = 0 represent a circle. If
coefficient of x 2 = coefficient of y 2
∴
Q
∴
⇒
and
∴
a1a 2 = 1
AM > GM
a1 + a 2
> a1a 2 = 1
2
a1 + a 2 > 2
a12 + a 22
> a12.a 22 = | a1a 2| = 1
2
a12 + a 22 > 2
(B) Let (a1 cosθ, a1 sin θ ) be any point on x 2 + y 2 = a12, then
chord of contact is
x (a1 cosθ ) + y (a1 sin θ ) = b 2
which is tangent of x 2 + y 2 = a 22
∴
| 0 + 0 − b 2|
{(a1 cosθ ) 2 + (a1 sin θ ) 2 }
= a 2 or b 2 = a1a 2
b = 1, a1a 2 = 1
AM > GM
a1 + a 2
⇒
> a1a 2 =| b|
2
∴
a1 + a 2 > 2| b|,
for
b = 1, a1 + a 2 > 2
and then, a12 + a 22 > 2
for
(C) 2 gg1 + 2 ff1 = c + c1
⇒
2 × a1 × a 2 + 0 = b + b
∴
a1a 2 = b for b = 1,a1a 2 = 1
and then,
a1 + a 2 > 2
Also,
a12 + a 22 > 2
81. Let S ≡ x 2 + y 2 − 1 = 0
∴ S (1,3 ) = 1 2 + 3 2 − 1 = 9 > 0
⇒ Point (1, 3) outside the circle x 2 + y 2 = 1
∴Two tangents can be drawn from (1, 3) to circle x 2 + y 2 = 1
⇒ Statement I is false
Also,
|3 − m|
(1 + m 2 )
=1
Squaring both sides, we get
9 + m 2 − 6m = 1 + m 2
4
or
6m = 8 ⇒m =
3
⇒ Statement II is true.
82. Q x 2 + y 2 + 2x − 4y = 0
⇒
( x + 1 ) 2 + (y − 2 ) 2 = 1 2 + 2 2
Let S ≡ ( x + 1 ) 2 + (y − 2 ) 2 − 1 2 − 2 2
Value of S(1, λ ) = (1 + 1 ) 2 + ( λ − 2 ) 2 − 1 2 − 2 2 < 0
or
(1 + 1 ) 2 + ( λ − 2 ) 2 < 1 2 + 2 2
∴Points (1, λ) inside the circle.
⇒ No tangents can be drawn from the point (1, λ) to the circle
x 2 + y 2 + 2 x − 4y = 0
and
(1 + 1 ) 2 + ( λ − 2 ) 2 < 1 2 + 2 2
⇒
(λ − 2)2 < 1
or
−1 < λ − 2 < 1
or
1 < λ <3
∴Statement II and Statement II are both true and statement II
is correct explanation for statement I.
83. Let A ≡ (1, 4), B ≡ (2,3) and C ≡ ( −1,6)
1 4 1
1
∴Area of ∆ABC = 2 3 1 = 0
2
−1 6 1
⇒ A, B, C are collinear
i.e. no circle is drawn.
Hence, Statement I is false and through three non-collinear
points in a plane only one circle can be drawn.
∴Statement II is true.
84. Q Locus of point of intersection of perpendicular tangents is
director circle.
Q x 2 + y 2 = 50 is director circle of x 2 + y 2 = 25
⇒ Statement I is false and statement II is true.
85. Here, C1 ≡ ( 0, 0), r1 = 2
and C 2 ≡ (3, 0 ), r2 = 2
∴
| C1C 2| = 3 and r1 + r2 = 4
and | r1 − r2| = 0
⇒ | r1 − r2| < | C1C 2| < r1 + r2
Hence, circles x 2 + y 2 = 4 and x 2 + y 2 − 6 x + 5 = 0 intersect
each other at two distinct points.
⇒ Statement I is true and Statement II is false.
Chap 04 Circle
86. Q Centre of circle (1, −1) lies on 3x − 4y = 7
351
For minimum radius, c must be equal to zero, then from
Eqs. (ii) and (iii),
1
1
and f = −
g=−
2
2
Equation of required circle, from Eq. (i), is
x2 + y 2 − x − y = 0
∴ 3 x − 4y = 7 is a diameter of circle
x 2 + y 2 − 2 x + 2y − 47 = 0
∴Statement I is true.
Statement II is always true but Statement I is not a correct
explanation of Statement I.
90. The given circle is
87. Statement I is true, line 4y − 3x − 5 = 0 passes through
A( −3, − 1 ) and its distance from O( 0, 0 ) is 1 unit = radius of circle
Centre and radius of this circle are (3, − 3 ) and 9 + 9 − 17 = 1
respectively. But given the required circle has normals
4y
–3
x=
5
Y
x 2 + y 2 − 6 x + 6y + 17 = 0
X¢
C2 (3, 1)
O
i
X
r
r
A(–3, –1)
l
B(0, –1)
C1 (3, – 3)
Y¢
The tangent at B( 0, − 1 ) is y = −1 and normal at A to line AB is
x = −3
 4
∴Equation of incident ray is y + 1 =  −  ( x + 3 )
 3
⇒
4 x + 3y + 15 = 0
Statement II is obviously true.
Hence, both statements are true and but Statement II is not
correct explanation of Statement I.
88. The given points are A(1,1), B(2,3) and C(3,5) which are
collinear as
Slope of AB = Slope of BC = 2
Hence, Statement II is true
The chords of contact are concurrent, then,
x1 y1 1
x2 y 2 1 = 0
x3 y 3 1
Radius of the circle Eq. (ii) = ( g 2 + f 2 − c )
1 + c 2
(1 + c ) 2 (1 + c ) 2
=
+
−c = 

4
4
 2 
or
( x − 3 ) ( x − 3y ) = 0
or
x = 3 and x − 3y = 0
…(i)
but point of intersection of normals is the centre of the circle.
Point of intersection of normals represented by Eq. (i) is
(3, 1) which is centre of the required circle. Since, given
circle and required circle touch each other externally, then
(if radius of required circles is r)
Sum of radii = Distance between the centres
r + 1 = (3 − 3 ) 2 + ( −3 − 1 ) 2 = 4
∴
r =3
∴Equation of required circle is
( x − 3 ) 2 + (y − 1 ) 2 = 3 2
or
x 2 + y 2 − 6 x − 2y + 1 = 0
91. Let OA = a and OB = b then the coordinates of A and B are
Hence, points ( x1, y1 ), ( x 2, y 2 ) and ( x 3, y 3 ) are collinear.
Therefore, Both statements are true and statement II is correct
explanation of statement I.
89. Let the equation of circle be
x 2 + y 2 + 2 gx + 2 fy + c = 0
…(i)
Since, circle Eq. (i) passes through (1,0) and (0,1) then
(1 + c )
1 + 2g + c = 0 ⇒ g = −
2
(1 + c )
and
1 + 2f + c = 0 ⇒ f = −
2
x 2 − 3 xy − 3 x + 9y = 0
(a , 0 ) and ( 0 , b ) respectively.
Since,
∠ AOB = π / 2
Hence, AB is the diameter of the required circle whose
equation is
( x − a ) ( x − 0 ) + (y − 0 ) (y − b ) = 0
or
x 2 + y 2 − ax − by = 0
Y
…(ii)
B (0, b)
…(iii)
n
M
O
A (a, 0)
m
L
X
... (i)
352
Textbook of Coordinate Geometry
93. Equation of chord of contact (QR) is 6x + 8y − r 2 = 0
Equation of tangent at (0,0) of Eq. (i) is
a
b
0 ⋅ x + 0 ⋅ y − (x + 0) − (0 + y ) = 0
2
2
or
ax + by = 0
a ⋅a + 0
∴
m = AL =
a2 + b2
or
and
∴
m=
a2
n=
PM =
and
OM =
| 6 ⋅6 + 8 ⋅8 − r 2 |
=
(6 2 + 8 2 )
| 0 + 0 − r2 |
(6 2 + 8 2 )
=
| 100 − r 2 |
10
r2
10
P (6, 8)
...(ii)
a2 + b2
n = BM =
∴
Q
a ⋅ 0 + b ⋅b
M
a2 + b2
r
b2
O (0
...(iii)
a2 + b2
, 0)
R
r
Adding Eqs. (ii) and (iii), we get
m + n = (a 2 + b 2 )
From Eqs. (ii) and (iii), we get
a = ± m (m + n )
and b = ± n (m + n )

r4 
= 2 r 2 −

100

From Eq. (i), equation of required circle is
x 2 + y 2 ± m (m + n ) x ± n(m + n ) y = 0
1
∴ Area of ∆QPR = ⋅ QR ⋅ PM
2
92. Solving the equations
then
or
∴
or
∴
∴
QR = 2 ⋅ QM = 2 {(OQ ) 2 − (OM ) 2 }
then
(2 + c ) x + 5c 2y = 1 and 3 x + 5y = 1
 1 − 3x 
(2 + c ) x + 5c 2 
 =1
 5 

1
r 4  | 100 − r 2 |
∆ ( say ) = ⋅ 2 r 2 −
 .
2
100
10

r 2 (100 − r 2 ) 3
(say)
=z
1000
dz
1
∴
=
{r 2 . 3 (100 − r 2 ) 2 . ( −2r ) + (100 − r 2 ) 3 ⋅ 2r }
dr 1000
2r (100 − r 2 ) 2
=
{100 − r 2 − 3r 2 }
1000
dz
For maximum or minimum
= 0, then we get r = 5, ( r ≠ 10 as
dr
P is outside the circle)
d 2z
 = − ve
and
dr 2r = 5
∴ ∆2 =
(2 + c ) x + c 2 (1 − 3 x ) = 1
1 − c2
2 + c − 3c 2
(1 + c ) (1 − c )
1+c
x=
=
(3 c + 2 ) (1 − c ) 3c + 2
1+c
x = Lim
c → 1 3c + 2
2
x=
5
6
1−
1 − 3x
5 =− 1
y =
=
5
5
25
x=
∴ ∆ is also maximum at r = 5 .
1
2
Therefore, the centre of the required circle is  , −  but
 5 25
circle passes through (2, 0)
2
2

 1

∴ Radius of the required circle =  − 2 +  − − 0
5

 25

94. Since, A ( 4, 6) and B (1, 9) do not lie on 2x − y + 4 = 0.
 5 15
Let M be the mid-point of AB, then coordinates of M is  , 
2 2 
B (1, 9)
2
M
64
1
1601
+
=
25 625
625
Hence, the required equation of the circle is
=
or
4=
y+
2
0
A (4, 6)
k) P
(h,
C
2x –
2
2
1
1601


 x −  + y +  =


5
25
625
5 ,15
2 2
25 x 2 + 25y 2 − 20 x + 2y − 60 = 0
D
Chap 04 Circle
Slope of AB =
9 −6
= −1
1−4
⇒
Equation of PM is,
5
(k + 3 )
2
∴ k = − 7 − 2 5 or k = − 7 + 2 5
…(i)
2
and
k+2=±
Since, radius from Eq. (ii),
4|k + 2|
r=
2
4 | −7 − 2 5 + 2 |
2
= 10 2 + 4 10 = 26.79
(radius)at
…(ii)
5
15


AD = 2 PM = 2 1 −  + 6 − 



2
2
=2
2|k + 2| = 5 |k + 3|
∴
15
5

y −
= 1 . x − 

2
2
which passes through P (h, k ), then
15
5
or
k−
=h −
k −h =5
2
2
and (h, k ) lie on 2 x − y + 4 = 0
∴
2h − k + 4 = 0
Solving Eqs. (i) and (ii), we get
h = 1 and k = 6
Now,
4|k + 2| =2 5 |k + 3|
⇒
∴ Slope of PM = 1
353
k = −7 − 2 5
=
4 | −7 + 2 5 + 2 |
2
= 10 2 − 4 10 = 1.5
2
(radius)at k = − 7 + 2
9 9
+ =3 2
4 4
5
=
Hence, radius of smaller circle is 1.5 units.
96. Let the centre of the circle C 2 is Q (h, k ), equation of the circle
AB = ( 4 − 1 ) 2 + (6 − 9 ) 2 = 3 2
∴ Area of rectangle
ABCD = AB × AD = 3 2 × 3 2 = 18 sq units.
95. Let C (h, k ) be the centre of the circle. Let AB and C ′ D be the
lines represented by
3 x − y = 6 and y = x respectively.
Clearly, the circle touches AB at A (1, − 3 ).
C 2 is
( x − h ) 2 + (y − k ) 2 = 5 2
or x + y − 2 xh − 2yk + h 2 + k 2 − 25 = 0
2
2
C 2 ≡ x 2 + y 2 − 2 xh − 2yk + h 2 + k 2 − 25 = 0
and equation of circle C1 ≡ x 2 + y 2 − 16 = 0
C2
B
Y
5
A
Q(h, k)
D
C1
P
(0, 0)
M
y=x
O
C′
∴ Equation of common chord is
C1 − C 2 = 0
⇒
−2 xh − 2yk + h 2 + k 2 − 9 = 0
X
N
3x –y = 6
C
(h, k)
A (1, –3)
= ( −3k − 8 − 1 ) + (k + 3 )
⇒
4|k + 2|
= | k + 3 | 10
2
2hx + 2ky − (h 2 + k 2 − 9 ) = 0
or
Equation of line ⊥ to 3 x − y = 6 is x + 3y = λ which passes
through (1, − 3 ).
then
1 −9 = λ
∴
λ = −8
∴ ⊥ line is x + 3y + 8 = 0
which passes through C (h, k )
then
…(i)
h + 3k + 8 = 0
Now, centre C (h, k ) ≡ C ( −3k − 8, k )
| − 3k − 8 − k |
Radius CN =
= CA
1+1
2
B
…(i)
Slope of this line = − h / k
But, it is given that its slope = 3 / 4
h 3
∴
− =
k 4
or
3k + 4h = 0
…( ii )
Let p be the length perpendicular from P ( 0, 0 ) on chord (1),
then
h2 + k 2 − 9
p=
( 4h 2 + 4k 2 )
or
p=
(h 2 + k 2 − 9 )
2 (h 2 + k 2 )
…( iii )
Length of this chord AB = 2 AM
= 2 (16 − p 2 )
2
…( ii )
This chord has maximum length, then p = 0, then from Eq. (iii),
h2 + k 2 − 9 = 0
…( iv )
354
Textbook of Coordinate Geometry
On solving Eqs. (ii) and (iv) we get
9
12
h = m ,k = ±
5
5
 9 12
 9 12
∴ Centre of C 2 is  , −  or  − , 
 5 5
5
5
Equation of circle with centre at Q (1, 0 ) and radius r is
…(ii)
( x − 1 ) 2 + (y − 0 ) 2 = r 2
(0 < r < 2)
Solving Eqs. (i) and (ii), we get
2 − r2
2
but R above the X-axis.
2
⇒
(2 x − y ) ( x − y ) = 0
⇒
y = 2x, y = x
are the equations of straight lines passing through origin.
Now, let the angle between tangents is 2α,
y = 2x
So, SQ = r
and MR =
=
x
3
Ay
r 4 − r2
1
∆ = ⋅r ⋅
2
2
αα
⇒
⇒
⇒
r 2 (4 − r 2 )
4
r4
2
∆ = (4 − r 2 ) = A
16
∆ ( say ) =
X
O
tan ( 45 ° + 2α ) = 2
tan 45 ° + tan 2α
=2
1 − tan 45 ° tan 2α
1 + tan 2α 2
2 tan 2α 1
=
= ⇒
1 − tan 2α 1
2
3
(By componendo and dividendo rule )
2 tan α
1
=
1 − tan 2 α 3
tan α =
−6 ± (36 + 4 )
= − 3 ± 10
2
= − 3 + 10
π

Q0 < α < 

4
3
= ( 10 − 3 )
OA
3
( 10 + 3 )
OA =
= 3 (3 + 10 )
( 10 − 3 ) ( 10 + 3 )
Now, in ∆OAC, tan α =
∴
98. The given circle is x 2 + y 2 = 1
with centre at O ( 0,0 ) and radius 1. It cuts X-axis at the points
when y = 0 then x = ± 1 i.e., at P ( −1, 0 ) and Q (1, 0 ) .
R
X
X′ (–1, 0) P
S M
O
T
Y′
Q (1, 0)
A=
r=
3
∴ A is maximum. Hence, ∆ is also maximum.
1 8 1 
8
∴ Maximum value of ∆ = × ×
4 − 

2 3 2
3
2
2
×
3
3
4 3
sq units.
=
9
=
…(i)
Y
(say)
1
( 4r 4 − r 6 )
16
dA 1
d 2A 1
∴
= (16r 3 − 6r 5 ) and
= ( 48r 2 − 30r 4 )
dr 16
dr 2 16
dA
For maximum and minimum area,
=0
dr
8
r2 =
3
 d 2A 
1 
8
64
∴  2
=  48 × − 30 ×  < 0
16 
3
9
8
 dr 
∴
⇒ tan 2 α + 6 tan α − 1 = 0
∴
r 4 − r2
2
1
∴ Area of ∆QSR = ⋅QS ⋅ MR
2
3 C
45°
r 4 − r2
2
2 −r 2 r 4 − r 2 

,
R≡

 2
2


∴
Y
then,
and y = ±
x=
97. Q 2x + y − 3xy = 0
2
99. The equation of any curve passing through
a1x + b1y + c1 = 0
a 2x + b2y + c 2 = 0
y = 0 and x = 0 is
(a1x + b1y + c1 )(a 2x + b2y + c 2 ) + λxy = 0
where, λ is a parameter.
This curve will represent a circle. If the
coefficient of x 2 = coefficient of y 2,
i.e.
a1a 2 = b1b2
and if the coefficient of xy = 0
then
a1 b2 + a 2b1 + λ = 0
∴
λ = − (a1b2 + a 2b1 )
…(i)
…(ii)
355
Chap 04 Circle
102. Let A ≡ ( −a , 0) and B ≡ (a , 0) be two fixed points.
Y
a2x+b2y+c2=0
x=0
Let one line which rotates about B an angle θ with the X -axis
at any time t and at that time the second line which rotates
about A make an angle 2θ with X-axis.
a1x+b1y+c1=0
O
2θ
X
y=0
A (– a, 0)
Substituting the value of λ in Eq. (i) then
(a1x + b1y + c1 ) (a 2x + b2y + c 2 ) − (a1b2 + a 2b1 ) xy = 0
⇒ a1a 2x 2 + b1b2y 2 + (a1c 2 + a 2c1 ) x + (b1c 2 + b2c1 ) y = 0
Now, equation of lines through B and A are respectively
y − 0 = tan θ ( x − a )
and
y − 0 = tan 2θ ( x + a )
2 tan θ
From Eq. (ii), y =
(x + a )
1 − tan 2 θ
From Eq. (ii), b1b2 = a1a 2
∴ Equation of required circle is
a1a 2( x 2 + y 2 ) + (a1c 2 + a 2c1) x + (b1c 2 + b2c1 ) y = 0
100. Let circle be
S ≡ x 2 + y 2 + 2 gx + 2 fy + c = 0
…( i )
Since, centre of this circle ( − g, − f ) lie on 2 x − 2y + 9 = 0
∴
−2 g + 2 f + 9 = 0
…( ii )
and the circle S = 0 and x 2 + y 2 − 4 = 0 cuts orthogonally.
∴
2g × 0 + 2 f × 0 = c − 4
∴
c=4
…( iii )
Substituting the values of g and c from Eqs. (ii) and (iii) in
Eq. (i), then
x 2 + y 2 + (2 f + 9 ) x + 2 fy + 4 = 0
⇒


2y


 (x − a ) 
=
 (x + a )
2

1 − y

( x − a ) 2 
2y ( x − a ) ( x + a )
y =
(x − a )2 − y 2
...(i)
...(ii)
[from Eq. (i)]
⇒ (x − a )2 − y 2 = 2 (x 2 − a 2 )
or x 2 + y 2 + 2ax − 3a 2 = 0 which is the required locus.
103. Let the coordinate of C is ( x1, y1 )
and let the coordinates of A and B are (0, 0) and (a, 0 )
C (x1, y1)
(x 2 + y 2 + 9x + 4) + 2 f (x + y ) = 0
or
θ
B (a, 0)
O (0, 0)
Hence, the circle S = 0 passes through fixed point
(Q form S ′ + λP = 0)
∴ x 2 + y 2 + 9 x + 4 = 0 and x + y = 0
 1 1
After solving we get ( −4, 4 ) or  − ,  .
 2 2
(0, 0)
A
101. Chords are bisected on the line y = x. Let ( x1, x1 ) be the
mid-point of the chord, then equation of the chord is T = S1
∴ xx1 + yx1 − a ( x + x1 ) − b (y + x1 ) + a 2 + b 2 − c 2
= x12 + x12 − 2ax1 − 2bx1 + a 2 + b 2 − c 2
⇒ ( x1 − a ) x + ( x1 − b ) y + ax1 + bx1 − 2 x12 = 0
given
⇒
⇒
(a, 0)
B
a
sin A BC
=
sin B AC
( BC ) 2 = k 2 ( AC ) 2
k=
( x1 − a ) 2 + y12 = k 2 ( x12 + y12 )
This chord passes through (a , b + c )
⇒ ( x1 − a ) a + ( x1 − b ) (b + c ) + ax1 + bx1 − 2 x12 = 0
⇒ (1 − k 2 ) x12 + (1 − k 2 ) y12 − 2ax1 + a 2 = 0
⇒
⇒
2 x12 − (2a + 2b + c ) x1 + a 2 + b 2 + bc = 0
which is quadratic in x1. Since, it is given that two chords are
bisected on the line y = x, then x1 must have two real roots,
B 2 − 4 AC > 0
⇒ (2a + 2b + c ) 2 − 4 ⋅ 2 (a 2 + b 2 + bc ) > 0
Hence, locus of C is
x2 + y 2 −
2ax
a2
+
=0
2
1 −k
1 − k2
 a

, 0

1 − k2 
− 8b 2 − 8bc > 0
⇒ 4a + 4b − 8ab + 4bc − 4ac − c < 0
2
2ax1
a
+
=0
1 − k2 1 − k2
This is a circle whose centre is
⇒ 4a 2 + 4b 2 + c 2 + 8ab + 4bc + 4ac − 8a 2 − 8b 2
2
x12 + y12 −
2
2
Hence, the condition on a , b , c is
4a 2 + 4b 2 − c 2 − 8ab + 4bc − 4ca < 0
and
radius =
a2
a2
ak
−
=
2 2
(1 − k )
(1 − k 2 ) (1 − k 2 )
(Qk ≠ 1)
356
Textbook of Coordinate Geometry
104. Let P be ( x1, y1 ) and line through P ( x1, y1 ) makes an angle θ
with X-axis, then
x − x1 y − y1
=
=r
cosθ
sin θ
Coordinates of any point on the curve is
( x1 + r cosθ, y1 + r sin θ ). This point must lie on
ax 2 + 2hxy + by 2 = 1
Q
θ
P (x1, y1)
∴ a ( x1 + r cosθ ) 2 + 2h ( x1 + r cosθ )(y1 + r sin θ )
+ b (y1 + r sin θ ) 2 = 1
∴ (a cos2 θ + h sin 2θ + b sin 2 θ ) r 2
+ 2 (ax1 cosθ + hx1 sin θ + hy1 cosθ ) r
+ ax12 + 2hx1y1 + by12 = 0
It is quadratic equation in r. Let roots of this equation are r1 and
r2 then
ax12 + 2hx1y1 + by12
r1r2 =
(a cos2 θ + h sin 2θ + b sin 2 θ )
∴
PQ ⋅ PR =
ax12 + 2hx1 y1 + by12
for a = b, h = 0
a cos2 θ + h sin 2θ + b sin 2 θ
PQ ⋅ PR =
+0+
= x12 + y12
a cos θ + 0 + a sin 2 θ
ax12
2
Now if y > 0, it becomes x 2 = 4y
and if y ≤ 0, it becomes x = 0
∴Combining the two, the required locus is
{( x,y ) : x 2 = 4y } ∪ {( 0,y ):y ≤ 0 }
106. s1 = x 2 + y 2 + 2ax + cy + a = 0
s 2 = x 2 + y 2 − 3ax + dy − 1 = 0
R
∴
∴ Locus of (h, k ) is, x 2 = 2y + 2 | y |
Equation of common chord of circles s1 and s 2 is given by
s1 − s 2 = 0
⇒
5ax + (c − d )y + a + 1 = 0
Given, that 5 x + by − a = 0 passes through P and Q
∴ The two equations should represent the same line
a c −d a + 1
⇒
=
=
1
b
−a
⇒
a + 1 = −a 2
a2 + a + 1 = 0
No real value of a.
107. Equation of circle with centre (0, 3) and radius 2 is
x 2 + (y − 3 ) 2 = 4
Let centre of the variable circle is (α , β )
Q It touches X-axis.
∴It’s equation is ( x − α ) 2 + (y + β ) 2 = β 2
Y
ay12
c1
which is independent of θ.
Then curve ax 2 + 2hxy + by 2 = 1 becomes ax 2 + 0 + ay 2 = 1
1
⇒
x2 + y 2 =
a
1
.
is a circle with centre (0, 0) and radius
a
105. Let the centre of circle C be (h,k). Then as this circle touches
axis of x, its radius =| k|
r2 c2
(a, b)
X¢
X
O
Y¢
Circle touch externally
⇒
∴
Y
r1
c1c 2 = r1 + r2
α 2 + (β − 3 ) 2 = 2 + β
α 2 + (β − 3 ) 2 = β 2 + 4 + 4β
C
k
A (0,1)
⇒
B(h, k)
1

∴Locus is x 2 = 10 y −  which is a parabola.

2
k
X
O
Also, it touches the given circle x + (y − 1 ) = 1, centre (0, 1)
radius 1, externally
Therefore, the distance between centres = sum of radii
2
⇒
(h − 0 ) 2 + (k − 1 ) 2 = 1 + | k |
⇒
h 2 + k 2 − 2k + 1 = (1 + | k | ) 2
⇒
h 2 + k 2 − 2k + 1 = 1 + 2 | k | + k 2
⇒
h 2 = 2k + 2 | k |
α 2 = 10(β − 1 / 2 )
2
108. Let the centre be (α ,β)
Q It cuts the circle x 2 + y 2 = p 2 orthogonally
∴Using 2 g1g 2 + 2 f1 f 2 = c1 + c 2, we get
2 ( −α ) × 0 + 2 ( −β ) × 0 = c1 − p 2
⇒
c1 = p 2
Let equation of circle is
x 2 + y 2 − 2αx − 2βy + p 2 = 0
Chap 04 Circle
357
T2T3 = latusrectum of parabola
It passes through
(a, b ) ⇒a 2 + b 2 − 2αa − 2βb + p 2 = 0
Y
∴ Locus of (α , β ) is
∴
2ax + 2by − (a 2 + b 2 + p 2 ) = 0
T2
109. Without loss of generally it we can assume the square ABCD
A (1, 1)
(–1, 1) B
with its vertices A(1, 1 ), B( −1,1 ), C( −1, − 1 ), D (1, − 1 )
P to be the point ( 0, 1 ) and Q as ( 2, 0 ).
X¢
P(0, –1)
X¢
C2
D
(1, –1)
Y¢
1
=2 2
2
1 1
1
× 2 2 = = 1 sq units.
∴ Area ( ∆T1T2T3 ) = ×
2
2
2
= 4×
C1
D(1, –1)
C
(–1, –1)
112. Point of intersection of 3x − 47 − 7 = 0 and 2x − 3y − 5 = 0 is
Y¢
PA 2 + PB 2 + PC 2 + PD 2
Then,
QA 2 + QB 2 + QC 2 + QD 2
1+1+5+5
=
2[( 2 − 1 ) 2 + 1 ] + 2(( 2 + 1 ) 2 + 1 )
12
=
= 0.75
16
110. Let C ′ be the circle touching circle C1 and L, so that C1 and C ′
are on the same side of L. Let us draw a line T parallel to L at a
distance equal to the radius of circle C1, on opposite side of L.
Then, the centre of C ′ is equidistant from the centre of C1 and
from line T .
⇒ locus of centre of C ′ is a parabola.
L
O
C
(–1, –1)
A(1, 1)
Q(Ö2, 0)
L
X
O
X
O
Y
B
(–1, 1)
T3
T1
(1, − 1 ) which is the centre of the circle and radius = 7
∴Equation is ( x − 1 ) 2 + (y + 1 ) 2 = 49
⇒
113. Let M(h,k ) be the mid-point of chord AB where
O (0, 0)
3 p/3
A
T
Also,
C1
2π
3
∠AOB =
∴
C¢
x 2 + y 2 − 2 x + 2y − 47 = 0
⇒
⇒
M (h, k)
B
π
.
3
π 3
OM = 3 cos =
3 2
3
2
h + k2 =
2
9
2
2
h +k =
4
∠AOM =
9
4
114. Equation of director circle of the given circle x 2 + y 2 = 169 is
x 2 + y 2 = 2 × 169 = 338.
∴ Locus of (h, k ) is x 2 + y 2 =
111. Since, S is equidistant from A and line BD, it traces a parabola.
 1 1
Clearly, AC is the axis, A(1,1 ) is the focus and T1  ,  is the
 2 2
vertex of parabola.
1
.
AT1 =
2
We know from every point on director circle, the tangents
drawn to given circle are perpendicular to each other.
Here, (17, 7) lies on director circle.
∴ The tangent from (17, 7) to given circle are mutually
perpendicular.
358
Textbook of Coordinate Geometry
Let coordinates of R be (α ,β ), then using the formula for
centriod of ∆ we get
3 +2 3 +α
3+ 0+β
= 3 and
=1
3
3
⇒
α = 0 and β = 0
∴Coordinates of R = ( 0, 0 )
 3 3
Now, coordinates of E = mid point of QR =  ,  and
 2 2
115. Equation of circle whose centre is (h,k )
( x − h ) 2 + (y − k ) 2 = k 2
Y
(h, k)
X¢
(–1, 1)
X
Y¢
(radius of circle = k because circle is tangent to x-axis)
Equation of circle passing through ( −1, + 1 )
∴
( −1 − h ) 2 + (1 − k ) 2 = k 2
⇒
⇒
∴
1 + h 2 + 2h + 1 + k 2 − 2k = k 2
h 2 + 2h − 2k + 2 = 0 D ≥ 0
(2 ) 2 − 4 × 1.( −2 K + 2 ) ≥ 0
⇒
4 − 4( −2k + 2 ) ≥ 0
1
⇒
1 + 2k − 2 ≥ 0 ⇒ k ≥
2
1
116. Slope of CD =
3
∴ Parametric equation of CD is
3 3
3
x−
y−
2 =
2 = ±1
1
3
2
2
∴ Two possible coordinates of C are
 − 3 3 3 1 3
 3 3 3 1 3
, +  or 
,− + 
+
+

2
2 2
 2
 2
2 2 2
i.e. (2 3, 2 ) or ( 3, 1 )
As (0, 0) and C lie on the same side of PQ
∴( 3, 1 ) should be the coordinates of C.
Remark : Remember ( x1, y1 ) and ( x 2, y 2 ) lie on the same or
opposite side of a line ax + by + c = 0 according as
ax1 + by1 + c
> 0 or < 0.∴ Equation of the circle is
ax 2 + by 2 + c
( x − 3 ) 2 + (y − 1 ) 2 = 1
coordinates of F = mid-point of PR = ( 3, 0 )
118. Equation of side QR is y = 3x and equation of side RP is
y =0
119. The given circle is x 2 + y 2 + 6x − 10y + 30 = 0 Centre ( −3,5),
radius = 2
L1 : 2 x + 3y + ( p − 3 ) = 0;
L2 : 2 x + 3y + p + 3 = 0
Clearly, L1|| L2
Distance between L1 and L2
=
p+3−p+3
 − 3 3 3 3 3
and coordinates of Q = 
, +  = ( 3,3)
+
2 2 2
 2
=
6
<2
13
⇒ If one line is a chord of the given circle, other line may
or may not the diameter of the circle.
Statement I is true and statement II is false.
120. The given circle is x 2 + y 2 + 2x + 4y − 3 = 0
P (1, 0)
C(–1, –2)
Q (a, b)
Centre ( −1, − 2 )
Let Q(α , β ) be the point diametrically opposite to the point
P(1, 0 ),
1+α
0+β
then,
= −1 and
= −2
2
2
⇒
α = −3, β = −4, So, Q is ( −3, − 4 )
121. Tangents PA and PB are drawn from the point P(1,3) to
circle x 2 + y 2 − 6 x − 4y − 11 = 0 with centre C(3,2 )
117. ∆PQR is an equilateral triangle, the incentre C must coincide
with centroid of ∆PQR and D, E, F must coincide with the mid
points of sides PQ, QR and RP respectively.
Also,
∠CPD = 30 ° ⇒ PD = 3
Writing the equation of side PQ in symmetric form we get,
3 3
3
x−
y−
2 =
2 =m 3
1
3
−
2
2
 3 3 3 −3 3 
+
+  = (2 3, 0 )
,
∴Coordinates of P = 
 2
2 2 2
22 + 32
A
P(1, 8)
C
(3, 2)
B
Clearly the circumcircle of ∆PAB will pass through C and as
∠A = 90 °, PC must be a diameter of the circle.
∴ Equation of required circle is
( x − 1 )( x − 3 ) + (y − 2 ) = 0
⇒
x 2 + y 2 − 4 x − 10y + 19 = 0
359
Chap 04 Circle
Y
122. Let r be the radius of required circle.
Clearly, in ∆C1CC 2, C1C = C 2C = r + 1 and P is mid-point of C1C 2
∴
CP ⊥C1C 2
Also,
PM ⊥CC1
Now, ∆PMC1 ~ ∆CPC1 (by AA similarity)
MC1 PC1
∴
=
PC1 CC1
C1
P
1
r
X¢
(0, 2)
(–1, 0) O
1 C
2
1
3
1
M
(h, 2)
X
Y¢
−5
2
5
 −5 
Centre  , 2 and r =
∴
 2 
2
5
Distance of centre from (−4, 0) is
2
∴ It lies on the circle.
126. The smaller region of circle is the region given by
x2 + y 2 ≤ 6
r
C
1
3
=
⇒ r + 1 = 9 ⇒ r = 8.
3 r +1
123. The given circles are
S1 ≡ x 2 + y 2 + 3 x + 7y + 2 p − 5 = 0
…(i)
S 2 ≡ x 2 + y 2 + 2 x + 2y − p 2 = 0
…(ii)
⇒
⇒
( −1 − h ) 2 + 4 = h 2 ⇒h =
and
2 x − 3y ≥ 1
…(i)
…(ii)
Y
∴ Equation of common chord PQ is S1 − S 2 = 0
⇒
L ≡ x + 5y + p 2 + 2 p − 5 = 0
1
y=
–3
x
2
x2+y2=6
⇒ Equation of circle passing through P and Q is S1 + λL = 0
⇒ ( x 2 + y 2 + 3 x + 7y + 2 p − 5 ) + λ ( x + 5y + p 2 + 2 p − 5 ) = 0
X¢
X
O
As it passes through (1, 1), therefore
⇒
(7 + 2 p ) + λ (2 p + p 2 + 1 ) = 0
2p + 7
λ=−
⇒
(p + 1)2
which does not exist for p = −1
124. Circle x 2 + y 2 − 4x − 8y − 5 = 0
Centre = (2, 4 )
Radius = 4 + 16 + 5 = 5
If circle is intersecting line 3 x − 4y = m, at two distinct points.
⇒ length of perpendicular from centre to the line < radius
|6 − 16 − m|
<5
⇒
5
⇒
|10 + m| < 25
⇒
−25 < m + 10 < 25
⇒
−35 < m < 15
125. Let centre of the circle be (h, 2) then radius =| h|
∴ Equation of circle becomes ( x − h ) 2 + (y − 2 ) 2 = h 2
As it passes through (−1, 0)
Y¢
 3
 1 1
We observe that only two points 2,  and  , −  satisfy
 4
 4 4
both the inequations Eqs. (i) and (ii), we get
∴2 points in S lie inside the smaller part.
127. As centre of one circle is (0, 0) and other circle passes through
(0, 0), therefore
a 
Aslo,
C1  , 0 C 2( 0, 0 )
2 
r1 =
|a |
r2 = C
2
C1C 2 = r1 − r2 =
|a |
2
|a | |a |
=
⇒ c =| a |
2
2
If the two circles touch each other, then they must touch each
other internally.
⇒
c−
360
Textbook of Coordinate Geometry


128. Any point P on line 4x − 5y = 20 is α ,
4α − 20
.
5 
130.
Y
T1
Equation of chord of contact AB to the circle x 2 + y 2 = 9
T2
2
X¢
A
P α, 4α – 20
5
M (h, k)
C (0, p)
Y¢
4x–5y=20
B
 4α − 20
drawn from point P α ,
 is

5 
 4α − 20
x. α + y . 
 =9
 5 
1
C1(0, 0) C3(3, 0)
…(i)
Also, the equation of chord AB whose mid-point is (h, k ) is
…(ii)
hx + ky = h 2 + k 2
Q Eqs. (i) and (ii) represent the same line, therefore
h
k
h2 + k 2
=
=
α 4α − 20
9
5
⇒
5kα = 4hα − 20h
and
9h = α(h 2 + k 2 )
20h
9h
and α = 2
⇒
α=
4h − 5k
h + k2
20h
9h
=
⇒
4h − 5k h 2 + k 2
⇒
20(h 2 + k 2 ) = 9( 4h − 5k )
P
X
x=2
From the figure it is clear that the intersection point of two
direct common tangents lies on X -axis.
Aslo
∆PT1C1 ~ ∆PT2C 2
⇒
PC1 : PC 2 = 2 :1
or P divides C1C 2 in the ratio 2 :1 externally
∴Coordinates of P are (6, 0 )
Let the equation of tangent through P be
y = m( x − 6 )
As it touches x 2 + y 2 = 4
∴
⇒
⇒
6m
m2 + 1
=2
36m 2 = 4(m 2 + 1 )
1
m=±
2 2
1
(x − 6)
2 2
Also x = 2 is the common tangent to the two circles.
131. Let centre of the circle ne (1, 4)
∴Equations of common tangents are y = ±
[Q circle touches x-axis at (1, 0)]
Y
∴ Locus of (h, k ) is 20 ( x 2 + y 2 ) − 36 x + 45y = 0
Sol. (Q. Nos. 129 to 130)
Equation of tangent PT to the circle x 2 + y 2 = 4 at the point
P( 3, 1 ) is x 3 + y = 4
Let the line L, perpendicular to tangent PT be
x −y 3 + λ = 0
X¢
…(i)
As it is tangent to the circle ( x − 3 ) + y = 1
2
Equation of L can be
x − 3y = 1
or
x − 3y = 5
(1, h)
B (2, 3)
A (1, 0)
X
Y¢
2
∴ length of perpendicular from centre of circle to the tangent
= radius of circle.
3+ λ
⇒
= 1 ⇒ λ = −1 or −5
2
129. From Eq. (i)
C
Let the circle passes through the point B(2,3 )
∴
CA = CB
⇒
CA 2 = CB 2
⇒
(1 − 1 ) 2 + (h − 0 ) 2 = (1 − 2 ) 2 + (h − 3 ) 2
⇒
h 2 = 1 + h 2 + 9 − 6h
10 5
⇒
h=
=
6 3
10
Thus, diameter is 2h = .
3
(radius)
Chap 04 Circle
Equation of circle C ≡ ( x − 1 ) 2 + (y − 1 ) 2 = 1
132. Since, circle touches X-axis at (3, 0)
∴ The equation of circle be
( x − 3 ) 2 + (y − 0 ) 2 + λy = 0
Radius of T = | y |
T touches C externally
therefore,
Distance between the centres = sum of their radii
Y
A (3, 0)
X¢
X
A
(1,–2)
Y¢
As it passes through (1, −2)
∴ Put x = 1, y = −2
⇒
(1 − 3 ) 2 + ( −2 ) 2 + λ ( −2 ) = 0 ⇒ λ = 4
∴Equation of circle is ( x − 3 ) 2 + y 2 − 8 = 0
Now, from the options (5, −2) satisfies equation of circle.
133. There can be two possibilities for the given circle as shown in
the figure
Y
⇒
( 0 − 1 ) 2 + (y − 1 ) 2 = 1 + | y |
⇒
( 0 − 1 ) 2 + (y − 1 ) 2 = (1 + | y | ) 2
⇒
1 + y 2 + 1 − 2y = 1 + y 2 + 2 | y |
2 | y | = 1 − 2y
1
If y > 0 then, 2y = 1 − 2y ⇒y =
4
If y < 0 then, −2y = 1 − 2y ⇒ 0 = 1 (not possible)
1
y =
∴
4
135. Let the equation of circle be
x 2 + y 2 + 2 gx + 2 fy + c = 0
It passes through (0, 1)
∴
1 + 2f + c = 0
This circle is orthogonal to ( x − 1 ) 2 + y 2 = 16
i.e.
Ö7
(3, 4)
4
4
X¢
X
…(i)
x 2 + y 2 − 2 x − 15 = 0
x2 + y 2 −1 = 0
and
3
361
∴We should have
2 g( −1 ) + 2 f ( 0 ) = c − 15
or
2 g + c − 15 = 0
and
2 g( 0 ) + 2 f ( 0 ) = c − 1 or c = 1
Solving Eqs. (i), (ii) and (iii), we get
c = 1, g = 7, f = −1
∴ Required circle is
x 2 + y 2 + 14 x − 2y + 1 = 0
…(ii)
With centre ( −7, 1 ) and radius = 7
(3, –4)
136. Intersection point of 2x − 3y + 4 = 0 and x − 2y + 3 = 0 is (1, 2)
A (2, 3)
Y¢
∴ The equations of circle can be
( x − 3 ) 2 + (y − 4 ) 2 = 4 2
P
(1, 2)
or
( x − 3 ) + (y + 4 ) = 4
i.e.
x 2 + y 2 − 6 x + 8y + 9 = 0
or
x 2 + y 2 − 6 x + 8y + 9 = 0
2
134.
2
2
B(a, b)
Y
Since, P is the fixed point for given family of lines
So, PB = PA
(α − 1 ) 2 + (β − 2 ) 2 = (2 − 1 ) 2 + (3 − 2 ) 2
C
X¢
(α − 1 ) 2 + (β − 2 ) 2 = 1 + 1 = 2
(1, 1)
(0, y)
T
( x − 1 ) 2 + (y − 2 ) 2 = ( 2 ) 2
X
Y¢
( x − a ) 2 + (y − b ) 2 = r 2
Therefore, given locus is a circle with centre (1, 2 ) and radius 2.
362
Textbook of Coordinate Geometry
137. x 2 + y 2 − 4x − 6y − 12 = 0
Y
…(i)
Centre C1 = (2, 3 ) and Radius, r1 = 5 units
x 2 + y 2 + 6 x + 18y + 26 = 0
P
…(ii)
Centre, C 2 = ( −3, − 9 )
and radius, r2 = 8 units
Q (1, tan q/2)
E
X¢
| C1C 2| = (2 + 3 ) 2 + (3 + 9 ) 2 = 13 units
r1 + r2 = 5 + 8 = 13
∴ | C1C 2| = r1 + r2
Therefore, there are three common
tangents.
138. For the given circle, centre : (4, 4), radius =6
O
R
S(1, 0)
C2
Y¢
∴ Equation of line through Q and parallel to RS is y = tan
∴ Intersection point E of normal and y = tan
6
tan
k
θ
= x tanθ
2
P(h , k )
k
⇒
X
O
X
C1
C(4, 4)
X¢
(cos q, sin q)
x=
1 − tan 2
θ
2
θ
2
θ
2
2
1 −y 2
or y 2 = 1 − 2 x
2
 1 −1 
1 1 
It is satisfied by the points  ,  and  ,  .
3 3
3 3
∴
6 + k = (h − 4 ) 2 + (k − 4 ) 2
(h − 4 ) 2 = 20k + 20
∴ locus of (h, k ) is ( x − 4 ) 2 = 20(y + 1 ), which is parabola.
139. Centre of S :O ( −3,2) and centre of given circle is A(2, − 3)
Locus of E : x =
141. (2) Equation of circle can be written as
( x + 1 ) 2 + (y + 2 ) 2 = p + 5
...(i)
Y
X′
(–2, 0)
O
X
O (–3, 2)
B
5
5Ö2
(0, –4)
Y′
A (2,–3)
Case I. For p = 0, circle passes through origin and cuts x-axis
and y-axis at ( − 2, 0 ) and ( 0, − 4 ) respectively.
Case II. If circle touch X-axis, then
Y
OA = 5 2
⇒
Also,
(Q AB = radius of the given circle)
AB = 5
Now, in ∆OAB,
(OB ) 2 = ( AB ) 2 + (OA ) 2 = 25 + 50 = 75
∴
OB = 5 3
X′
O
X
(0, –2+√3)
From Eq. (i), we get
( x + 1 ) 2 + (y + 2 ) 2 = 2 2
Cut off Y -axis at (put x = 0)
(y + 2 ) 2 = 3
(0, –2–√3)
Y′
⇒ y = −2 ± 3
140. Circle : x 2 + y 2 = 1
Equation of tangent at P(cosθ,sin θ )
x cosθ + y sin θ = 1
Equation of normal at P
y = x tanθ
Equation of tangent at S is x = 1
θ

 1 − cosθ 
∴
Q 1,
 = Q 1, tan 


2
sin θ 
(1 ) 2 = − p ⇒ p = − 1
…(i)
…(ii)
or (0, -2 ± 3)
Case III. If circle touch Y -axis, then
X′
(2 ) 2 = − p
⇒ p=−4
From Eq. (i), we get
( x + 1 ) 2 + (y + 2 ) 2 = 1
Cut off X -axis at (put y = 0)
( x + 1 ) 2 = − 3 (impossible)
Y
O
1
(–1, –2)
Y′
X
CHAPTER
05
Parabola
Learning Part
Session 1
● Introduction
● Section of a Right Circular Cone by Different Planes
● Equation of Conic Section
● How to Find the Centre of Conics
● Standard Equation of Parabola
● Other forms of Parabola with Latusrectum 4a
● General Equation of a Parabola
●
The Generalised form (y - k ) 2 = 4 a ( x - h)
Session 2
● Position of a Point ( x , y ) with respect to a Parabola
1
1
y 2 = 4 ax
●
●
●
●
Intersection of a Line and a Parabola
Point of Intersection of Tangents at any Two Points
on the Parabola
Point of Intersection of Normals at any Two Points
on the Parabola
Co-normal Points
Session 3
2
● Pair of Tangents SS
1 =T
●
●
●
Equation of the Chord Bisected at a Given Point
Lengths of Tangent, Subtangent, Normal and Subnormal
Reflection Property of a Parabola
●
●
●
●
●
●
●
●
Conic Section
Conic Section : Definition
Recognisation of Conics
Parabola : Definition
Some Terms Related to Parabola
Smart Table
Equation of Parabola if Equation of axis, Tangent
at Vertex and Latusrectum are given
Parabolic Curve
●
Parametric Relation between the Coordinates of
the Ends of a Focal Chord of a Parabola
●
Equation of Tangent in Different Forms
Equation of Normals in Different Forms
●
●
Relation Between ’ t1 ’ and ‘ t 2 ’ if Normal at ‘ t1 ’
meets the Parabola Again at ’t 2 ’
●
Circle Through Co-normal Points
●
Chord of Contact
●
Diameter
Some Standard Properties of the Parabola
Study of Parabola of the Form
(ax + by ) 2 + 2 gx + 2 fy + c = 0
●
●
Practice Part
●
JEE Type Examples
●
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
Session 1
Introduction, Conic Section, Section of a Right Circular
Cone by Different Planes, Conic Section : Definition, Equation
of Conic Section, Recognisation of Conics, How to Find the
Centre of Conics, Parabola : Definition, Other forms of Parabola
with Latusrectum 4a, General Equation of a Parabola, The
Generalised form (y-k)2 = 4a (x-h), Parabolic Curve
Introduction
The famous Greek mathematician Euclid, the father of
creative Geometry, near about 300 BC considering various
plane sections of a right circular cone found many curves,
which are called conics or conic sections.
Section of a Right Circular
Cone by Different Planes
1. Section of a right circular cone by a plane which is
passing through its vertex is a pair of straight lines,
lines always passes through the vertex of the cone.
Conic Section
V Plane
Let l 1 be a fixed vertical line and l 2 be another line
intersecting it at a fixed point V and inclined to it at an
angle a.
P
Axis
Q
2. Section of a right circular cone by a plane which
parallel to its base is a circle.
O′
Upper
nappe
l1
α
l2
Generator
V
Plane
V
Lower
nappe
O
Circle
Circular base
O
Suppose we rotate the line l 2 around the line l 1 in such a
way the angle remains constant then, the surface
generated is a double-napped right circular hollow cone.
The point V is called the vertex, the line l 1 is the axis of
the cone. The rotating line l 2 is called a generator of the
cone. The vertex separates the cone into two parts called
nappes. The constant angle a is called the semi-vertical
angle of the cone.
3. Section of a right circular cone by a plane which is
parallel to a generator of the cone is a parabola.
Plane
V
Parabola
O
Chap 05 Parabola
4. Section of a right circular cone by a plane which is
not parallel to any generator and not parallel or
perpendicular to the axis of the cone is an ellipse.
V
Plane
365
The fixed point is called the focus of the conic and this
fixed line is called the directrix of the conic. Also, this
constant ratio is called the eccentricity of the conic and is
denoted by e.
SP
In the figure,
= constant = e
PM
Þ
SP = e PM
Ellipse
O
Equation of Conic Section
5. Section of a right circular cone by a plane which is
parallel to the axis of the cone is a hyperbola.
O′
If the focus is (a,b) and the directrix is ax + by + c = 0, then
the equation of the conic section whose
eccentricity = e is SP = e PM
P(x, y)
Hyperbola
ax + by + c = 0
M
V
Axis
S (α, β)
O
3D View
Circle
Ellipse
|ax + by + c |
Þ
( x - a ) 2 + (y - b) 2 = e ×
Þ
( x - a ) 2 + (y - b) 2 = e 2 ×
(a 2 + b 2 )
(ax + by + c ) 2
(a 2 + b 2 )
.
Important Terms
Parabola
Hyperbola
Conic Section : Definition
The locus of a point which moves in a plane such that the
ratio of its distance from a fixed point to its perpendicular
distance from a fixed straight line is always constant, is
known as a conic section or a conic.
Directrix
M
P
Axis The straight line passing through the focus and
perpendicular to the directrix is called the axis of the conic
section.
Vertex The points of intersection of the conic section
and the axis is (are) called vertex (vertices) of the conic
section.
Focal Chord Any chord passing through the focus is
called focal chord of the conic section.
Double Ordinate A straight line drawn perpendicular to
the axis and terminated at both end of the curve is a
double ordinate of the conic section.
Latusrectum The double ordinate passing through the
focus is called the latusrectum of the conic section.
Centre The point which bisects every chord of the conic
passing through it, is called the centre of the conic section.
S (focus)
Remark
Parabola has no centre but circle, ellipse and hyperbola have
centre.
366
Textbook of Coordinate Geometry
y Example 1 Find the locus of a point, which moves
such that its distance from the point (0, –1) is twice its
distance from the line 3x + 4 y + 1 = 0.
Sol. Let P ( x 1,y1 ) be the point, whose locus is required.
Its distance from (0, - 1) = 2 ´ its distance from the line
3x + 4y + 1 = 0.
|3x + 4y1 + 1 |
Þ
( x 1 - 0) 2 + ( y 1 + 1) 2 = 2 ´ 1
( 32 + 4 2 )
Þ
5
x 12
Condition
or
+
e >1 ; D = 0, h > ab
The lines will be real and
distinct intersecting at S.
e = 1 ; D = 0, h 2 = ab
The lines will coincident
2
+ ( y 1 + 1) = 2 | 3x 1 + 4y 1 + 1 |
=4
39y12
Nature of Conic
2
Squaring and simplifying, we have
25 ( x 12 + y12 + 2y1 + 1)
11x 12
The nature of the conic section depends upon the position
of the focus S with respect to the directrix and also upon
the value of the eccentricity e. Two different cases arise.
Case I (When the focus lies on the directrix)
In this case Eq. (i) represents the Degenerate conic
whose nature is given in the following table :
(9 x 12
+
16y12
2
The lines will be imaginary.
e < 1 ; D = 0, h < ab
+ 24 x 1y1 + 6x 1 + 8y1 + 1)
+ 96x 1y1 + 24 x 1 - 18y1 - 21 = 0
Hence, the locus of ( x 1,y1 ) is
11x 2 + 39y 2 + 96xy + 24 x - 18y - 21 = 0
Case II (When the focus does not lie on the
directrix)
In this case Eq. (i) represents the Non-degenerate conic
whose nature is given in the following table :
Condition
y Example 2 What conic does the equation
25 ( x 2 + y 2 - 2x + 1) = (4 x - 3y + 1) 2 represent ?
Sol. Given equation is
25 ( x 2 + y 2 - 2x + 1) = ( 4 x - 3y + 1)2
2
2
0 < e < 1; D ¹ 0, h < ab
Write the right hand side of this equation, so that it appears
in perpendicular distance form, then
æ 4 x - 3y + 1 ö
÷
( 4 x - 3y + 1)2 = 25 ç
ç ( 4 2 + 32 ) ÷
è
ø
2
or
( x - 1) 2 + ( y - 0) 2 =
rectangular hyperbola
e > 1; D ¹ 0, h > ab; a + b = 0
Remark
Sol. Compare the given equation with
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
\
a = 13, h = - 9, b = 37, g = 1, f = 7, c = - 2 ,
then, D = abc + 2 fgh - af 2 - bg 2 - ch 2
( 4 2 + 32 )
Here, e = 1
Thus, the given equation represents a parabola. It may
noted that (10
, ) is the focus and 4 x - 3y + 1 = 0 is the
directrix of the parabola.
= (13) (37) ( -2) + 2 (7) (1) ( -9)
- 13 (7 )2 - 37(1)2 + 2 ( -9 )2
= - 962 - 126 - 637 - 37 + 162 = - 1600 ¹ 0
and also h 2 = ( -9 )2 = 81 and ab = 13 ´ 37 = 481
Here, h 2 < ab
Recognisation of Conics
So, we have h 2 < ab and D ¹ 0.
Hence, the given equation represents an ellipse.
The equation of conics represented by the general
equation of second degree
...(i)
can be recognised easily by the condition given in the
tabular form. For this, first we have to find discriminant of
the equation. We know that the discriminant of above
equation is represented by D, where
D = abc + 2 fgh - af 2 - bg 2 -ch 2
a hyperbola
2
e > 1; D ¹ 0, h > ab
y Example 3 What conic does
13x 2 - 18 xy + 37 y 2 + 2x + 14 y - 2 = 0 represent?
2
| 4 x - 3y + 1 |
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
an ellipse
2
1. If conic represents an empty set, then D ¹ 0, h2 < ab.
2. If conic represents a single point, the D = 0, h2 < ab.
then, Eq. (i) can be re-written as
é 4 x - 3y + 1 ù
25 [( x - 1)2 + (y - 0)2 ] = 25 ê
ú
êë ( 4 2 + 32 ) úû
a parabola
e =1 ; D ¹ 0, h = ab
…(i)
Nature of Conic
y Example 4 What conic is represented by the
equation ax + by = 1 ?
Sol. Given conic is ax + by = 1
On squaring both sides, we get
ax + by + 2 abxy = 1
Þ
ax + by - 1 = - 2 abxy
Chap 05 Parabola
Again, on squaring both sides, then
(ax + by - 1)2 = 4abxy
a 2 x 2 + b 2y 2 + 1 + 2abxy - 2by - 2ax = 4abxy
Þ
Þ
a 2 x 2 + b 2y 2 - 2abxy - 2ax - 2by + 1 = 0
Þ
a 2 x 2 - 2abxy + b 2y 2 - 2ax - 2by + 1 = 0
…(i)
1
3
5
then, a = 2, h = , b = 3 , g = - , f = , c = l.
2
2
2
1
2
h = , ab = 6
Q
4
2
\
h < ab
and
D = abc + 2 fgh - af 2 - bg 2 - ch 2
Comparing the Eq. (i) with the equation
A = a 2 , H = - ab, B = b 2 , G = - a , F = -b, C = 1
then, D = ABC + 2FGH - AF 2 - BG 2 - CH 2
= a 2b 2 - 2a 2b 2 - a 2b 2 - a 2b 2 - a 2b 2
= -4a 2b 2 ¹ 0 and H 2 = a 2b 2 = AB
So, we have D ¹ 0 and H 2 = AB.
Hence, the given equation represents a parabola.
y Example 5 If the equation x 2 - y 2 - 2x + 2y + l = 0
represents a degenerate conic, find the value of l.
Sol. For degenerate conic D = 0
Comparing the given equation of conic with
ax 2 + 2hxy + by 2 + 2gx + 2 f y + c = 0
\
\
a = 1, b = - 1, h = 0, g = -1, f = 1, c = l
D = abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
Þ ( 1) ( - 1) ( l ) + 0 - 1 ´ ( 1) 2 + 1 ´ ( - 1) 2 - l ( 0) 2 = 0
Þ
- l -1+1=0 Þ l =0
y Example 6 If the equation x 2 + y 2 - 2x - 2y + c = 0
represents an empty set, then find the value of c.
Sol. For empty set D ¹ 0 and h 2 < ab.
Now, comparing the given equation of conic with
ax 2 + 2hxy + by 2 + 2gx + 2 f y + c ¢ = 0
then a = 1, h = 0, b = 1, g = -1, f = -1, c ¢ = c
Q
h 2 < ab
\
and
0 < 1 which is true
D = abc ¢ + 2 fgh - af 2 - bg 2 - c ¢ h 2 ¹ 0
Þ
(1)(1)(c ) + 0 - 1 ´ ( -1)2 - 1 ´ ( -1)2 - 0 ¹ 0
Þ
\
Hence,
c -2 ¹ 0
c ¹2
c Î R~ (2)
y Example 7 If the equation of conic
2x 2 + xy + 3y 2 - 3x + 5y + l = 0
represent a single point, then find the value of l.
Sol. For single point,
h 2 < ab and D = 0
Comparing the given equation with
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
5
3 1
´- ´
2
2 2
9
1
25
-2´
-3´ - l ´
4
4
4
15 25 27 l
= 6l 4
2
4
4
23l
=
- 23 = 0
4
l=4
= ( 2) ( 3) ( l ) + 2 ´
Ax 2 + 2Hxy + By 2 + 2Gx + 2Fy + C = 0
\
367
\
y Example 8 For what value of l the equation of conic
2xy + 4 x - 6 y + l = 0 represents two intersecting
straight lines, if l = 17, then this equation represents?
Sol. Comparing the given equation of conic with
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
\
a = 0 , b = 0, h = 1 , g = 2 , f = -3, c = l
For two intersecting lines,
h 2 > ab, D = 0
Q
\
ab = 0, h = 1
h 2 > ab
and
D = abc + 2 fgh - af
2
- bg 2 - ch 2
= 0 + 2 ´ - 3 ´ 2 ´ 1 - 0 - 0 - l ( 1) 2
= -12 - l = 0
\
l = - 12
For l = 17, then the given equation of conic
2xy + 4 x - 6y + 17 = 0 according to the first system but
here c = 17.
\
a = 0, b = 0, h = 1, g = 2 , f = -3, c = 17,
\
D = abc + 2 f gh - af 2 - bg 2 - ch 2
= 0 + 2 ´ - 3 ´ 2 ´ 1 - 0 - 0 - 17 ´ (1)2
\
= - 12 - 17 = -29 ¹ 0
D ¹ 0 and h 2 > ab
So, we have D ¹ 0 and h 2 > ab.
Hence, the given equation represents a hyperbola.
How to Find the Centre of Conics
If S º ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0.
Partially differentiating w.r.t.x and y, we get
¶S
¶S
= 2ax + 2hy + 2 g ;
= 2hx + 2by + 2 f
¶x
¶y
(Treating y as constant)
(Treating x as constant)
368
Textbook of Coordinate Geometry
For centre,
¶S
= 0 and
¶x
¶S
=0
¶y
\ 2ax + 2hy + 2 g = 0 and 2hx + 2by + 2 f = 0
Þ
ax + hy + g = 0 and
hx + by + f = 0
Solving these equations we get the centre
æ hf - bg gh - af ö
( x ,y ) = ç
,
÷.
è ab - h 2 ab - h 2 ø
æ ( -2)( -29 ) - (11)( -22) ( -22) ( -2) - (14 ) ( -29 ) ö
,
=ç
÷
è (14 )(11) - ( -2)2
(14 ) (11) - ( -2)2
ø
or
= (2, 3)
A parabola is the locus of a point which moves in a plane
such that its distance from a fixed point (i.e. focus) is
always equal to its distance from a fixed straight line
(i.e., directrix).
a
h
(Repeat Ist member)
or
Standard Equation of Parabola
Let S be the focus and ZM be the directrix of the parabola.
Draw SZ perpendicular to ZM, let A be the mid point of
SZ, then as
AS = AZ
ab - h 2 , hf - bg, gh - af
æ hf - bg gh - af ö
or points ç
,
÷
è ab - h 2 ab - h 2 ø
or
æ C 13 C 23 ö
,
ç
÷.
è C 33 C 33 ø
Y
OR
According to first two rows,
ax + hy + g = 0 and hx + by + f = 0.
After solving we get find the centre of conic.
Z
y Example 9 Find the centre of the conic
14 x 2 - 4 xy + 11y 2 - 44 x - 58 y + 71 = 0
2
Differentiating partially w.r.t. x and y, then
¶f
¶f
= 28x - 4y - 44 and
= - 4 x + 22y - 58
¶x
¶y
\
or
and
or
¶f
= 0 and
¶x
¶f
= 0,
¶y
28x - 4y - 44 = 0
7 x - y - 11 = 0
Then,
¼(i)
-4 x + 22y - 58 = 0
-2x + 11y = 29
On solving Eqs. (i) and (ii) we get,
x = 2 and y = 3
\ Centre is (2, 3).
Aliter : Comparing the given conic with
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
\ a = 14, h = - 2, b = 11, g = - 22, f = - 29, c = 71
A
S
(a, 0)
N
X
So, A lies on the parabola. Take A as the origin and a line
AY through A perpendicular to AX as Y-axis.
Let
AS = AZ = a > 0
then, coordinate of S is (a,0 ) and the equation of ZM is
x = - a or x + a = 0
Now, take P ( x , y ) be any point on the parabola. Join SP
and from P draw PM perpendicular to the directrix ZM.
2
Sol. Let f ( x , y ) º 14 x - 4 xy + 11y - 44 x - 58y + 71 = 0
For centre,
P (x, y)
M
x+a=0
\
æ hf - bg gh - af ö
Centre ç
,
÷
è ab - h 2 ab - h 2 ø
Parabola : Definition
Remembering Method
½a h g½
Since,
D =½
½h b f½
½
½g f c½
Write first two rows,
a
h
g
i.e.
h b
f
\
¼ (ii)
SP = ( x - a ) 2 + (y - 0 ) 2 = ( x - a ) 2 + y 2
and
PM = ZN = AZ + AN = a + x
Now, for the parabola SP = PM
Þ
(SP ) 2 = ( PM ) 2 Þ ( x - a ) 2 + y 2 = (a + x ) 2
Þ
y 2 = (a + x ) 2 - ( x - a ) 2 = 4ax
\
y 2 = 4ax ,
which is required equation of the parabola.
Remark
A parabola has two real foci situated on its axis one of which
is the focus S and the other lies at infinity. the corresponding
directrix is also at infinity.
Chap 05 Parabola
1. Axis The axis of the parabola is the straight line
which is passes through focus and perpendicular to
the directrix of the parabola.
Y
L
P Q (h, 2Öah)
Focal
chord
Double
ordinate
X
S (a, 0) N
x+a=0
M
Z
Latus
rectum
(0, 0)
A
Focus
P¢
L¢
Directrix
Axis
Q¢ (h, – 2Öah)
For the parabola y 2 = 4ax , X-axis is the axis.
Here, all powers of y are even in y 2 = 4ax then,
parabola y 2 = 4ax is symmetrical about its axis (i.e.
X-axis).
or
2
If the point ( x , y ) lie on the parabola y = 4ax , then
the point ( x , - y ) also lies on it. Hence, the parabola is
symmetrical about X-axis (i.e. axis of parabola).
2. Vertex The point of intersection of the parabola and
its axis is called the vertex of the parabola. For the
parabola y 2 = 4ax .
A (0, 0 ) i.e. the origin is the vertex.
3. Double ordinate If Q be the point on the parabola,
draw QN perpendicular to the axis of parabola and
produced to meet the curve again at Q ¢, then QQ ¢ is
called a double ordinate.
If abscissa of Q is h, then ordinate of Q,
y 2 = 4ah or y = 2 ah
(for first quadrant)
and ordinate of Q ¢ is y = -2 ah (for fourth quadrant)
Hence, coordinates of Q and Q ¢ are (h, 2 ah ) and
(h, -2 ah ), respectively.
4. Latusrectum The double ordinate LL¢ passes through
the focus is called the latusrectum of the parabola.
Since focus S (a, 0 ) the equation of the latusrectum of
the parabola is x = a, then solving
x =a
5. Focal chord A chord of a parabola which is passing
through the focus is called a focal chord of the
parabola. In the given figure, PP ¢ and LL¢ are the focal
chords.
Remarks
1. In objective questions use LL¢ as focal chord and in
subjective questions use PP¢ as focal chord.
2. Length of smallest focal chord of the parabola 4a. Hence, the
latusrectum of a parabola is the smallest focal chord.
6. Focal distance The focal distance of any point P on
the parabola is its distance from the focus S i.e. SP
Also, SP = PM = Distance of P from the directrix.
If
P º (x , y )
then,
SP = PM = x + a
7. Parametric equations From the equation of the
y 2x
parabola y 2 = 4ax , we can write
=
=t
2a y
where ‘t ’ is a parameter.
Then, y = 2at
and x = at 2
The equations x = at 2 and y = 2at are called
parametric equations. The point (at 2 , 2at ) is also
referred to as the point ‘t ’.
Remarks
1. Coordinates of any point on the parabola y 2 = 4 ax, may be
taken as ( at 2, 2at ).
2. Equation of chord joining t1 and t2 is 2x - ( t1 + t2 ) y + 2at1t2 = 0.
3. If the chord joining t1, t2 and t3, t4 pass through a point ( c, 0 )
c
on the axis, then t1t2 = t3t4 = - .
a
Other forms of Parabola with
Latusrectum 4a
(1) Parabola opening to left (i.e. y 2 = - 4ax ) : (a > 0 )
(i) Vertex is A (0, 0 ).
(ii) Focus is S ( -a, 0 ).
(iii) Equation of the directrix MZ is x - a = 0.
(iv) Equation of the axis is y = 0 i.e. X-axis.
(v) Equation of the tangent at the vertex is x = 0 i.e. Y-axis.
Y
L
and y 2 = 4ax
then, we get
y = ± 2a
Hence, the coordinates of the extremities of the
latusrectum are L (a, 2a ) and L ¢ (a, - 2a ), respectively.
Since,
LS = L ¢ S = 2a
\ Length of latusrectum LL ¢ = 2 ( LS ) = 2 ( L ¢ S ) = 4a.
X′
P
M
S (–a,0) A
L′
Y′
(vi) Length of latusrectum = LL ¢ = 4a.
x– a=0
Some Terms Related to Parabola
369
Z
X
370
Textbook of Coordinate Geometry
(ix) Parametric coordinates is (2at , at 2 ).
(vii) Ends of latusrectum are L ( -a , 2a ) and L ¢ ( -a , - 2a ).
(viii) Equation of latusrectum is x = - a i.e. x + a = 0.
(ix) Parametric coordinates is ( -at 2 , 2at ).
(3) Parabola opening downwards (i.e.
x 2 = - 4ay ) : (a > 0 )
(2) Parabola opening upwards (i.e. x 2 = 4ay ) : (a > 0 )
(i) Vertex is A (0, 0 ).
(ii) Focus is S (0, a ).
(iii) Equation of the directrix MZ is y + a = 0.
(i)
(iii)
(iv)
(v)
Y
Vertex is A (0, 0 ). (ii) Focus is S (0, - a ).
Equation of the directrix MZ is y - a = 0.
Equation of the axis is x = 0 i.e.Y-axis.
Equation of the tangent at the vertex is y = 0 i.e.
X-axis.
Y
S (0, a)
L′
L
My–a=0
Z
P
X′
X
A
A
X′
X
P
M
y+a=0 Z
L′
S(0,–a)
L
Y′
(iv) Equations of the axis is x = 0 i.e.Y-axis.
(v) Equation of the tangent at the vertex is y = 0 i.e. X-axis.
(vi) Ends of latusrectum are L (2a, a ) and L ¢ ( -2a, a ).
(vii) Length of latusrectum = LL ¢ = 4a.
(viii) Equation of latusrectum is y = a i.e. y - a = 0.
Y′
(vi) Length of latusrectum = LL ¢ = 4a
(vii) Ends of latusrectum are L (2a, -a ) and L ¢ ( -2a, - a )
(viii) Equation of latusrectum is y = - a i.e. y + a = 0.
(ix) Parametric coordinates are (2at , - at 2 ).
Smart Table : The Study of Standard Parabolas
Equation and Graph of
the parabola
y 2 = 4ax, a > 0
y 2 = - 4ax, a > 0
x 2 = 4ay , a > 0
x 2 = - 4ay , a > 0
Y
Z
Y
Y
Y
L
M
X′
Z
A
S
L′
Y′
P
L
X X′
P
A
S
S
L′
M
Z
X
P
X′
A
A
X′
X
Z M
Y′
S
Y′
Vertex
( 0, 0 )
( 0, 0 )
( 0, 0 )
( 0, 0 )
Focus
(a, 0 )
( - a, 0 )
( 0, a )
( 0, - a )
Equation of the axis
y=0
y =0
x =0
x =0
Equation of tangent at
vertex
Equation of directrix
x =0
x =0
y =0
y =0
x +a =0
x -a = 0
y +a =0
y -a = 0
Length of latusrectum
4a
4a
4a
4a
Ends points of
latusrectum
Equation of
latusrectum
Focal distance of a
point P (x , y )
(a, ± 2a )
( - a, ± 2a )
( ± 2a, a )
( ± 2a, - a )
x -a = 0
x +a =0
y -a = 0
y +a =0
x +a
a -x
y +a
a -y
Parametric coordinates
(at 2, 2at )
( - at 2, 2at )
(2at, at 2 )
(2at, - at 2 )
1
1
1
1
Eccentricity (e)
X
P
L′
L′
Y′
M
L
L
Chap 05 Parabola
General Equation of a Parabola
Let S (a,b ) be the focus, and lx + my + n = 0 is the equation
of the directrix. Let P ( x ,y ) be any point on the parabola.
Then by definition SP = PM
Y
y)
(x,
P
+n
=0
my
S (a, b)
lx +
( x - a ) 2 + (y - b ) 2 =
Þ
( x - a ) 2 + (y - b ) 2 =
= ( x 2 + 4y 2 - 4 xy + 6x - 12y + 9 )
\
4 x 2 + y 2 + 4 xy + 4 x + 32y + 16 = 0
y Example 11 Find the equation of the parabola whose
focus is (4, - 3) and vertex is (4, - 1) .
-3 + 1
=¥
4-4
which is parallel to Y-axis.
\ Directrix parallel to X-axis.
\
L
A
L′
X
O
Þ
5 ( x 2 + y 2 + 2x + 4y + 5)
Sol. Let A ( 4, - 1) be the vertex and S ( 4, - 3) be the focus.
M
Z
Þ
371
Slope of AS =
Y
Z
y –1 = 0
O
X′
|lx +my + n |
M
X
A (4, –1)
(l 2 +m 2 )
(lx +my + n ) 2
P (x, y)
(l 2 +m 2 )
Þ m 2 x 2 + l 2 y 2 - 2lmxy + x term+ y term + constant = 0
S (4, –3)
2
This is of the form (mx - ly ) + 2 gx + 2 fy + c = 0.
This equation is the general equation of parabola.
Remark
Second degree terms in the general equation of a parabola
forms a perfect square.
y Example 10 Find the equation of the parabola whose
focus is at ( -1,-2) and the directrix is the straight line
x - 2y + 3 = 0.
Sol. Let P ( x ,y ) be any point on the parabola whose focus is
S( -1,-2) and the directrix x - 2y + 3 = 0 . Draw PM
perpendicular from P ( x ,y ) on the directrix x - 2y + 3 = 0 .
Y
x–
X′
2
y+
3=
0
P
(x, y)
Þ
æ| y - 1 |ö
( x - 4 ) + ( y + 3) = ç
÷
è 12 ø
Þ
( x - 4 ) 2 + ( y + 3) 2 = ( y - 1) 2
X
O
M
Y′
Let Z ( x 1,y1 ) be any point on the directrix, then A is the
mid-point of SZ .
x +4
\
4= 1
Þ x1 = 4
2
y -3
and
-1 = 1
Þ y1 = 1
2
\
Z = ( 4, 1)
Also, directrix is parallel to X-axis and passes through
Z ( 4,1) , so equation of directrix is
y = 1 or y - 1 = 0
Now, let P ( x ,y ) be any point on the parabola. Join SP and
draw PM perpendicular to the directrix. Then, by definition
SP = PM
Þ
(SP )2 = ( PM )2
\
2
2
x 2 - 8x + 8y + 24 = 0
Aliter :
Here a = AS = 2
\ Length of latusrectum = 4a = 8
Equation of parabola with vertex (0, 0) and open downward
is x 2 = - 8y .
S
(–1,–2)
Y′
Then, by definition
SP = PM
Þ
(SP ) 2 = ( PM )2
æ | x - 2y + 3|
Þ ( x + 1) 2 + ( y + 2) 2 = ç
ç ( 1) 2 + ( - 2) 2
è
2
ö
÷
÷
ø
2
Shifting ( 4, - 1) on (0, 0), we get required parabola
( x - 4 )2 = - 8(y + 1)
\
x 2 - 8x + 8y + 24 = 0
372
Textbook of Coordinate Geometry
Since, R divide QQ ¢ in 1 : 2
and T divide QQ ¢ in 2 : 1
For locus, let R (h ,k ), then
y Example 12 The focal distance of a point on a
parabola y 2 = 8 x is 8. Find it.
y 2 = 8x
Sol. Comparing
Y
M
x + 2 = 0 Directrix
with y 2 = 4ax
Z
y1
= k or y1 = 3k
3
On substituting the values of x 1 and y1in Eq. (ii), then
(3k )2 = 4a (h ) or 9k 2 = 4ah
x1 = h
P (x1, y1)
S
A
Focal
distance = 8
X
then,
or
…(i)
Q
SP = 8
Þ
PM = 8
Þ
x1 + 2 = 8
or
x1 = 6
From Eq. (i), y12 = 8 ´ 6
[QSP = PM ]
y1 = ± 4 3
\ The required points are (6, 4 3 ) and (6, - 4 3 ).
y Example 13 QQ ¢ is a double ordinate of a parabola
y 2 = 4ax . Find the locus of its point of trisection .
Sol. Let the double ordinate QQ' meet the axis of the parabola
y 2 = 4ax
…(i)
Let coordinates of Q be ( x 1, y1 ), then coordinates of Q ¢ be
( x 1 - y1 ) since, Q and Q ¢ lies on Eq. (i), then
…(ii)
y12 = 4ax 1
Let R and T be the points of trisection of QQ'. Then, the
coordinates of R and T are
æ 1 × x 1 + 2 × x 1 1 × ( -y 1 ) + 2 × y 1 ö
æ y1 ö
,
ç
÷ or ç x 1, ÷
è 3ø
è 1+2
1+2
ø
æ 2 × x 1 + 1 × x 1 2 × ( -y 1 ) + 1 × y 1 ö
and ç
,
÷
ø
è 2+1
2+1
respectively.
y ö
æ
or ç x 1,- 1 ÷
è
3ø
Y
Q
A
L
T
Q′
y1
= k'
3
Located
parabola
X
9k ¢ 2 = 4ah '
\ The required locus is 9y 2 = 4ax .
Hence, the locus of point of trisection is
9y 2 = 4ax .
Aliter : Let R and T be the points of trisection of double
ordinates QQ'. Let (h ,k ) be the coordinates of R,
then,
AL = h and RL = k
RT = RL + LT = k + k = 2k .
Since,
RQ = TR = Q ¢ T = 2k
\
LQ = LR + RQ = k + 2k = 3k
Thus, the coordinates of Q are (h , 3k ).
Since, (h , 3k ) lies on y 2 = 4ax
Þ
9k 2 = 4ah
Hence, the locus of (h , k ) is 9y 2 = 4ax .
y Example 14 Prove that the area of the triangle
inscribed in the parabola y 2 = 4ax is
1
( y 1 ~ y 2 ) ( y 2 ~ y 3 ) ( y 3 ~ y 1 ), where y 1 , y 2 , y 3 are
8a
the ordinates of the vertices.
Sol. Let the vertices of the triangle be ( x 1, y1 ) , ( x 2 , y 2 ) and
( x 3 , y 3 ).
Q ( x 1, y1 ) is a point on the parabola y 2 = 4ax .
\
y12 = 4ax 1
y12
4a
y2
Similarly,
x2 = 2
4a
y2
and
x3 = 3
4a
Now , vertices of triangle are
ö æy 2
ö
ö
æ y12
æy 2
ç ,y1 ÷ , ç 2 ,y 2 ÷ and ç 3 ,y 3 ÷.
ø è 4a ø
è 4a
è 4a ø
\
R
x 1 = h ' and -
or
y1 = - 3k '
On substituting the values of x 1 and y1 in Eq. (ii), then
( - 3k ' )2 = 4a (h ' )
y12 = 8x 1
\
and
\ The required locus is 9y 2 = 4ax similarly, let T (h ',k ' )
\
4a = 8 Þ a = 2
\ Equation of directrix is x + 2 = 0.
Let P ( x 1, y1 ) on the parabola
y 2 = 8x
\
(internally)
(internally).
x1 =
Chap 05 Parabola
y12
y1 1
4a
2
1 y
\ Required area of the triangle = | 2 y 2 1 |
2 4a
y 32
y3 1
4a
y12 y1 1
1 2
1
= | y2 y2 1 | =
(y 1 ~ y 2 ) (y 2 ~ y 3 ) (y 3 ~ y 1 )
8a 2
8a
y3 y3 1
y Example 15 Find the length of the side of an
equilateral triangle inscribed in the parabola y 2 = 4ax ,
so that one angular point is at the vertex.
Sol. Let ABC be the inscribed equilateral triangle, with one
angular point at the vertex A of the parabola
…(i)
y 2 = 4ax
Let the length of the side of equilateral triangle = l
\
AB = BC = CA = l
Y
Þ
æ | x - y + 2| ö
( x - 0) 2 + ( y - 0) 2 = ç
÷
ø
è
2
2 ( x 2 + y 2 ) = ( x - y + 2) 2
2x 2 + 2y 2 = x 2 + y 2 - 2xy + 4 x - 4y + 4
\ The coordinates of B is (l cos 30° , l sin 30° )
æl 3 l ö
, ÷.
ç
è 2 2ø
2
æl 3 ö
æl ö
Since, B lies on Eq. (i), then ç ÷ = 4a ç
÷ or l = 8a 3
è2ø
è 2 ø
y Example 16 Prove that the equation of the parabola
whose focus is (0,0) and tangent at the vertex is
x - y + 1 = 0 is x 2 + y 2 + 2xy - 4 x + 4 y - 4 = 0.
Equation of Parabola if
Equation of axis, Tangent at
Vertex and Latusrectum are
given
Let equation of axis is ax + by + c = 0 and equation of
tangent at vertex is bx - ay + d = 0.
Equation of parabola is
tio
at n of
ve ta
rte ng
en
x
t
C
Y′
Y
x
M
x–
A
X′
+
–y
2=
y+
Eq
ua
Sol. Let focus is S (0, 0) and A is the vertex of the parabola take
any point Z such that AS = AZ given tangent at vertex is
x - y + 1 = 0, since directrix is parallel to the tangent at
the vertex.
Z
2
X
l
i.e.,
2
=
2
2
\
l = ±2
Þ
l =2
[Q l is positive since directrix in this
case always lies in II quadrant]
\ Equation of directrix is x - y + 2 = 0.
Now, take P ( x , y ) be any point on the parabola, draw
PM ^ ZM , then from definition,
SP = PM
Þ
(SP )2 = ( PM )2
\ x 2 + y 2 + 2xy - 4 x + 4y - 4 = 0
30°
30°
A
|l |
Þ
Þ
l
X′
\ Equation of directrix is x - y + l = 0
where, l is constant.
Q A is the mid-point of SZ .
\
SZ = 2SA
|0 - 0 + l |
| 0 - 0 + 1|
=2´
Þ
2
2
( 1 + ( - 1) )
( 12 + ( - 1) 2 )
Þ
B
N
P(x,y)
A
0
1=
0
M
axis
P (x, y)
X
S
( PM ) 2 = (Latusrectum) (PN )
2
Y′
373
æ ax + by + c ö
æ
ö
÷ =(Latusrectum) ç bx - ay + d ÷
Þ ç
ç (b 2 + a 2 ) ÷
ç (a 2 + b 2 ) ÷
è
ø
ø
è
374
Textbook of Coordinate Geometry
y Example 17 Find the equation of the parabola whose
latusrectum is 4 units, axis is the line 3x + 4 y - 4 = 0 and
the tangent at the vertex is the line 4 x - 3y + 7 = 0 .
0
Sol. Let P ( x , y ) be any point on the parabola and let PM and PN
are perpendiculars from P on the axis and tangent at the
vertex respectively, then
Another form is ( x - h ) 2 = 4a (y - k ) axis parallel to
7=
Y
This is called generalised form of the parabola Eq. (i)
and axis A ¢ X ¢¢|| AX with its vertex at A ¢ (h, k ). Its focus
is at (a + h, k ) and length of latusrectum = 4a, the
equation of the directrix is
x =h -a Þ x +a -h =0
Y′′
4x
–3
y+
Y
N
S
P (x, y)
a
A
M
X′
y=k–a
X
O
3x
+
A′ (h, k)
Directrix
Z′
4y
–4
X′
=
X
A
0
Y′
Y′
Y-axis with its vertex (h,k ) its focus is at (h,a + k ) and
length of latusrectum = 4a, the equation of the directrix
is
y = k - a Þ y + a - k = 0.
( PM )2 = (latusrectum) ( PN )
2
Þ
æ
ö
æ 3x + 4y - 4 ö
ç
÷ = 4 ç 4 x - 3y + 7 ÷
ç 4 2 + ( - 3) 2 ÷
ç 32 + 4 2 ÷
è
ø
è
ø
\
(3x + 4y - 4 )2 = 20 ( 4 x - 3y + 7 )
Remark
The parametric equation of ( y - k )2 = 4a(x - h) are x = h + at 2
and y = k + 2at.
which is required parabola.
The Generalised form
Parabolic Curve
(y - k)2 = 4 a(x - h)
The parabola
y 2 = 4ax
…(i)
2
can be written as (y - 0 ) = 4a ( x - 0 ).
Now,
The vertex of this parabola is A(0,0 )
x=h–a
2
ìï æ
B ö
B 2 C üï
+ ý
= Aí ç x + ÷ 2A ø
4 A 2 A ïþ
ïî è
a
A′ (h, k)
S
X′′
2
ìï æ
B ö ( B 2 - 4 AC ) üï
= Aí ç x + ÷ ý
2A ø
4A2
ïþ
ïî è
Directrix
X′
A
2
B ö
1æ
B 2 - 4 AC ö
æ
or ç x + ÷ = ç y +
÷
è
2A ø
Aè
4A
ø
X
Y′
Now, when origin is shifted at A' (h, k ) without changing the
direction of axes, its equation becomes
(y - k ) 2 = 4a ( x - h )
y = Ax 2 + Bx + C
B
Cü
ì
= Aí x 2 + x + ý
A
Aþ
î
Y
Z′
The equations y = Ax 2 + Bx + C and x = Ay 2 + By + C
are always represents parabolas generally called
parabolic curve.
…(ii)
Comparing it with ( x - h ) 2 = 4a (y - k ) it represent a
æ B B 2 - 4 AC ö
parabola with vertex at (h,k ) = ç ,÷
4A
è 2A
ø
Chap 05 Parabola
and axis parallel to Y-axis and latusrectum =
1
| A|
and the curve opening upwards and downwards depending
upon the sign of A and B.
The optimum distance of its vertex V from OY is
B 2 - 4 AC
.
4A
Remarks
Y
1. The optimum distance of vertex from OX or OY can be easily
obtained using calculus Method.
2. Equation of the parabola with axis parallel to the X-axis is of
the form x = Ay 2 + By + C.
V (h, k)
3. Equation of the parabola with axis parallel to the Y-axis is of
the form
y = Ax 2 + Bx + C.
k
h
M
O
Method to Make Perfect Square
X
The optimum distance of its vertex V from OX is
B 2 - 4 AC
4A
and
x = Ay 2 + By + C
B
Cü
ì
= A íy 2 + y + ý
A
Aþ
î
2
ìïæ
B ö
B2
C üï
= A íç y +
+ ý
÷ 2A ø
4 A 2 A ïþ
ïîè
2
ìïæ
B 2 - 4 AC üï
B ö
= A íç y +
÷
ý
2A ø
4 A 2 ïþ
ïîè
x = ay 2 ± by + g
If
first make the coefficient of y 2 is unity
b
gü
ì
x = a íy 2 ± y + ý
a
aþ
î
Now, in braces write y and put the sign after y which
between y 2 and y i.e. ± and after this sign write the half
b
the coefficient of y i.e. .
2a
i.e.,
b ö
æ
Now, write in braces ç y ±
÷
è
2a ø
2
2
ìï æ
bö
b2
g üï
+ ý
x = aí ç y ± ÷ è
2a ø
4a 2 a ïþ
îï
\
Comparing it with (y - k ) 2 = 4a( x - h ), it represent a
parabola with vertex at
æ B 2 - 4 AC
B ö
(h, k ) = ç ,- ÷
4A
2A ø
è
and axis parallel to X-axis and latusrectum =
1
| A|
and the curve opening left and right depending upon the
sign of A and B.
Y
h
V (h , k )
k
O
X
2
b2
æ bö
ç ÷ = 2
è 2a ø
4a
and always subtract
2
B 2 - 4 AC ö
1 æ
B ö
æ
Þ çy +
÷
÷ = çx +
è
2A ø
4A
Aè
ø
N
375
2
ìï æ
b ö (b2 - 4 ga ) üï
= aí ç y ± ÷ ý
è
2a ø
4a 2 þï
îï
y Example 18 Find the vertex, focus, latusrectum, axis
and the directrix of the parabola x 2 + 8 x + 12y + 4 = 0.
Sol. The equation of parabola is
x 2 + 8x + 12y + 4 = 0
…(i)
2
Þ
( x + 4 ) - 16 + 12y + 4 = 0
Þ
( x + 4 )2 - 12 + 12y = 0
Þ
( x + 4 )2 = - 12y + 12
Þ
( x + 4 )2 = - 12 (y - 1)
Let
x + 4 = X,y -1 =Y
\
X 2 = -12Y
Comparing it with X 2 = -4aY
\
a=3
\ Vertex of Eq. (iii) is (0, 0)
i.e.
X = 0 ,Y = 0
…(ii)
…(iii)
376
Textbook of Coordinate Geometry
From Eq. (ii),
Hence, axis of parabola parallel to X-axis.
x + 4 = 0,y -1 = 0
\
x = - 4, y = 1
\ Vertex of Eq. (i) is ( -4, 1).
Foucs of Eq. (iii) is (0, –3)
i.e.
X = 0, Y = - 3
From Eq. (ii),
x + 4 = 0, y - 1 = - 3
\
x = -4 , y = -2
\ Focus of Eq. (i) is (–4, –2).
and latusrectum = 4a = 12.
Equation of axis of Eq. (iii) is X = 0
\ Equation of axis of Eq. (i) is x + 4 = 0
Equation of directrix of Eq. (iii) is
Y = 3 or y - 1 = 3
\
y-4 =0
\ Equation of directrix of Eq. (i) is
y - 4 = 0.
Y
(3, 2)
A
The equation is of the form
( y - k ) 2 = 4a ( x - h )
as (h,k) is the vertex (3,2)
a = distance between the focus and the vertex
= ( 5 - 3) 2 + ( 2 - 2) 2 = 2
(y + b )2 = - 2ax + b 2 - c
æ
b2 - c ö
(y + b )2 = - 2a ç x ÷
2a ø
è
…(i)
y +b =Y, x -
b -c
=X
2a
From Eq. (i) ,
Y 2 = -2aX
axis of its parabola is Y = 0
or
y + b = 0,
which is parallel to X-axis
and vertex of Eq. (ii) is X = 0, Y = 0
or
y 2 - 8x - 4y - 28 = 0.
y Example 21 Find the equation of the parabola with
latusrectum joining the points (3, 6) and (3, –2).
-2-6
= ¥, since latusrectum
3-3
is perpendicular to axis. Hence, axis parallel to X-axis. The
equation of the two possible parabolas will be of the form
…(i)
( y - k ) 2 = ± 4a ( x - h )
Since,
2
Let
Hence, the required equation is
( y - 2) 2 = 8( x - 3)
Sol. Slope of (3,6) and (3, –2) is
2
(y + b ) - b + 2ax + c = 0
Þ
( y - 2) 2 = 4a ( x - 3)
or
Sol. The equation of parabola is
y 2 + 2ax + 2by + c = 0
Þ
…(ii)
latusrectum = (3 - 3)2 + (6 + 2)2 = 8
\
4a = 8
Þ
a=2
\ From Eq. (i),
(y - k )2 = ± 8 ( x - h )
Since, (3,6) and (3, –2) lie on the parabola, then
(6 - k )2 = ± 8 (3 - h )
and
2
Y
x-
y Example 20 Find the equation of the parabola with
its vertex at (3, 2) and its focus at (5, 2).
L (3, 6)
A
(1, 2)
S (3, 2) A′ (5,2)
X
O
L′ (3, –2)
Sol. Let Vertex A (32
, ) and focus is S(52
, )
Slope of AS =
2-2
= 0, which is parallel to X-axis.
5-3
On solving Eqs. (ii) and (iii), we get
k =2
…(ii)
…(iii)
( -2 - k ) = ± 8 ( 3 - h )
b2 - c
= 0, y + b = 0
2a
b2 - c
, y = -b
x=
Þ
2a
æb2 - c
ö
\ Vertex of given parabola is ç
, - b ÷.
è 2a
ø
Þ
X
O
y Example 19 Prove that the equation
y 2 + 2ax + 2by + c = 0 represents a parabola whose axis
is parallel to the axis of x. Find its vertex.
2
S (5, 2)
Chap 05 Parabola
From Eq. (ii),
Þ
\
and
Þ
\
16 = ± 8 (3 - h ),
\
h = 3 ±2
\
h = 5,1
Hence, values of (h, k) are (5,2) and (1, 2).
The required parabolas are
( y - 2) 2 = 8( x - 5)
and
(y - 2)2 = -8( x - 1).
y Example 22 Find the equation to the parabola whose
axis parallel to the Y-axis and which passes through
the points (0, 4) (1, 9) and (4, 5) and determine its
latusrectum .
Sol. The equation of parabola parallel to Y-axis is
y = Ax 2 + Bx + C
The points (0,4) , (1,9) and (4,5) lie on Eq. (i), then
4 = 0+0+C Þ C = 4
Þ
9 = A + B +C
…(i)
9 = A+B+4
[QC = 4 ]
…(iii)
A+B=5
5 = 16A + 4 B + C
[QC = 4 ]
5 = 16A + 4 B + 4
16A + 4 B = 1
1
…(iv)
Þ
4A + B =
4
On solving Eqs. (iii) and (iv), we get
19
79
…(v)
A = - ,B =
12
12
On substituting the values of A , B and C from Eqs. (ii) and
Eq. (v) in Eq. (i), then equation of parabola is
19
79
y = - x2 + x + 4
12
12
1
12
Hence, length of latusrectum =
= .
19
19
12
…(ii)
Exercise for Session 1
1.
The vertex of the parabola y 2 + 6x - 2y + 13 = 0 is
(a) (-2, 1)
(c) (1, 1)
2.
1
3
2
3
4
(d)
3
(b)
(c) 1
3.
The value of p such that the vertex of y = x 2 + 2px + 13 is 4 units above the X-axis is
(a) ± 2
(c) ± 3
4.
(b) 4
(d) 5
The length of the latusrectum of the parabola whose focus is (3,3) and directrix is 3 x - 4y - 2 = 0, is
(a) 1
(c) 4
5.
6.
(b) 2
(d) 8
If the vertex and focus of a parabola are (3,3) and (–3, 3) respectively , then its equation is
(a) x 2 - 6x + 24y - 63 = 0
(b) x 2 - 6x + 24y + 81 = 0
(c) y 2 - 6y + 24x - 63 = 0
(d) y 2 - 6y - 24x + 81 = 0
If the vertex of the parabola y = x 2 - 8x + c lies on X-axis, then the value of c is
(a) 4
(c) 16
7.
(b) (2, - 1)
(d) (1 , –1)
If the parabola y 2 = 4 ax passes through (3, 2), then the length of latusrectum is
(a)
(b) – 4
(d) –16
The parabola having its focus at (3,2) and directrix along the Y-axis has its vertex at
3
(a) æç , 1ö÷
è2 ø
3
(b) æç , 2ö÷
è2 ø
3
(c) æç ,
è2
3
(d) æç , è2
1ö
÷
2ø
377
1ö
÷
2ø
378
Textbook of Coordinate Geometry
8.
The directrix of the parabola x 2 - 4x - 8y + 12 = 0 is
(a) y = 0
(c) y = -1
9.
The equation of the latusrectum of the parabola x 2 + 4x + 2y = 0 is
(a) 3y - 2 = 0
(c) 2y - 3 = 0
10.
12.
(b) 3y + 2 = 0
(d) 2y + 3 = 0
The focus of the parabola x 2 - 8x + 2y + 7 = 0 is
1
(a) æç 0, - ö÷
è
2ø
9
(c) æç 4, ö÷
è 2ø
11.
(b) x = 1
(d) x = -1
(b) (4,4)
9
(d) æç -4, - ö÷
è
2ø
The equation of the parabola with the focus (3,0) and directrix x + 3 = 0 is
(a) y 2 = 2x
(b) y 2 = 3x
(c) y 2 = 6x
(d) y 2 = 12x
Equation of the parabola whose axis is parallel to Y-axis and which passes through the points (1,0) , (0,0) and
( -2,4), is
(a) 2x 2 + 2x = 3y
(b) 2x 2 - 2x = 3y
(c) 2x 2 + 2x = y
(d) 2x 2 - 2x = y
13.
Find the equation of the parabola whose focus is (5,3) and directrix is the line 3x - 4y + 1 = 0 .
14.
Find the equation of the parabola is focus is at (–6, –6) and vertex is at (–2, 2).
15.
Find the vertex, focus, axis, directrix and latusrectum of the parabola 4y 2 + 12x - 20y + 67 = 0.
16.
æxö
Find the name of the conic represented by ç ÷ +
èa ø
17.
Determine the name of the curve described parametrically by the equations
æyö
ç ÷ = 1.
èb ø
x = t 2 + t + 1, y = t 2 - t + 1.
18.
Prove that the equation of the parabola whose vertex and focus are on the X-axis at a distance a and a ¢ from
the origin respectively is y 2 = 4 (a ¢ - a ) ( x - a ).
19.
Find the equation of the parabola whose axis is parallel to X-axis and which passes through the points (0, 4) ,
(1, 9) and (–2, 6). Also, find its latusrectum.
20.
The equation ax 2 + 4xy + y 2 + ax + 3y + 2 = 0 represents a parabola, then find the value of a.
Session 2
Position of a Point (x1, y1) with respect to a Parabola y2 = 4ax,
Parametric Relation between the Coordinates of the Ends of a Focal
Chord of a Parabola, Intersection of a Line and a Parabola, Equation of
Tangent in Different Forms, Point of Intersection of Tangents at any
Two Points on the Parabola, Equation of Normals in Different Forms,
Point of Intersection of Normals at any Two Points on the Parabola,
Circle Through Co-normal Points
Position of a Point (x 1 , y 1 ) with
Respect to a Parabola y 2 = 4 ax
Theorem The point ( x 1 , y 1 ) lies outside, on or inside the
parabola y 2 = 4ax according as
y 12
- 4ax 1 >, =, or < 0 .
Proof Let P ( x 1 , y 1 ) be a point. From P draw PM ^ AX
(on the axis of parabola) meeting the parabola y 2 = 4ax at
Q let the coordinate of Q be ( x 1 , y 2 ).
P (x1, y1)
Y
A
M
We have, k 2 - 4k + 9h + 13 = (2)2
Hence,
K(i)
2
Now, P will be outside, on or inside the parabola y = 4ax
according as
PM >, =, or <QM
y 12 > , = ,or < y 22
Þ
Hence,
k 2 - 3h > 0
Sol. Let the point (h ,k ) = ( -2,2)
y 22 = 4ax 1
Þ
Sol. Let the point (h , k ) = (2, 3)
with respect to the parabola y 2 - 4 y + 9 x + 13 = 0.
y 12 = 4ax
( PM ) >, = ,or <(QM )
y Example 23 Show that the point (2, 3) lies outside
the parabola y 2 = 3x .
y Example 24 Find the position of the point ( -2,2)
Since, Q ( x 1 , y 2 ) lies on the parabola
Þ
3. The point ( x1, y1 ) lies inside, on or outside x 2 = - 4 ay
according as x12 + 4 ay1 <, = ,or > 0
This shows that (2,3) lies outside the parabola y 2 = 3x .
X
Y′
2
2. The point ( x1, y1 ) lies inside, on or outside x 2 = 4 ay according
as x12 - 4 ay1 <, = ,or > 0
\
y 2 = 4ax
then,
1. The point ( x1, y1 ) lies inside, on or outside y 2 = - 4 ax
according as y12 + 4 ax1 <, =,or > 0
We have, k 2 - 3h = 32 - 32
. = 9 -6= 3 > 0
Q (x1, y2)
X′
Remarks
2
y 12 >, = , or < 4ax 1
y 12 - 4ax 1 > , = , or < 0
[from Eq. (i)]
- 4 (2) + 9 ( -2) + 13 = 4 - 8 - 18 + 13 = -9 < 0
k 2 - 4k + 9h + 13 < 0
Therefore, the point ( -2,2) lies inside the parabola
y 2 - 4y + 9 x + 13 = 0.
Parameteric Relation between
the Coordinates of the Ends of
a Focal Chord of a Parabola
Let y 2 = 4ax be a parabola, if PQ be a focal chord.
Then, P º (at 12 , 2at 1 ) and Q º (at 22 , 2at 2 )
Since, PQ passes through the focus S (a ,0 ).
380
\
Textbook of Coordinate Geometry
Q ,S , P are collinear.
Remark
Y
2
)
P (at1 , 2at1
t+
Q
1
³ 2for all t ¹ 0
t
[Q AM ³ GM]
2
1
a æçt + ö÷ ³ 4a
è tø
\
X
S (a , 0 )
Q (a
t2 2,
Þ Length of focal chord ³ latusrectum i.e. The length of
smallest focal chord of the parabola is 4a . Hence, the
latusrectum of a parabola is the smallest focal chord.
2at
y Example 26 Prove that the semi-latusrectum of the
parabola y 2 = 4ax is the harmonic mean between the
segments of any focal chord of the parabola.
2)
\ Slope of PS = Slope of QS
2at 1 - 0 0 - 2at 2
2t
2t 2
=
Þ 1 =
Þ
at 12 - a
a - at 22
t 12 - 1 t 22 - 1
Þ
t 1 (t 22 - 1) = t 2 (t 12 - 1)
Þ
Þ
t 1 t 2 (t 2 - t 1 ) + (t 2 - t 1 ) = 0
t 2 - t 1 ¹ 0 or t 1 t 2 + 1 = 0
If PQ be the focal chord, if
æ a -2a ö
P º (at 2 , 2at ), then Q º ç 2 ,
÷
èt
t ø
1
t 1 t 2 = -1 or t 2 = - ,
t1
Þ
Sol. Let parabola be y 2 = 4ax
…(i)
which is required relation.
Remark
If one extremity of a focal chord is (at 12 , 2at1 ) then the other
æ a 2a ö
extremity (at22 , 2at2 ) becomes çç 2 ,- ÷÷ by virtue of relation Eq. (i).
t1 ø
è t1
\ Length of latusrectum LL ' = 4a.
1
\ Semi-latusrectum = ( 4a ) = 2a.
2
If sections of focal chord are k1 and k 2 ,
then,
k1 = SP = PM = a + at 2 = a(1 + t 2 )
and
k 2 = SQ = QN = a +
y Example 25 If the point (at 2 , 2at ) be the extremity of
a focal chord of parabola y 2 = 4ax then show that the
æ 1ö
length of the focal chord is a çt + ÷ .
è tø
Y
at)
P (at 2
x+a=0
N
Q
a,
t2
and
k2 =
X
=
–
2a
t
\ Length of focal chord = PQ
[QSP = PM and SQ = QN ]
= SP + SQ
= PM + QN
a
= at 2 + a + 2 + a
t
2
1ö
ö
æ 2 1
æ
= a çt + 2 + 2÷ = a çt + ÷
ø
è
è
tø
t
a(1 + t 2 )
t2
d
Q
a,
P (at2, 2at)
K1
S (a, 0)
–
X
L′
2a
t
\ Harmonic mean of k1
=
S (a, 0)
(0,0)
A
t2
M
A
=
L
N
2,
t
2
M
Z
2
Sol. Since, one extremity of focal chord is P (at 2 ,2at ), then the
æ a 2a ö
other extremity is Q ç 2 ,- ÷
[Replacing t by –1/t]
èt
t ø
a
Y
x+a=0
A
2k1k 2
k1 + k 2
2
2
=
2
1
1
t
1
+
+
k 2 k 1 a (1 + t 2 ) a (1 + t 2 )
2
= 2a = Semi-latusrectum.
1
a
Remarks
1. The length of focal chord having parameters t1 and t2 for its
end points is a ( t2 - t1 ) 2.
2. If l1 and l2 are the length of segments of a focal chord of a
4l l
parabola, then its latusrectum is 1 2 .
l1 + l2
381
Chap 05 Parabola
y Example 27. Show that the focal chord of
parabola y 2 = 4ax makes an angle a with the
X-axis is of length 4a cosec 2a.
Sol. Let P (at 12 , 2at 1 ) and Q (at 22 , 2at 2 ) be the end points of a
focal chord PQ which makes an angle a with the axis of
the parabola. Then,
PQ = a (t 2 - t 1 )2
2
2
tana = slope of PQ
2at 2 - 2at 1
=
at 22 - at 12
Þ
tana =
Þ
…(ii)
Þ
t 2 + t 1 = 2 cota
On substituting the value of t 2 + t 1 from Eq. (ii) in Eq. (i),
then
PQ = a( 4 cot 2 a + 4 )
x+a=0
2,
P (at 2at)
M
Z
(0,0)
A
d
N
S (a, 0)
Q
a,
t2
–
X
2a
t
\ Equation of focal chord is (i.e. equation of PS)
2at - 0
y -0= 2
( x - a)
at - a
y=
Þ
1ö
l æ
= çt + ÷
a è
tø
\
l=
2t
2
( t - 1)
( x - a)
2
Þ
(t - 1)y = 2tx - 2at
Þ
2tx - (t 2 - 1)y - 2at = 0
If d be the distance of this focal chord from the vertex (00
, )
of the parabola y 2 = 4ax , then
d=
| 0 - 0 - 2at |
(2t )2 + (t 2 - 1)2
1
t
1ö
æ
çt + ÷
è
tø
…(i)
2
2
…(ii)
4a 3
2
4a 2
4a 3
=
l
(l / a )
Þ lµ
1
d
d2
i.e. the length of the focal chord varies inversely as the
square of its distance from vertex.
Intersection of a Line and a
Parabola
Let the parabola be
Y
t +
4a 2
From Eqs. (i) and (ii), d 2 =
= 4a cosec 2 a.
Sol. Let P (at 2 , 2at ) be one end of a focal chord of the parabola
y 2 = 4ax . The focus of its parabola is S (a ,0).
2a
=
If length of focal chord = PQ = (l say )
\
l = PQ = PS + SQ = PM + QN
a
l = at 2 + a + 2 + a
\
t
2
1 ö
ö
æ 2 1
æ
Þ
l = a çt + 2 + 2÷ Þ l = a çt + ÷
ø
è
è
t ø
t
2
t 2 + t1
y Example 28 Prove that the length of a focal chord
of a parabola varies inversly as the square of its
distance from the vertex.
d2 =
( t 2 + 1)
2a ö
æa
The other end of the focal chord is Q ç 2 , - ÷
èt
t ø
[Q t 1t 2 = - 1]
= a[(t 2 + t 1 ) + 4 ]
\
Þ
…(i)
= a [(t 2 + t 1 ) - 4t 1t 2 ]
|2at |
=
y 2 = 4ax
and the given line be
y = mx + c
On eliminating x from Eqs. (i) and (ii), then
æy -c ö
y 2 = 4a ç
÷
è m ø
Þ
my 2 - 4ay + 4ac = 0
…(i)
…(ii)
…(iii)
This equation being quadratic in y, gives two values of y,
shows that every straight line will cut the parabola in two
points may be real, coincident or imaginary according as
discriminant of Eq. ( iii ) > , = < 0
i.e. ( -4a ) 2 - 4 × m × 4ac >, =, < 0 or a - mc >, =, < 0
or
a >, =, < mc
…(iv)
Condition of tangency
If the line Eq. (ii) touches the parabola Eq. (i), then Eq. (iii)
has equal roots
\ Discriminant of Eq. (iii) = 0
Þ
( -4a ) 2 - 4m × 4ac = 0
382
Textbook of Coordinate Geometry
Þ
c=
a
,m ¹ 0
m
…(v)
So, the line y = mx + c touches the parabola y 2 = 4ax if
a
c = (which is condition of tangency).
m
Substituting the value of c from Eq. (v) in Eq. (ii), then
a
y = mx + , m ¹ 0
m
a
Hence, the line y = mx + will always be a tangent to the
m
parabola y 2 = 4ax .
a
The point of contact Substituting c = in Eq. (iii), then
m
a
æ ö
my 2 - 4ay + 4a ç ÷ = 0
èm ø
Þ
m 2 y 2 - 4amy + 4a 2 = 0
Þ
(my - 2a ) 2 = 0
Þ
a
m
2a
a
= mx +
\
m
m
a
a
or x =
Þ
mx =
m
m2
æ a 2a ö
Hence, the point of contact is ç , ÷ , (m ¹ 0 ) this
èm 2 m ø
known as m-point on the parabola.
ln = am 2
Aliter :
Given line lx + my + n = 0
and the parabola
y 2 = 4ax
Substituting the value of x from Eq. (i) i.e. x = -
…(i)
…(ii)
n + my
in
l
Eq. (ii), then
(we should not substituting the value of y from Eq. (i), in
Eq. (ii) since y is quadratic, substituting the value of x since
x is linear).
æ n + my ö
y 2 = 4a ç ÷
ø
è
l
\
ly 2 + 4amy + 4an = 0
…(iii)
( 4am )2 = 4 × l 4an
am 2 = ln
ln = am 2
y Example 30 Show that the line x cos a + y sin a = p
touches the parabola y 2 = 4ax , if p cos a + a sin 2 a = 0
and that the point of contact is (a tan 2 a , - 2a tan a ) .
0 - 4 ay + 4 ac = 0
y=c
which gives only one value of y and so every line parallel to
X-axis cuts the parabola only in one real point.
y Example 29 Prove that the straight line
lx + my + n = 0 touches the parabola y 2 = 4ax , if
ln = am 2 .
Sol. The given line is
x cos a + y sin a = p
Þ
y = - x cota + p coseca
Comparing this line with y = mx + c .
\
m = - cota and c = p coseca
since, the given line touches the parabola
a
\
c = or cm = a
m
Þ
( p coseca )( - cot a ) = a
Þ
a sin 2 a + p cos a = 0
æ a 2a ö
and point of contact is ç 2 , ÷
èm m ø
Sol. The given line is lx + my + n = 0
n
l
xm
m
Comparing this line with y = Mx + c
\
Þ
If m = 0, then Eq. (iii) gives
y =-
Þ
i.e.,
Remark
or
a
Þ cM = a
M
æ n öæ l ö
ç- ÷ç- ÷ = a
è m øè m ø
Since, Eq. (i) touches the parabola Eq. (ii), then roots of
Eq. (iii) must be coincident and condition for the same is
B 2 = 4 AC ,
Substituting this value of y in y = mx +
Þ
M=-
c =
Þ
my - 2a = 0
2a
y=
m
or
l
n
and c = m
m
The line Eq. (i) will touch the parabola y 2 = 4ax , if
\
…(i)
i.e.
æ a
2a ö
,÷
ç
2
è cot a
cot a ø
or
(a tan 2 a , - 2a tan a )
Chap 05 Parabola
x y
+ = 1 touches the
l m
parabola y 2 = 4a( x + b ), if m 2 (l + b ) + al 2 = 0.
y Example 31 Prove that the line
Sol. The given parabola is
y 2 = 4a( x + b )
reduces to
Þ
m2 æ
ç1 +
l è
bö
÷
lø
Þ
Þ
al 2
+ (l + b ) = 0
m2
al 2 + m 2 (l + b ) = 0
2
2
m (l + b ) + al = 0
…(ii)
y = 8ax
y 2 = 4(2a )x
or
…(iv)
\ Equation of tangent of Eq. (ii) is
2a
y = mx +
m
or
m 2 x - my + 2a = 0
…(iii)
It is also tangent of Eq. (i), then the length of perpendicular
from centre of Eq. (i) i.e. (0, 0) to Eq. (iii) must be equal to
the radius of Eq. (i) i.e. a 2.
\
| 0 - 0 + 2a |
2 2
( m ) + ( -m )
…(i)
…(ii)
ì æ
yö
ü
y 2 = 4aíl ç1 - ÷ + bý
ø
è
m
î
þ
4al
2
…(iii)
or
y - 4a(l + b ) = 0
y +
m
Since, the line Eq. (i) touches the parabola Eq. (ii), then the
roots of Eq. (iii) are equal.
Þ
…(i)
2
The parabola Eq. (ii) is y = 8ax
respectively substituting the value of x from Eq. (i)
yö
æ
x = l ç1 - ÷
è
mø
2
X
2
Aliter :
The given line and parabola are
x y
+
=1
l m
and
y 2 = 4a( x + b )
æ 4al ö
\ ç
÷ - 4 × 1 { -4a(l + b )} = 0
èm ø
ent
x 2 + y 2 = 2a 2
and
bö
÷ = -a
lø
in Eq. (ii), then
n tang
…(ii)
…(iii)
m 2 (l + b ) + al 2 = 0
\
o
Comm
O
The line Eq. (iv) will touch the parabola Eq. (iii), if
bö
a
æ
m ç1 + ÷ =
è
l ø æ mö
ç- ÷
è l ø
Þ
Sol. The given curves are
Y
x y
+
=1
l m
X -b Y
+
=1
l
m
X - bö
æ
Y = m ç1 ÷
è
l ø
æ
æ mö
Y = ç - ÷ X + m ç1 +
è
è l ø
y Example 32 Find the equations of the straight lines
touching both x 2 + y 2 = 2a 2 and y 2 = 8ax .
…(i)
Vertex of this parabola is ( -b, 0).
Now, shifting (0, 0) at ( -b, 0),
then, x = X + ( -b ) and y = Y + 0
or
x + b = X and y = Y
From Eq. (i),Y 2 = 4aX
and the line
383
2
=a 2 Þ
Þ
m4 + m2 - 2 = 0
Þ
( m 2 + 2) ( m 2 - 1) = 0
Q
m2 + 2 ¹ 0
\
4a 2
m4 + m2
= 2a 2
[gives the imaginary values]
2
m -1=0
Þ
m = ±1
Hence, from Eq. (iii) the required tangents are
x ± y + 2a = 0.
Equation of Tangent in
Different Forms
1. Point Form :
To find the equation of the tangent to the parabola
y 2 = 4ax at the point ( x 1 , y 1 ).
(First Principal Method) Equation of parabola is
y 2 = 4ax
Let P º ( x 1 , y 1 ) and Q º ( x 2 , y 2 ) be any two points on
parabola (i), then
…(i)
384
Textbook of Coordinate Geometry
)
, y1
P (x 1
2. Parametric Form :
To find the equation of tangent to the parabola y 2 = 4ax
T
at the point (at 2 , 2at ) or ‘t’.
Q (x
2 , y2 )
and
y 12 = 4ax 1
…(ii)
y 22 = 4ax 2
…(iii)
y (2at ) = 2a( x + at 2 ) Þ ty = x + at 2
Remark
Subtracting Eq. (ii) from Eq. (iii), then
The equations of tangent of all standard parabolas at ‘t’.
y 22 - y 12 = 4a( x 2 - x 1 )
y2 - y1
4a
=
x2 - x1 y2 + y1
Þ
Since, the equation of tangent of the parabola y 2 = 4ax at
…(i)
( x 1 , y 1 ) is yy 1 = 2a( x + x 1 )
2
replacing x 1 by at and y 1 by 2at, then Eq. (i) becomes
Equations of
Parabolas
…(iv)
Equation of PQ is
y - y1
y - y1 = 2
(x - x 1 )
x2 - x1
From Eqs. (iv) and (v), then
4a
(x - x 1 )
y - y1 =
y2 + y1
…(v)
…(vi)
Now, for tangent at P , Q ® P , i.e. x 2 ® x 1 and y 2 ® y 1 ,
then Eq. (vi) becomes
4a
( x - x 1 ) Þ yy 1 - y 12 = 2ax - 2ax 1
y - y1 =
2y 1
yy 1 = 2ax
Þ
+ y 12
- 2ax 1
[from Eq. (ii)]
Þ
yy 1 = 2ax + 4ax 1 - 2ax 1
Þ
yy 1 = 2ax + 2ax 1
\
yy 1 = 2a( x + x 1 ),
which is the required equation of tangent at ( x 1 , y 1 ).
Parametric
coordinates ‘t’
y 2 = 4ax
(at 2, 2at )
ty = x + at 2
y 2 = - 4ax
( -at 2, 2at )
ty = - x + at 2
x 2 = 4ay
(2at, at 2 )
tx = y + at 2
x 2 = - 4ay
(2at, - at 2 )
tx = - y + at 2
3. Slope Form :
To find the equation of tangent and point of contact in
terms of m(slope) to the parabola y 2 = 4ax .
The equation of tangent to the parabola y 2 = 4ax at
( x 1 , y 1 ) is yy 1 = 2a( x + x 1 ).
Since, m is the slope of the tangent, then
2a
2a
m=
Þ y1 =
y1
m
Equations of Parabolas
Tangent at( x1, y1 )
2
yy1 = 2 a( x + x1 )
2
y = - 4 ax
yy1 = - 2 a( x + x1 )
x 2 = 4 ay
xx1 = 2 a( y + y1 )
y = 4 ax
2
x = - 4 ay
xx1 = - 2 a( y + y1 )
…(i)
Since, ( x 1 , y 1 ) lies on y 2 = 4ax , therefore
y 12 = 4ax 1 Þ
Remarks
1. The equation of tangent at ( x1, y1 ) can also be obtained by
x + x1
y + y1
and xy by
replacing x 2 by xx1, y 2 by yy1, x by
, y by
2
2
xy1 + x1 y
and without changing the constant (if any) in the
2
equation of curve. This method is apply only when the
equations of parabola is polynomial of second degree in x
and y.
2. Equation of tangents of all standard parabolas at ( x1, y1 ).
Tangent
at ‘t’
\
x1 =
4a 2
m2
= 4ax 1
a
m2
Substituting the values of x 1 and y 1 in Eq. (i), we get
a
…(ii)
y = mx +
m
a
Thus, y = mx + is a tangent to the parabola y 2 = 4ax ,
m
where, m is the slope of the tangent.
æ a 2a ö
The coordinates of the point of contact are ç , ÷ .
èm 2 m ø
Comparing Eq. (ii) with y = mx + c ,
a
m
which is condition of tangency.
c=
when, y = mx + c is the tangent of y 2 = 4ax .
Chap 05 Parabola
385
Remark
The equation of tangent, condition of tangency and point of contact in terms of slope ( m) of all standard parabolas.
Equation of parabolas
Point of contact in terms of
slope (m)
Equation of tangent in terms of
slope (m)
Condition of tangency
y 2 = 4ax
æ a 2a ö
ç 2, ÷
èm m ø
y = mx +
a
m
c=
y 2 = - 4ax
æ a 2a ö
ç- 2 , ÷
è m mø
y = mx -
a
m
c =-
x 2 = 4ay
(2am, am 2 )
y = mx - am 2
c = - am 2
x 2 = - 4ay
(2am, - am 2 )
y = mx + am 2
c = am 2
(y - k ) 2 = 4a( x - h )
2a ö
a
æ
çh + 2 , k + ÷
è
mø
m
y = mx - mh + k +
a
m
c + mh = k +
a
m
(y - k ) 2 = - 4a( x - h )
a
2a ö
æ
çh - 2 , k + ÷
è
mø
m
y = mx - mh + k -
a
m
c + mh = k -
a
m
( x - h ) 2 = 4a(y - k )
(h + 2am, k + am 2 )
y = mx - mh + k - am 2
c + mh = k - am 2
( x - h ) 2 = - 4a(y - k )
(h + 2am, k - am 2 )
y = mx - mh + k + am 2
c + mh = k + am 2
Point of Intersection of
Tangents at any two Points on
the Parabola
Let the parabola be y 2 = 4ax
let two points on the parabola are
P º (at 12 , 2at 1 ) and Q º (at 22 , 2at 2 ).
Equation of tangents at P (at 12 , 2at 1 )
and
Q (at 22 , 2at 2 )
are
t 1 y = x + at 12
and
t 2 y = x + at 22
…(i)
a
m
a
m
Remarks
1. The geometric mean of the x-coordinates of P and Q
(i.e. at12 ´ at22 = at1t2 ) is the x-coordinate of the point of
intersection of tangents at P and Q on the parabola. If P and
Q are the ends points of focal chord, then x-coordinate of
point of intersection of tangents at P and Q is (– at1 t2 ).
2. The arithmetic mean of the y-coordinates of P and Q
æi.e. 2at1 + 2at2 = a ( t + t ) ö is the y-coordinate of the point of
ç
1
2 ÷
è
ø
2
intersection of tangents at P and Q on the parabola.
Remembering Method :
G
O
A
GM of at 12 and at 22
AM of 2at 1 and 2at 2
[GOA rule]
…(ii)
Y
2,
at 1
P(
t 1)
2a
i.e.
i.e. at 1 t 2
X′
O
R
(at1t2, a(t1 + t2))
X
Q (a
t2 2,
2at
2)
Y′
On solving these equations, we get x = at 1 t 2 , y = a(t 1 + t 2 )
Thus, the coordinates of the point of intersection of
tangents at
(at 12 , 2at 1 ) and (at 22 , 2at 2 ) are
(at 1 t 2 , a(t 1 + t 2 )).
,
2at 1 + 2at 2
= a(t 1 + t 2 )
2
y Example 33 Find the equation of the common
tangents to the parabola y 2 = 4ax and x 2 = 4by .
Sol. The equation of any tangent in terms of slope (m) to the
parabola y 2 = 4ax is
a
…(i)
m
If this line is also tangent to the parabola x 2 = 4ay , then Eq. (i)
meets x 2 = 4by in two coincident points.
y = mx +
386
Textbook of Coordinate Geometry
i.e.,
Þ
or
\
and
t 1y = x + at 12
…(i)
at 22
…(ii)
t 2y = x +
Y
æ -4ab ö
( -4bm ) = 4 × 1 ç
÷
è m ø
2
16b 2m 3 + 16ab = 0, m ¹ 0
R
3
P
X
90°
m = -a /b
m = - a1/ 3 / b1/ 3
Q
Substituting the value of m in Eq. (i) the required equation is
ab1/ 3
a1/ 3
y = - 1/ 3 x - 1/ 3
a
b
1/ 3
a
Þ
y = - 1/ 3 x - a 2 / 3b1/ 3
b
\
a1/ 3 x + b1/ 3y + a 2 / 3b 2 / 3 = 0
y Example 34 The tangents to the parabola y 2 = 4ax
make angle q 1 and q 2 with X-axis. Find the locus of
their point of intersection, if cotq 1 + cotq 2 = c .
Sol. Let the equation of any tangent to the parabola y 2 = 4ax
is
…(i)
y = mx + (a / m )
Let ( x 1, y1 ) be the point of intersection of the tangents to
y 2 = 4ax , then Eq. (i) passes through ( x 1, y1 ).
\
or
Sol. Let the points P (at 12 , 2at 1 ) and Q (at 22 , 2at 2 ) on the parabola
y 2 = 4ax tangents at P and Q are
Directrix
Substituting the value of y from Eq. (i) in x 2 = 4by , we get
aö
æ
x 2 = 4b çmx + ÷
è
mø
4ab
Þ
x 2 - 4bmx =0
m
The roots of this quadratic are equal provided
B 2 = 4 AC
y1 = mx 1 + (a / m )
m 2 x 1 - my1 + a = 0
Let m1 and m 2 be the roots of this quadratic equation, then
m1 + m 2 = y1 / x 1 and m1m 2 = a / x 1
or
tan q 1 + tan q 2 = y1 / x 1
and
…(ii)
tan q 1 tan q 2 = a / x 1
Now,
(given)
cot q 1 + cot q 2 = c
1
1
Þ
+
=c
tan q 1 tan q 2
Þ
tan q 1 + tan q 2
=c
tan q 1 tan q 2
Þ
y1 / x 1
=c
a / x1
[from Eq. (ii)]
Þ
y1 = ac
The required locus is y = ac ,
which is a line parallel to X-axis.
y Example 35. Show that the locus of the points of
intersection of the mutually perpendicular tangents to
a parabola is the directrix of the parabola.
Q Point of intersection of these tangents is (at 1t 2 , a(t 1 + t 2 ))
Let this point is (h , k ),
then,
…(iii)
h = at 1t 2
and
…(iv)
k = a (t 1 + t 2 )
1
1
Slope of tangents Eqs. (i) and (ii) are and , respectively.
t1
t2
Since, tangents are perpendicular, then
1
1
´
= -1
t1 t 2
or
t 1t 2 = - 1
From Eqs. (iii) and (v), we get
h = - a or h + a = 0
\ Locus of the point of intersection of tangents is
x +a=0
which is directrix of y 2 = 4ax .
…(v)
Aliter :
Let the equation of any tangent to the parabola y 2 = 4ax is
…(i)
y = mx + a / m
2
Let the point of intersection of the tangents to y = 4ax
then, Eq (i) passes through ( x 1, y1 ) .
\
y1 = mx 1 + a / m
or
m 2 x 1 - my1 + a = 0
Let m1, m 2 be the roots of this quadratic equation then
m1m 2 = a / x 1 = - 1
[since, tangents are perpendiculars]
Þ
a + x1 = 0
\ Locus of the point of intersection of tangents is
x + a = 0 which is directrix of y 2 = 4ax .
Remark
Locus of the point of intersection of the perpendicular tangents
to the parabola y 2 = 4ax is called the director circle. Its equation
is x + a = 0, which is parabola’s own directrix.
Chap 05 Parabola
y Example 36 The tangents to the parabola y 2 = 4ax
at P (at 12 , 2at 1 ) and Q (at 22 , 2at 2 ) intersect at R. Prove
1
that the area of the DPQR is a 2 |(t 1 - t 2 )| 3 .
2
387
, y1)
P (x 1
Tang
ent
al
Norm
T
Sol. Equations of tangents at P (at 12 , 2at 1 ) and Q (at 22 , 2at 2 ) are
t 1y = x + at 12
and
t 2y = x +
Y
…(ii)
2 , 2a
P
X′
…(i)
at 22
t 1)
(at 1
X
A
The slope of the tangent at ( x 1 , y 1 ) = 2a / y 1
Since, the normal at ( x 1 , y 1 ) is perpendicular to the
tangent at ( x 1 , y 1 ).
\ Slope of normal at ( x 1 , y 1 ) = - y 1 / 2a
Hence, the equation of normal at ( x 1 , y 1 ) is
y - y1 = -
R
(at1t2, a(t1+ t2)
Y′
Q (a
t2 2,
y1
( x - x 1 ).
2a
Remarks
2at
2)
Since, point of intersect of Eqs. (i) and (ii) is
R(at 1t 2 , a(t 1 + t 2 )).
at 12
2at 1
1
1
2at 2
1
at 22
\ Area of DPQR =
2 at t a(t + t ) 1
1
2
1 2
Applying R 2 ® R 2 - R1 and R 3 ® R 3 - R1
at 12
2at 1
1
1
a(t 22 - t 12 ) 2a(t 2 - t 1 ) 0
=
2 at (t - t ) a(t - t ) 0
2
1
1 2
1
Expanding with respect to first row
1 a(t 22 - t 12 ) 2a(t 2 - t 1 )
=
2 at 1(t 2 - t 1 ) a(t 2 - t 1 )
1 2
t + t1 2
a (t 2 - t 1 )2 2
t1
1
2
1
= a 2 (t 1 - t 2 )2 |(t 2 - t 1 )|
2
1
= a 2 (t 1 - t 2 )2 |(t 1 - t 2 )|
2
1
= a 2 |(t 1 - t 2 )3 |.
2
=
1. The equation of normal at ( x1, y1 ) can also be obtained by
this method
x - x1
y - y1
…(i)
=
a¢ x1 + hy1 + g hx1 + by1 + f
a¢, b, g, f , h are obtained by comparing the given parabola with
…(ii)
a¢ x 2 + 2hxy + by 2 + 2gx + 2fy + c = 0
and denominators of Eq. (i) can easily remembered by the
first two rows of this determinant
½a¢ h g½
i.e.
½
½
½h b f ½
½g f c ½
Since, first row a¢( x1 ) + h( y1 ) + g( 1)
and second row, h( x1 ) + b( y1 ) + f ( 1)
Here, parabola
y 2 = 4 ax
y 2 - 4 ax = 0
or
Comparing Eqs. (ii) and (iii), then we get
a¢ = 0, b = 1, g = - 2 a, h = 0, f = 0
From Eq. (i), equation of normal of Eq. (iii) is
x - x1
y - y1
=
0 + 0 - 2a 0 + y1 + 0
y1
( x - x1 )
2a
2. Equations of normals of all standard parabolas at ( x1, y1 ).
or
y - y1 = -
Equations of Parabola
Equations of Normals in
Different Forms
1. Point Form: To find the equation of the normal to
the parabola y 2 = 4ax at the point ( x 1 , y 1 ).
Since, the equation of the tangent to the parabola y 2 = 4ax
at ( x 1 , y 1 ) is
yy 1 = 2a( x + x 1 )
…(iii)
2
Normal at ( x1, y1 )
y = 4 ax
y - y1 = -
y 2 = - 4 ax
y - y1 =
x 2 = 4 ay
y - y1 = -
x 2 = - 4 ay
y - y1 =
y1
( x - x1 )
2a
y1
( x - x1 )
2a
2a
( x - x1 )
x1
2a
( x - x1 )
x1
388
Textbook of Coordinate Geometry
2. Parametric form :
To find the equation of normal to the parabola y 2 = 4ax
at the point (at 2 , 2at ) or ‘t’.
Since, the equation of normal of the parabola y 2 = 4ax at
( x 1 , y 1 ) is
y
…(i)
y - y 1 = - 1 (x - x 1 )
2a
Replacing x 1 by at 2 and y 1 by 2at, then Eq. (i) becomes
The equation of normal to the parabola y 2 = 4ax at
( x 1 , y 1 ) is
y
y - y 1 = - 1 (x - x 1 )
2a
Since, m is the slope of the normal,
y
then,
m = - 1 Þ y 1 = - 2am
2a
Since, ( x 1 , y 1 ) lies on y 2 = 4ax , therefore
y 12 = 4ax 1
y - 2at = - t ( x - at 2 )
y + tx = 2at + at 3
or
Remark
Parametric
coordinates ‘t’
( at 2, 2 at )
y + tx = 2 at + at 3
y 2 = - 4 ax
( -at 2, 2 at )
y - tx = 2 at + at 3
x 2 = 4 ay
(2 at, at 2 )
x + ty = 2 at + at 3
x = - 4 ay
(2 at, - at 2 )
x - ty = 2 at + at
4a 2m 2 = 4ax 1
\
x 1 = am 2
y + 2am = m( x - am 2 )
Normals
at ‘t’
y 2 = 4 ax
2
Þ
On substituting the values of x 1 and y 1 in Eq. (i) we get
The equations of normals of all standard parabolas at ‘t’
Equations of
Parabolas
3
3. Slope form :
To find the Equation of normal, condition for
normality and point of contact in terms of m (slope) to
the parabola y 2 = 4ax
y = mx - 2am - am 3
\
…(ii)
3
Thus, y = mx - 2am - am is a normal to the parabola
y 2 = 4ax , where m is the slope of the normal. The
coordinates of the point of contact are (am 2 , - 2am )
On comparing Eq. (ii) with
y = mx + c
c = - 2am - am 3
\
which is condition of normality when y = mx + c is the
normal of y 2 = 4ax .
Remark
The equations of normals, point of contact and condition of normality in terms of slope ( m) of all standard parabolas.
Equation of parabolas
Point of contact in terms of
slope (m)
Equation of normals in
terms of slope (m)
Condition of
normality
y 2 = 4 ax
( am2, - 2 am)
y = mx - 2 am - am3
c = - 2 am - am3
y 2 = - 4 ax
( -am2, 2 am)
æ 2a a ö
ç- , 2÷
è m m ø
y = mx + 2 am + am3
a
y = mx + 2 a + 2
m
c = 2 am + am3
a
c = 2a + 2
m
x 2 = - 4 ay
a ö
æ 2a
ç ,- 2÷
èm m ø
y = mx - 2 a -
( y - k )2 = 4 a( x - h)
( h + am2, k - 2 am)
y - k = m( x - h)
- 2 am - am3
c = k - mh - 2 am - am3
( y - k )2 = - 4 a( x - h)
( h - am2, k + 2 am)
y - k = m( x - h)
+ 2 am + am3
c = k - mh + 2 am + am3
( x - h)2 = 4 a( y - k )
2a
a ö
æ
çh - , k + 2÷
è
m
m ø
y - k = m( x - h)
c = k - mh + 2 a +
x 2 = 4 ay
…(i)
a
m2
a
+ 2a + 2
m
c = - 2a -
a
m2
a
m2
Chap 05 Parabola
Y
Point of Intersection of
Normals at any Two Points on
the Parabola
P (a
X′
2 2a
t1 ,
P
and Q
Y′
= (at 22 , 2at 2 ).
P
R
X
2at 2 = - at 1 t 22 + 2at 1 + at 13
Þ
2a(t 2 - t 1 ) + at 1 (t 22 - t 12 ) = 0
Þ
Q
a(t 2 - t 1 )[2 + t 1 (t 2 + t 1 )] = 0
a(t 2 - t 1 ) ¹ 0
[Qt 1 and t 2 are different]
2 + t 1 (t 2 + t 1 ) = 0
2
t 2 = -t1 t1
2
t2
(a
,2
)
at 2
Equations of normals
y = -t1x
and
y = -t2x
\
at P (at 12 , 2at 1 )
+ 2at 1 + at 13
+ 2at 2 + at 23
and Q (at 22 , 2at 2 )
are
…(i)
…(ii)
If R is the point of intersection, then
R º [2a + a (t 12 + t 22 + t 1 t 2 ), - at 1 t 2 (t 1 + t 2 )]
(Remember)
Point of intersection of normals at t 1 and t 2
y2 = 4 ax
Equation of normal
at any point ‘t’
y + tx = 2at + at 3
y = - 4 ax y - tx = 2at + at
3
Point of intersection of
normals at t 1 and t 2
( 2a + a ( t12 + t1 t 2 + t 22 ),
( -2a -
a ( t12
+ t1 t 2 + t 22 ),
æ 1 1ö
( t1 - t2 ) = 2 ç - ÷
è t2 t1 ø
or
t1t2 = 2
[Qt1 ¹ t2 ]
2. If the normals to the parabola y 2 = 4 ax at the points t1 and t2
intersect again on the parabola at the point t3, then
t3 = - ( t1 + t2 ) and the line joining t1 and t2 passes through a
fixed point ( - 2a, 0 ).
x + ty = 2at + at 3
( - at1 t 2 ( t1 + t 2 ), 2a
y Example 37 Show that normal to the parabola
y 2 = 8 x at the point (2, 4) meets it again at (18, –12).
Find also the length of the normal chord.
3
( at1 t 2 ( t1 + t 2 ), - 2a
Sol. Comparing the given parabola (i.e. y 2 = 8x ) with
y 2 = 4ax .
at1t2 (t1 + t2 ))
x2 = 4 ay
Remarks
Þ
- at1t2 (t1 + t2 ))
2
\
1. If normals at ‘t1’ and ‘t2’ meets the parabola y 2 = 4 ax at
same point, then t1t2 = 2.
Proof Suppose normals meet at ‘ T ’, then
2
2
T = - t1 - = - t2 t1
t2
On solving Eqs. (i) and (ii), we get
x = 2a + a(t 12 + t 22 + t 1 t 2 ) and y = - at 1 t 2 (t 1 + t 2 )
Parabola
2)
\
Q
Y′
2at
Eq. (i) passes through Q (at 22 , 2at 2 ).
(at 1
A
X′
t2 2,
Since, it meet the parabola again at Q (at 22 , 2at 2 ), then
t 1)
2 , 2a
Y
X
Q (a
Let the points on the parabola are
º (at 12 , 2at 1 )
t 1)
A
Let the parabola be y 2 = 4ax .
389
2
x = - 4 ay x - ty = 2at + at
+ a(t12 + t1t2 + t22 ))
- a(t12 + t1t2 + t22 ))
Relation between ‘t 1 ’ and ‘t 2 ’ if Normal
at ‘t 1 ’ meets the Parabola Again at ‘t 2 ’
Let the parabola be y 2 = 4ax , equation of normal at
P (at 12 , 2at 1 ) is
y = - t 1 x + 2at 1 + at 13
…(i)
\
4a = 8 Þ a = 2
Since, normal at ( x 1, y1 ) to the parabola y 2 = 4ax is
y1
( x - x1 )
2a
Here, x 1 = 2 and y1 = 4.
\ Equation of normal is
4
y - 4 = - ( x - 2)
4
y - y1 = -
390
Textbook of Coordinate Geometry
it meet the curve again Q say (at 22 , 2at 2 ).
Y
P(
X′
)
2, 4
\
t 2 = - t1 -
2,
P
Q (1
8, –
Y′
1 2)
Þ
y - 4 = -x +2
Þ
x +y -6=0
On solving Eq. (i) and y 2 = 8x ,
X′
)
2at 1
X
A
θ
…(i)
Q (at 2, 2at )
2
2
y = 8(6 - y )
Y′
2
y + 8y - 48 = 0
Þ (y + 12)(y - 4 ) = 0
\
y = - 12 and y = 4
then, x = 18 and x = 2.
Hence, the point of intersection of normal and parabola are
(18, - 12) and (2, 4 ), therefore normal meets the parabola at
(18, - 12) and length of normal chord is distance between
their points
= PQ = (18 - 2)2 + ( -12 - 4 )2 = 16 2
Now, angle between the normal and parabola
= Angle between the normal and tangent at Q
(i.e. t 2y = x + at 22 )
If q be the angle, then
1
-t 1 m1 - m 2
tt +1
t2
tanq =
=
=- 12
t 2 - t1
1 + m1m 2
æ1ö
1 + ( -t 1 ) ç ÷
èt 2 ø
æ
2ö
t 1 ç -t 1 - ÷ + 1
t1 ø
è
=2
-t 1 - - t 1
t1
y Example 38 Prove that the chord
y - x 2 + 4a 2 = 0 is a normal chord of the
parabola y 2 = 4ax . Also, find the point on the parabola
=-
when the given chord is normal to the parabola.
Sol. We have,
i.e.,
y - x 2 + 4a 2 = 0
y = x 2 - 4a 2
…(i)
Comparing the Eq. (i) with the equation y = mx + c , then
m = 2, c = - 4a 2
Since,
(a t 1
2
then,
Þ
…(ii)
Y
X
A
2
t1
-2am - am 3 = - 2a 2 - a( 2 )3
= - 2a 2 - 2a 2 = - 4a 2 = c
Hence, the given chord is normal to the parabola y 2 = 4ax .
The coordinates of the points are (am 2 , - 2am ) i.e.
=
\
[from Eq. (ii)]
t
-t 12 - 1
=- 1
2
æ 1 + t 12 ö
- 2ç
÷
è t1 ø
tan f
2
[from Eq. (i)]
ö
æ1
q = tan -1 ç tan f ÷
ø
è2
y Example 40 Prove that the normal chord to a
parabola y 2 = 4ax at the point whose ordinate is equal
to abscissa subtends a right angle at the focus.
Sol. Let the normal at P (at 12 , 2at 1 ) meet the curve at Q (at 22 , 2at 2 ).
(2a, - 2 2 a ) .
\ PQ is a normal chord
y Example 39 If the normal to a parabola y 2 = 4ax ,
makes an angle f with the axis. Show that it will cut
æ1
ö
the curve again at an angle tan -1 ç tan f ÷ .
è2
ø
Sol. Let the normal at P (at 12 , 2at 1 ) be
y = - t 1x + 2at 1 + at 13 .
\ tan f = - t 1 = slope of the normal,
…(i)
and
t 2 = - t1 -
2
.
t1
By given condition, 2at 1 = at 12
\ t 1 = 2 from Eq. (i), t 2 = - 3
then, P ( 4a, 4a ) and Q (9a, - 6a )
but focus S(a, 0).
4a - 0 4a 4
Slope of SP =
=
=
\
4a - a 3a 3
…(i)
Chap 05 Parabola
Y
X′
2
t1
P (a
t 1)
X
S
A
Q (at 2, 2at )
2
2
Y′
and slope of SQ =
, 2a
-6a - 0
6a
3
==9a - a
8a
4
Q Slope of SP ´ Slope of SQ =
4
3
´ - = -1
3
4
\
ÐPSQ = p / 2
i.e. PQ subtends a right angle at the focus S.
y Example 41 If the normal to the parabola y 2 = 4ax
at point t 1 cuts the parabola again at point t 2 , prove
that t 22 = 8.
Sol. A normal at point t 1 cuts the parabola again at t 2 . Then,
t 2 = - t1 -
2
Þ t 12 + t 1t 2 + 2 = 0
t1
Since, t 1 is real , so (t 2 )2 - 4 × 1 × 2 ³ 0
t 22 ³ 8
Þ
Co-normal Points
In general three normals can be drawn from a point to a
parabola and their feet, points where they meet the
parabola are called conormal points.
Let P (h, k ) be any given point and y 2 = 4ax be a parabola.
The equation of any normal to y 2 = 4ax is
y = mx - 2am - am 3
This is a cubic equation in m, so it has three roots, say
m 1 , m 2 and m 3 .
\
m 1 + m 2 + m 3 = 0,
(2a - h )
,
m 1m 2 + m 2m 3 + m 3m 1 =
a
k
…(ii)
m 1m 2m 3 = a
Hence, for any given point P (h, k ), Eq. (i) has three real or
imaginary roots. Corresponding to each of these three
roots, we have one normal passing through P (h, k ).
Hence, in total, we have three normals PA, PB and PC
drawn through P to the parabola.
Points A, B, C in which the three normals from P (h, k )
meet the parabola are called co-normal points.
Corollary 1 The algebraic sum of the slopes of three
concurrent normals is zero. This follows from Eq. (ii).
Corollary 2 The algebraic sum of ordinates of the feets of
three normals drawn to a parabola from a given point is
zero.
Let the ordinates of A, B, C be y 1 , y 2 , y 3 respectively, then
y 1 = - 2am 1 , y 2 = - 2am 2 and y 3 = - 2am 3
\ Algebraic sum of these ordinates is
y 1 + y 2 + y 3 = - 2am 1 - 2am 2 - 2am 3
= - 2a (m 1 + m 2 + m 3 )
[from Eq. (ii)]
= - 2a ´ 0
=0
Corollary 3 If three normals drawn to any parabola
y 2 = 4ax from a given point (h, k ) be real then h > 2a.
When normals are real, then all the three roots of Eq. (i)
are real and in that case
Þ
Y
Þ
m 12 + m 22 + m 23 > 0
(for any values of m 1 , m 2 , m 3 )
(m 1 + m 2 + m 3 ) 2 - 2
(m 1m 2 + m 2m 3 + m 3m 1 ) > 0
(0 ) 2 -
A
B
X′
\
O
X
P (h, k)
2 (2a - h )
> 0 Þ h - 2a > 0
a
h > 2a
Remark
For a = 1normals drawn to the parabola y 2 = 4 x from any point
( h, k ) are real, if h > 2.
C
Corollary 4 If three normals drawn to any parabola
y 2 = 4ax from a given point (h, k ) be real and distinct, then
Y′
If it passes through (h, k ), then
k = mh - 2am - am
3
391
Þ am + m (2a - h ) + k = 0
27ak 2 < 4 (h - 2a ) 3
3
Let
…(i)
f (m ) = am 3 + m (2a - h ) + k
392
Textbook of Coordinate Geometry
\
f ¢ (m ) = 3am 2 + (2a - h )
y = mx - 2am - am 3
Two distinct roots of f ¢ (m ) = 0 are
æ h - 2a ö
a= ç
÷
è 3a ø
æ h - 2a ö
and b = - ç
÷,
è 3a ø
Now,
f (a ) f (b) < 0 Þ f (a ) f ( -a ) < 0
Þ
3
3
(aa + a (2a - h ) + k ) ( -aa - a (2a - h ) + k ) < 0
Þ
k 2 - (aa 2 + (2a - h )) 2 a 2 < 0
Þ
æ h - 2a
ö (h - 2a )
k -ç
+ (2a - h ) ÷
<0
è 3
ø
3a
Þ
æ 4a - 2h ö æ h - 2a ö
k2 - ç
÷ ç
÷ <0
è 3 ø è 3a ø
2
2
2
Þ
\
k2 -
4 (h - 2a ) 3
< 0 Þ 27ak 2 - 4 (h - 2a ) 3 < 0
27a
27ak 2 < 4 (h - 2a ) 3
Corollary 5 The centroid of the triangle formed by the feet
of the three normals lies on the axis of the parabola.
If A ( x 1 , y 1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) be vertices of DABC,
then its centroid is
æ x1 + x2 + x 3 y1 + y2 + y 3 ö æ x1 + x2 + x 3 ö
, 0÷ .
,
ç
÷ =ç
è
ø è
3
3
3
ø
Since, y 1 + y 2 + y 3 = 0 (from corollary 2). Hence, the
centroid lies on the X-axis OX, which is the axis of the
parabola also.
x + x2 + x 3 1
Now, 1
= (am 12 + am 22 + am 23 )
3
3
a 2
= (m 1 + m 22 + m 23 )
3
a
= {(m 1 + m 2 + m 3 ) 2
3
- 2 (m 1m 2 + m 2m 3 + m 3m 1 )}
=
a
3
é 2
ì2a - h üù 2h - 4a
ê(0 ) - 2 í a ýú = 3
þû
î
ë
æ 2h - 4a ö
, 0÷ .
\ Centroid of DABC is ç
è 3
ø
y Example 42 Show that the locus of points such
that two of the three normals drawn from them to
the parabola y 2 = 4ax coincide is 27ay 2 = 4 ( x - 2a ) 3 .
Sol. Let (h , k ) be the point of intersection of three normals to
the parabola y 2 = 4ax . The equation of any normal to
y 2 = 4ax is
If it passes through (h , k ), then
k = mh - 2am - am 3
am 3 + m (2a - h ) + k = 0
Þ
Let the roots of Eq. (i) be m1, m 2 and m 3 .
Then, from Eq. (i), m1 + m 2 + m 3 = 0
(2a - h )
m1m 2 + m 2m 3 + m 3m1 =
a
k
and
m1m 2m 3 = a
But here, two of the three normals are given to be
coincident i.e. m1 = m 2 .
Putting m1 = m 2 in Eqs. (ii) and (iv), we get
2m1 + m 3 = 0
k
and
m12m 3 = a
Putting m 3 = - 2m1 from Eq. (v) in Eq. (vi), we get
k
- 2m13 = a
k
3
Þ
m1 =
2a
Since, m1 is a root of Eq. (i).
\
am13 + m1 (2a - h ) + k = 0
Þ
æk ö æk ö
aç ÷+ç ÷
è 2a ø è 2a ø
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
1/ 3
(2a - h ) + k = 0
1/ 3
é
æk ö ù
ê putting m1 = ç ÷ ú
è 2a ø ú
êë
û
Þ
Þ
Þ
æk ö
ç ÷
è 2a ø
1/ 3
(2a - h ) = -
3k
2
27k 3
k
(2a - h )3 = 8
2a
27ak 2 = 4 (h - 2a )3
Hence, the locus of (h , k ) is
27ay 2 = 4 ( x - 2a )3 .
y Example 43 Find the locus of the point through
which pass three normals to the parabola y 2 = 4ax
such that two of them make angles a and b
respectively with the axis such that tan a tan b = 2 .
Sol. Let (h , k ) be the point of intersection of three normals to
the parabola y 2 = 4ax .
The equation of any normal to y 2 = 4ax is
y = mx - 2am - am 3
If it passes through (h1,k ), then
k = mh - 2am - am 3
Þ
am 3 + m (2a - h ) + k = 0
…(i)
Chap 05 Parabola
Then, VA = 2a + am12 , VB = 2a + am 22
Let roots of Eq. (i) be m1, m 2 , m 3 then from Eq. (i)
k
m1m 2m 3 = a
Also m1 = tan a , m 2 = tan b and tan a tan b = 2
m1m 2 = 2
\
VC = 2a +
…(iii)
Given, VA , VB and VC are in AP.
\
2VB = VA + VC
Þ
4a + 2am 22 = 2a + am12 + 2a + am 32
3
Þ
Þ
2m 22 = m12 + m 32
Þ
2m 22 = (m1 + m 3 )2 - 2m1m 3
Þ
2m 22 = (m1 + m 2 + m 3 - m 2 )2 -
Þ
2m 22 = (0 - m 2 )2 -
Þ
m 23
k
æ kö
a ç- ÷ (2a - h ) + k = 0
è 2a ø
2a
-
Þ
k
3
8a
2
-k +
kh
+k =0
2a
y Example 44 If the three normals from a point to the
parabola y 2 = 4ax cut the axis in points whose
distance from the vertex are in AP, show that the point
lies on the curve 27ay 2 = 2 ( x - 2a ) 3 .
Sol. Let (h , k ) be the point of intersection of three normals to
the parabola y 2 = 4ax . The equation of any normal to
…(i)
y 2 = 4ax is y = mx - 2am - am 3
Y
B
A
C
…(vi)
Þ
Þ
Þ
Þ
- am 23 - k = m 2 (2a - h )
( - am 23 - k )3 = m 23 (2a - h )3
2k
( -2k - k )3 =
(2a - h )3 [from Eq. (vi)]
a
2k
(h - 2a )3
- 27k 3 = a
27ak 2 = 2 (h - 2a )3
Hence, locus of (h , k ) is
27ay 2 = 2 ( x - 2a )3 .
X
P (h , k )
y Example 45 The normals at P , Q , R on the parabola
y 2 = 4ax meet in a point on the line y = k. Prove that
the sides of the DPQR touch the parabola
x 2 - 2ky = 0.
Y′
If it passes through (h , k ) then
k = mh - 2am - am 3
Þ
2k
=
a
(2a - h )
a
m1m 2m 3 (2a - h )
=
Þ m 2 (m1 + m 2 + m 3 - m 2 ) +
m2
a
k
(2a - h )
Þ
m 2 (0 - m 2 ) =
am 2
a
Þ
V
2 æ kö
ç- ÷
m2 è a ø
Now, from Eq. (iv), m 2 (m1 + m 3 ) + m 3m1 =
\ Required locus of (h , k ) is y 2 - 4ax = 0.
X′
2m1m 2m 3
m2
[from Eqs. (iii) and (v)]
k 2 - 4ah = 0
Þ
and
am 32 .
…(ii)
k
From Eqs. (ii) and (iii), 2m 3 = a
k
or
m3 = 2a
Which being a root of Eq. (i) must satisfy it
i.e.
am 33 + m 3 (2a - h ) + k = 0
393
am 3 + m (2a - h ) + k = 0
Let roots of Eq. (ii) be m1, m 2 , m 3 then from Eq. (ii)
m1 + m 2 + m 3 = 0
(2a - h )
m1m 2 + m 2m 3 + m 3m1 =
a
k
and
m1m 2m 3 = a
Since, Eq. (i) cuts the axis of parabola viz. y = 0 at
(2a + am 2 , 0).
…(ii)
…(iii)
…(iv)
vertex of the parabola y 2 = 4ax .
y = mx - 2am - am 3
…(v)
…(i)
Also, any point on the line y = k is ( x 1, k ).
If Eq. (i) passes through ( x 1, k ) then k = mx 1 - 2am - am 3
or
\The normal through (h , k ) cut the axis at A (2a + am12 , 0),
B (2a + am 22 , 0) and C (2a + am 32 , 0) and let V (0, 0) be the
Sol. Any normal to the parabola y 2 = 4ax is
am 3 + m (2a - x 1 ) + k = 0
If the roots of this equation are m1, m 2 , m 3 then we get
…(ii)
m1 + m 2 + m 3 = 0
(2a - x 1 )
…(iii)
m1m 2 + m 2m 3 + m 3m1 =
a
k
…(iv)
and
m1m 2m 3 = a
394
Textbook of Coordinate Geometry
Also, coordinates of three points P , Q and R are
(am 12, - 2am1 ) (am 22 , - 2am 2 ) and (am 32 , - 2am 3 ),
respectively.
\ The equation of the line PQ is
( -2am 2 ) - ( -2am1 )
y - ( -2am1 ) =
( x - am12 )
am 22 - am12
Þ
Þ
Circle Through Co-normal Points
To find the equation of the circle passing through the three
(conormal) points on the parabola, normals at which pass
through a given point (a, b).
Y
B
2
y + 2am1 = ( x - am12 )
(m 2 + m1 )
A
(α,β)
y (m1 + m 2 ) + 2am1(m1 + m 2 ) = - 2x + 2am12
O
y (m1 + m 2 ) + 2am1m 2 = - 2x
2am1m 2m 3
Þ y (m1 + m 2 + m 3 - m 3 ) +
= - 2x
m3
Þ
Þ
Þ
Þ
y (0 - m 3 ) -
2k
= - 2x
m3
[from Eqs. (ii) and (iv)]
- ym 32 - 2k = - 2m 3 x
\
x 2 - 2ky = 0
am 3 + (2a - a ) m + b = 0
Sol. Given, parabola is 3y 2 + 4y - 6x + 8 = 0
Þ
4 ö
æ
3 çy 2 + y ÷ = 6x - 8
è
3 ø
Þ
2
ìï æ
2ö
4 üï
3 í çy + ÷ - ý = 6x - 8
3ø
9ï
ïî è
þ
2
4ö
2ö
æ
æ
3 çy + ÷ = ç6x - 8 + ÷
è
è
3ø
3ø
2
\
Let
Then,
Comparing with
\
10 ö
2ö
æ
æ
çy + ÷ = 2 ç x - ÷
è
è
9ø
3ø
2
10
y + =Y, x =X
3
9
Y 2 = 2X
m1 + m2 + m 3 = 0
(2a - a )
m 1m 2 + m 2m 3 + m 3m 1 =
a
b
and
m 1m 2m 3 = a
Let the equation of the circle through A, B, C be
x 2 + y 2 + 2 gx + 2 fy + c = 0
…(i)
1
2
2ö
æ
any point on the axis of parabola is ç x, - ÷
è
3ø
10
and X > 2a Þ x >1
9
19
Þ
x>
9
… (ii)
…(iii)
…(iv)
2
If the point (am , - 2am ) lies on it, then
(am 2 ) 2 + ( -2am ) 2 + 2 g (am 2 ) + 2 f ( -2 am ) + c = 0
Þ
a 2m 4 + ( 4a 2 + 2ag ) m 2 - 4afm + c = 0
…(v)
This is a biquadratic equation in m. Hence, there are four
values of m, say m 1 , m 2 , m 3 and m 4 such that the circle
pass through the points.
A (am 12 , - 2am 1 ), B (am 22 , - 2am 2 ), C (am 23 , -2am 3 ) and
D (am 24 , - 2am 4 ).
\ m1 + m2 + m 3 + m 4 = 0
Þ
0 +m4 = 0
Þ
m4 = 0
2
\
(am 4 , - 2am 4 ) = (0, 0 )
Y 2 = 4aX
a=
…(E)
\
y Example 46 Find the point on the axis of the
parabola 3y 2 + 4 y - 6 x + 8 = 0 from when three
distinct normals can be drawn.
Þ
Let A (am 12 , - 2am 1 ), B (am 22 , - 2am 2 ) and C (am 23 , - 2am 3 )
be the three points on the parabola
Since, point of intersection of normals is (a, b), then
which is a quadratic in m 3 .
Since, PQ will touch it, then
B 2 - 4 AC = 0
( -2x )2 - 4 × y × 2k = 0
C
y 2 = 4ax
ym 32 - 2m 3 x + 2k = 0,
Þ
X
…(F)
[from Eq. (i)]
Thus, the circle passes through the vertex of the parabola
y 2 = 4ax from Eq. (iv),
0 +0 +0 +0 +c =0
\
c =0
2 4
2
From Eq. (v), a m + ( 4a + 2ag ) m 2 - 4afm = 0
Þ
am 3 + ( 4a + 2 g ) m - 4 f = 0
…(vi)
395
Chap 05 Parabola
Now, Eqs. (E) and (vi) are identical.
4a + 2 g
4f
\
1=
=2a - a
b
Remark
This is likewise true for the pairs of chords AB, CD and AD, BC.
\
2 g = - (2a + a ), 2 f = - b/ 2
\ The equation of the required circle is
b
x 2 + y 2 - (2a + a ) x - y = 0
2
[from Eq. (iv)]
Corollary 1. The algebraic sum of the ordinates of the four
points of intersection of a circle and a parabola is zero.
Sum of ordinates
= -2am 1 - 2am 2 - 2am 3 - 2am 4
[from Eq. (F)]
= - 2a (m 1 + m 2 + m 3 + m 4 )
= - 2a ´ 0 = 0
Corollary 2. The common chords of a circle and a parabola
are in pairs equally inclined to the axis of the parabola.
Corollary 3. The circle through conormal point passes
through the vertex (0, 0 ) of the parabola.
Corollary 4. The centroid of four points; in which a circle
intersects a parabola, lies on the axis of the parabola.
4
ö
æ 4
2
ç åam i å( -2am i ) ÷
÷
ç i =1
i =1
,
Centroid = ç
÷
4
4
÷
ç
÷
ç
ø
è
a
æa
ö
= ç {( Sm 1 ) 2 - 2 Sm 1m 2 }, - ( Sm 1 ) ÷
è4
ø
2
æa æ
2 ( 4a 2 + 2ag ) ö ö
= çç ç 0 ÷ , 0 ÷÷
a2
ø ø
è4 è
= ( -2a - g, 0 )
ψ
Here y = 0, which is axis of the parabola y 2 = 4ax .
ψ
y Example 47 A circle cuts the parabola y 2 = 4ax at
right angles and passes through the focus, show that
its centre lies on the curve y 2 (a + 2x ) = a (a + 3x ) 2 .
Y
A
φ
B
φ
θ
θ
X
O
C
D
Let A, B, C, D be the points of intersection of the circle and
the parabola with A (am 12 , - 2am 1 ), B (am 22 , - 2am 2 )
C (am 23 , - 2am 3 ) and D (am 24 , - 2am 4 ) then equation of AC
and BD are
y (m 1 + m 3 ) = - 2 x - 2am 1m 3
and
y (m 2 + m 4 ) = - 2 x - 2am 2m 4 , respectively.
\ Slopes of the chords AC and BD are
2
2
and , respectively.
m1 + m 3
m2 + m 4
2
\ Slope of AC = m1 + m 3
2
[Qm 1 + m 2 + m 3 + m 4 = 0]
=
m2 + m 4
æ
ö
2
= - ç÷ = - Slope of BD
è m2 + m 4 ø
\ Their slopes are equal in magnitude and opposite in
sign.
\ The chords of AC and BD are equally inclined to the axis.
Sol. Let the circle x 2 + y 2 + 2gx + 2 fy + c = 0
…(i)
meet the parabola y 2 = 4ax at any point P (at 2 , 2at ) cutting
it at right angles.
We have to find locus of centre of circle Eq. (i),
i.e.
(-g, - f )
But given circle Eq. (i) passes through the focus (a , 0), then
a 2 + 0 + 2ga + 0 + c = 0
…(ii)
Þ
c = - a 2 - 2ag
90°
P
S(a,0)
Now, the circle and parabola intersect at P (at 2 , 2at ) at
right angles.
Since, tangent at P (at 2 , 2at ) to the parabola y 2 = 4ax is
ty = x + at 2
Hence, this tangent must pass through the centre ( - g , - f )
of the circle
\
- ft = - g + at 2
Þ
ft = g - at 2
2
Also the point P (at , 2at ) lies on the circle (i), then
…(iii)
396
Textbook of Coordinate Geometry
a 2t 4 + 4a 2t 2 + 2agt 2 + 4aft + c = 0
Hence, from Eqs. (ii) and (iii) when we put values for c and ft.
\ a 2t 4 + 4a 2t 2 + 2agt 2 + 4ag - 4a 2t 2 - a 2 - 2ag = 0
f
æ a - 2g ö
æ a - 2g ö
÷
÷ = g -aç
ç
è a ø
è a ø
f
æ a - 2g ö
ç
÷ = ( 3g - a )
è a ø
Þ
a 2t 4 + 2agt 2 + 2ag - a 2 = 0
Þ
Þ
at 4 + 2gt 2 + 2g - a = 0
Þ
f 2 ( a - 2g ) = a ( 3g - a ) 2
Þ
( - f )2 [a + 2 ( - g )] = a (a - 3g )2
Þ
( - f )2 [a + 2 ( - g )] = a [a + 3 ( - g )]2
Þ
Þ
Þ
4
2
a ( t - 1) + 2g ( t + 1) = 0
2
2
2
(t + 1) [a (t - 1) + 2g ] = 0 But t + 1 ¹ 0
a ( t 2 - 1) + 2g = 0 Þ t 2 =
a - 2g
a
Hence, locus of the centre ( - g , - f ) is the curve
y 2 ( a + 2x ) = a ( a + 3x ) 2 .
Hence, from Eq. (iii),
Exercise for Session 2
1.
If 2x + y + l = 0 is a normal to the parabola y 2 = - 8x , then the value of l is
(a) - 24
2.
1
2
2
(d) -
2
(b) x + y - a = 0
(c) x - y + a = 0
2
(d) x - y - a = 0
2
The circle x + y + 4lx = 0 which l ÎR touches the parabola y = 8x . The value of l is given by
(b) l Î (-¥, 0)
(c) l Î(1, ¥)
(d) l Î(-¥, 1)
2
If the normals at two points P and Q of a parabola y = 4ax intersect at a third point R on the curve, then the
product of ordinates of P and Q is
(b) 2a 2
(c) - 4a 2
(d) 8a 2
The normals at three points P , Q, R of the parabola y 2 = 4ax meet in (h, k ). The centroid of DPQR lies on
(a) x = 0
7.
1
2
The common tangent to the parabola y 2 = 4ax and x 2 = 4ay is
(a) 4a 2
6.
(c) -
(b) 2
(a) l Î (0, ¥)
5.
(d) 24
The slope of a chord of the parabola y = 4ax which is normal at one end and which subtends a right angle at
the origin is
(a) x + y + a = 0
4.
(c) - 8
2
(a)
3.
(b) - 16
(b) y = 0
(c) x = - a
(d) y = a
The set of points on the axis of the parabola y 2 - 4x - 2y + 5 = 0 from which all the three normals to the
parabola are real, is
(a) (l, 0) ; l > 1
(b) (l, 1) ; l > 3
(c) (l, 2) ; l > 6
(d) (l, 3) ; l > 8
8.
Prove that any three tangents to a parabola whose slopes are in harmonic progression enclose a triangle of
constant area.
9.
A chord of parabola y 2 = 4ax subtends a right angle at the vertex. Find the locus of the point of intersection of
tangents at its extremities.
10.
Find the equation of the normal to the parabola y 2 = 4x which is
(a) parallel to the line y = 2x - 5.
(b) perpendicular to the line 2x + 6y + 5 = 0.
11.
The ordinates of points P and Q on the parabola y 2 = 12x are in the ratio 1 : 2. Find the locus of the point of
intersection of the normals to the parabola at P and Q.
12.
The normals at P , Q, R on the parabola y 2 = 4ax meet in a point on the line y = c . Prove that the sides of the
DPQR touch the parabola x 2 = 2cy .
13.
The normals are drawn from (2l, 0) to the parabola y 2 = 4x . Show that l must be greater than 1. One normal is
always the X-axis. Find l for which the other two normals are perpendicular to each other.
Session 3
Pair of Tangents SS1 = T 2 , Chord of Contact, Equation of the Chord Bisected at
a Given Point, Diameter, Lengths of Tangent, Subtangent, Normal and Subnormal,
Some Standard Properties of the Parabola, Reflection Property of a Parabola,
Study of Parabola of the Form (ax 2 + by)2 + 2gx + 2fy + c = 0
Pair of Tangents SS1 = T 2
If y 12 - 4ax 1 > 0, then any point P ( x 1 , y 1 ) lies outside the
parabola and a pair of tangents PQ , PR can be drawn to it
from P. We find their equation as follows.
Let T (h, k ) be any point on the pair of tangents PQ or PR
drawn from any external point P ( x 1 , y 1 ) to the parabola
y 2 = 4ax .
k)
T (h,
Q
Let P ( x 1 , y 1 ) be any point outside the parabola. Let a
chord of the parabola through the point P ( x 1 , y 1 ) cut the
parabola at Q and let R (h, k ) be any arbitrary point on the
line PQ ( R inside or outside).
Let Q divide PR in the ratio l : 1, then coordinates of Q are
æ lh + x 1 lk + y 1 ö
,
[QPQ : QR = l : 1]
ç
÷
l +1 ø
è l +1
Since, Q lies on parabola Eq. (i), then
2
æ lk + y 1 ö
æ lh + x 1 ö
ç
÷ = 4a ç
÷
è l +1 ø
è l +1 ø
P (x
1,y)
1
R
R (h, k)
Equation of PT is
y - y1 =
Þ
k - y1
(x - x 1 )
h - x1
Q
æ k - y1 ö
æ hy 1 - kx 1 ö
y =ç
÷ x +ç
÷
èh - x1 ø
è h - x1 ø
P (x1 , y1 )
which is tangent to the parabola
y 2 = 4ax
a
c=
\
m
æ hy - kx 1 ö æ k - y 1 ö
Þ
cm = a
Þ ç 1
֍
÷ =a
è h - x1 ø èh - x1 ø
Þ
Þ
Þ (k 2 - 4ah ) l2 + 2 [ky 1 - 2a (h + x 1 )]l
+ (y 12 - 4ax 1 ) = 0 …(ii)
Line PR will become tangent to parabola Eq. (i), then roots
of Eq. (ii) are equal
(k - y 1 ) (hy 1 - kx 1 ) = a (h - x 1 ) 2
\ Locus of (h, k ), equation of pair of tangents is
(y - y 1 ) ( xy 1 - x 1 y ) = a ( x - x 1 ) 2
where
SS 1 = T 2
S = y 2 - 4ax , S 1 = y 12 - 4ax 1
4 [ky 1 - 2a (h + x 1 )]2 - 4 (k 2 - 4ah ) (y 12 - 4ax 1 ) = 0
or
{ky 1 - 2a (h + x 1 )} 2 = (k 2 - 4ah ) (y 12 - 4ax 1 )
i.e.
and
T = yy 1 - 2a ( x + x 1 ).
Aliter :
Let the parabola be y 2 = 4ax
\
Hence, locus of R (h, k ) i.e. equation of pair of tangents
from P( x 1 , y 1 ) is
{yy 1 - 2a ( x + x 1 )} 2 = (y 2 - 4ax ) (y 12 - 4ax 1 )
Þ (y 2 - 4ax ) (y 12 - 4ax 1 ) = {yy 1 - 2a ( x + x 1 )} 2
\
( lk + y 1 ) 2 - 4a ( lh + x 1 ) ( l + 1) = 0
T 2 = SS 1 or SS 1 = T 2
Remark
…(i)
S = 0 is the equation of the curve, S1 is obtained from S by
replacing x by x1 and y by y1 and T = 0 is the equation tangent at
( x1, y1 ) to S = 0.
398
Textbook of Coordinate Geometry
Chord of Contact
Let
P º ( x 1, y1 )
and the tangents from P touch the parabola at Q (at 12 , 2at 1 )
The chord joining the points of contact of two tangents
drawn from an external point to a parabola is known as the
chord of contact of tangents drawn from external point.
Theorem The chord of contact of tangents drawn from a
point ( x 1 , y 1 ) to the parabola y 2 = 4ax is
and R(at 22 , 2at 2 ) then P is the point of intersection of
tangents.
\
x 1 = at 1t 2 and y1 = a(t 1 + t 2 )
x
y
Þ
t 1t 2 = 1 and t 1 + t 2 = 1 …(ii)
a
a
yy 1 = 2a ( x + x 1 ).
QR = (at 12 - at 22 )2 + (2at 1 - 2at 2 )2
Now,
= a 2 (t 1 - t 2 )2 [(t 1 + t 2 )2 + 4 ]
Proof Let PQ and PR be tangents to the parabola y 2 = 4ax
drawn from any external point P ( x 1 , y 1 ), then QR is called
chord of contact of the parabola y 2 = 4ax .
Y
Q
X′
= | a || t 1 - t 2 | {(t 1 + t 2 )2 + 4 }
= | a | {(t 1 + t 2 )2 - 4t 1t 2 }
′)
,y
(x′
æy 2 4x ö æy 2
ö
= | a | ç 12 - 1 ÷ × ç 12 + 4 ÷
a ø èa
èa
ø
[from Eq. (ii)]
X
A
P
(x1, y1)
=|a|
Chord of contact
R (x′′,
Y′
y′′)
=
Let
Q º ( x ¢ , y ¢ ) and R º ( x ¢¢,y ¢¢ )
Equation of tangent PQ is
…(i)
yy ¢ = 2a ( x + x ¢ )
and equation of tangent PR is
…(ii)
yy ¢¢ = 2a ( x + x ¢¢ )
Since, lines Eqs. (i) and (ii) pass through ( x 1 , y 1 ), then
y 1 y ¢ = 2a ( x 1 + x ¢ ) and y 1 y ¢¢ =2a ( x 1 + x ¢¢ )
Hence, it is clear Q ( x ¢ , y ¢ ) and R ( x ¢¢,y ¢¢ ) lie on
yy 1 = 2a ( x + x 1 )
which is chord of contact QR.
Sol. Given parabola is
y 2 = 4ax
- 4ax 1 )
×
|a|
(y12
2
+ 4a )
|a|
1
(y12 - 4ax 1 )(y12 + 4a 2 )
|a|
y 2 = 2 (yy1 - 2ax 1 )
Þ
Þ
… (i)
2
y - 2yy1 + 4ax 1 = 0
k1 + k 2 = 2y1 and k1k = 4ax 1
\
(k 2 - k1 ) = (k1 + k 2 )2 - 4k1k 2
\
= ( 4y12 - 16ax 1 ) = 2 (y12 - 4ax 1 )
…(ii)
Y
Q (h1, k1)
X′
…(i)
Q (at12, 2at1)
X
A
R (h 2 , k 2 )
P (x
1 , y1 )
Y′
)
, y1
P (x 1
X′
(y12
Aliter :
Equation of QR is yy1 = 2a ( x + x 1 )
yy - 2ax 1
x= 1
Þ
2a
The ordinates of Q and R are the roots of the equation
æ yy - 2ax 1 ö
y 2 = 4a ç 1
÷
ø
è
2a
y Example 48 Tangents are drawn from the point
( x 1 , y 1 ) to the parabola y 2 = 4ax , show that the length
1
of their chord of contact is
( y 12 - 4ax 1 )( y 12 + 4a 2 ).
|a|
Y
{(t 1 + t 2 )2 + 4 }
A
X
R (at22, 2at2)
Y′
Since, Q (h1, k1 ) and R (h 2 , k 2 ) lie on the parabola y 2 = 4ax ,
therefore
k12 = 4ah1 and k 22 = 4ah 2
Þ
k 22 - k12 = 4a (h 2 - h1 )
(k 2 + k1 ) (k 2 - k1 ) = 4a (h 2 - h1 )
Chap 05 Parabola
Þ
2y1 (k 2 - k1 ) = 4a (h 2 - h1 )
y (k - k 1 )
(h 2 - h1 ) = 1 2
2a
Þ
Proof Since, equation of the parabola is
y12 (k 2 - k1 )2
2
4a
(k 2 - k 1 )
2
( y 1 + 4a 2 )
=
2|a |
=
=
2
(y12
- 4ax 1 )
2 |a |
y 2 = 4ax
…(iii)
Q (x2, y2)
[from Eq. (iii)]
P (x1, y1)
(y12 + 4a 2 )
[from Eq. (ii)]
1
(y12 - 4ax 1 ) (y12 + 4a 2 )
|a|
y Example 49 Prove that the area of the triangle
formed by the tangents drawn from ( x 1 , y 1 ) to y 2 = 4ax
and their chord of contact is ( y 12 - 4ax 1 ) 3/ 2 / 2a .
Sol. Equation of QR (chord of contact) is
yy1 = 2a ( x + x 1 )
Þ
yy1 - 2a ( x + x 1 ) = 0
Q PM = Length of perpendicular from P ( x 1, y1 ) on QR
| y y - 2a ( x 1 + x 1 ) | |(y12 - 4ax 1 )|
= 1 1
=
(y12 + 4a 2 )
(y12 + 4a 2 )
[Since, P ( x 1, y1 ) lies outside the parabola. So, y12 - 4ax 1 > 0]
R
(x3, y3)
Since, Q and R lie on parabola (i),
y 22 = 4ax 2 and y 23 = 4ax 3
\
y 23 - y 22 = 4a ( x 3 - x 2 )
Þ
y 3 - y2
4a
=
x 3 - x2 y 3 + y2
=
1,
Equation of QR is y - y 1 =
2a
(x - x 1 )
y1
Þ
yy 1 - y 12 = 2ax - 2ax 1
Þ
yy 1 - 2a ( x + x 1 ) = y 12 - 4ax 1
[subtracting 2ax 1 from both sides]
y1 )
R
Now, area of DPQR =
1
QR × PM
2
(y 2 - 4ax 1 )
1 1
(y12 - 4ax 1 ) (y12 + 4a 2 ) 1
=
2 |a|
(y12 + 4a 2 )
= (y12 - 4ax 1 )3 / 2 / 2a, if a > 0
Equation of the Chord Bisected
at a Given Point
Theorem The equation of the chord of the parabola
y 2 = 4ax which is bisected at ( x 1 , y 1 ) is
yy 1 - 2a ( x + x 1 ) = y 12 - 4ax 1
or
4a
[QP ( x 1 , y 1 ) is mid-point of QR]
2y 1
y 3 - y 2 2a
=
= Slope of QR
x 3 - x2 y1
\
Q
P(x
…(i)
Let QR be the chord of the parabola whose mid-point is
P ( x 1 , y 1 ).
Now, QR = (k 2 - k1 )2 + (h 2 - h1 )2
= (k 2 - k 1 )2 +
399
T = S1,
where, T = yy 1 - 2a ( x + x 1 ) and S 1 = y 12 - 4ax 1 .
\
T = S1,
where
T = yy 1 - 2a ( x + x 1 ) and S 1 = y 12 - 4ax 1 .
y Example 50 Find the locus of the mid-points of the
chords of the parabola y 2 = 4ax which subtend a right
angle at the vertex of the parabola.
Sol. Let P (h , k ) be the mid-point of a chord QR of the parabola
y 2 = 4ax then equation of chord QR is
or
T = S1
yk - 2a( x + h ) = k 2 - 4ah
Þ
yk - 2ax = k 2 - 2ah
…(i)
If A is the vertex of the parabola. For combined equation of
AQ and AR making homogeneous of y 2 = 4ax with the help
of Eq. (i).
\
y 2 = 4ax
Þ
æ yk - 2ax ö
y 2 = 4ax ç 2
÷
è k - 2ah ø
400
Textbook of Coordinate Geometry
Þ
y 2 (k 2 - 2ah ) - 4akxy + 8a 2 x 2 = 0
Diameter
Q
(0, 0) A
90°
P (h, k)
R
Since, ÐQAR = 90°
\ Coefficient of x 2 + Coefficient of y 2 = 0
k 2 - 2ah + 8a 2 = 0
Hence, the locus of P (h , k ) is y 2 - 2ax + 8a 2 = 0.
y Example 51 Show that the locus of the middle points
of normal chords of the parabola y 2 = 4ax is
y 4 - 2a ( x - 2a ) y 2 + 8a 4 = 0.
The locus of the middle points of a system of parallel chords
is called a diameter and in case of a parabola this diameter
is shown to be a straight line which is parallel to the axis of
the parabola.
Theorem The equation of the diameter bisecting chords
2a
of slope m of the parabola y 2 = 4ax is y = .
m
Proof Let y = mx + c be system of parallel chords to
y 2 = 4ax for different chords c varies, m remains constant.
Let the extremities of any chord PQ of the set be P ( x 1 , y 1 )
and Q ( x 2 , y 2 ) and let its middle point be M (h, k ).
Y
P
Sol. Equation of the normal chord at any point (at 2 , 2at ) of the
parabola y 2 = 4ax is
y + tx = 2at + at 3
t, 2
R (a
2
X′
…(i)
Þ
y 2 = 4ax and y = mx + c .
yy1 - 2ax = y12 - 2ax 1
2a
y1
…(ii)
…(iii)
2at + at 3
t
= 2
-2a y1 - 2ax 1
y12 - 2ax 1
= 2a + at 2
-2a
2
æ -2a ö
y12 - 2ax 1
= 2a + a ç
÷
-2a
è y1 ø
Þ
y12 - 2ax 1 2ay12 + 4a 3
=
-2a
y12
Þ
y14 - 2ax 1y12 = - 4a 2y12 - 8a 4
Þ
X
On solving equations
Q From Eqs. (i) and (ii) are identical, comparing, them
t
1
2at + at 3
=
= 2
y1 -2a y1 - 2ax 1
Þ
A
Q (x2, y2)
Q
From last two relations,
y = mx + c
Y′
But if M ( x 1, y1 ) be its middle point its equation must be
also
T = S1
Þ
yy1 - 2a ( x + x 1 ) = y12 - 4ax 1
From first two relations, t = -
M
at)
M (x1 , y1)
Þ
(x1 , y1 )
y14 - 2a ( x 1 - 2a ) y12 + 8a 4 = 0
Hence, the locus of middle point ( x 1, y1 ) is
y 4 - 2a ( x - 2a ) y 2 + 8a 4 = 0.
[from Eq. (iii)]
æy -c ö
y 2 = 4a ç
÷
è m ø
\
my 2 - 4ay + 4ac = 0
y + y 2 2a
4a
or 1
y1 + y2 =
=
\
m
m
2
[Q(h, k ) is the mid-point of PQ]
2a
Hence, locus of M (h, k ) is y = .
m
Aliter :
Let (h, k ) be the middle point of the chord y = mx + c of the
parabola y 2 = 4ax then
T = S 1 Þ ky - 2a( x + h ) = k 2 - 4ah
2a
2a
slope =
=m Þ k =
k
m
2a
Hence, locus of the mid-point is y = .
m
\
Remarks
1. The point in which any diameter meets the curve is called
the extremity of the diameter.
2. Any line which is parallel to the axis of the parabola drawn
through any point on the parabola is called diameter and its
equation is y-coordinate of that point.
If point on diameter ( x1, y1 ), then diameter is y = y1.
Chap 05 Parabola
Corollary 1. The tangent at the extremity of a diameter of
a parabola is parallel to the system of chords it bisects.
y = mx + c
Diameter
a , 2a
m2 m
Corollary 3. To find the equation of a parabola when the
axes are any diameter and the tangent to the parabola at the
point where this diameter meets the curve.
Let the equation of the parabola be
y 2 = 4ax
Let y = mx + c (c variable) represents the system of parallel
2a
chords, then the equation of diameter of y 2 = 4ax is y = .
m
æ a 2a ö
The diameter meets the parabola y 2 = 4ax at ç , ÷
èm 2 m ø
y = mx + c
R
(x1, y1)
2 2at 1)
,
P (at 1
y=
2a
m
Diameter
Q
(at22, 2at2)
\ Equation of diameter is
y = 2a / m Þ y = a(t 1 + t 2 )
(at 12 , 2at 1 )
(at 22 , 2at 2 )
…(i)
and Q
meet at a
Now, tangents at P
point [at 1 t 2 , a (t 1 + t 2 )] which lies on Eq. (i).
Aliter
Let equation of any chord PQ be y = mx + c .
If tangents at P and Q meet at R ( x 1 , y 1 ), then PQ is the
chord of contact with respect to R ( x 1 , y 1 ) .
\ Equation of PQ is
2ax 1
2a
yy 1 = 2 x ( x + x 1 ) or y =
x+
y1
y1
which is identical to y = mx + c
2a
2a
or y 1 = .
m=
y1
m
2a
Hence, locus of R ( x 1 , y 1 ) is y = , which bisects the
m
chord PQ.
…(i)
Let AB be the diameter of the parabola Eq. (i), then its
2a
equation is y =
m
Y
a , 2a
m2 m
A
a
and tangent is y = mx + which is parallel to y = mx + c .
m
Corollary 2. Tangents at the end of any chord meet on the
diameter which bisects the chords.
If extremities of the chord be P (at 12 , 2at 1 ) and Q (at 22 , 2at 2 )
then its slope
2at 2 - 2at 1
2
.
Þ m=
m=
2
2
(t 2 + t 1 )
at 2 - at 1
401
T P
θ
θ
Q
S
M
N
B
X
Since, A is the extremity of the diameter
æ a 2a ö
Coordinates of A is ç , ÷
\
èm 2 m ø
where, m = tanq
then, the equation of tangent AT at A is
a
y = mx + .
m
Now, let P be any point on the parabola Eq. (i), whose
coordinates referred to Vx and Vy are ( x , y ) and referred
to diameter AB and tangent AT are ( X , Y ).
then
[since, PQ || AT ]
X = AQ and Y = QP
Now, VN = VL + LN = VL + AM = VL + AQ + QM
a
=
+ X + QP cosq
m2
a
or
…(ii)
+ X + Y cosq
x=
m2
and
PN = PM + MN = PM + AL
2a
= QP sinq +
m
2a
… (iii)
\
y = Y sinq +
m
From Eqs. (ii) and (iii) coordinates of P are
2a ö
æ a
ç 2 + X + Y cos q, Y sinq + ÷
èm
mø
Now, P lies on Eq. (i).
2
\
2a ö
ö
æ a
æ
çY sinq + ÷ = 4a ç 2 + X + Y cos q ÷
è
ø
ø
è
m
m
402
Textbook of Coordinate Geometry
Þ
Y 2 sin2 q +
=
4a
m
4a 2
m
2
+
It meets the diameter through P i.e. X -axis or y = 0, then
Eq. (iii) reduces
0 = x + lt 12
4a
Y sinq
m
2
+ 4aX + 4aY cosq
2
Similarly, -
Y 2 sin2 q + 4a 2 cot 2 q + 4a cos qY
Þ
2
2
= 4a cot q + 4aX + 4aY cos q
2
x = - lt 12 = PB
Þ
\
lt 22
2
( PA )
[Qm = tanq ]
2
Þ Y sin q = 4aX
…(iv)
…(v)
= PC
2
= ( -lt 1t 2 ) = l2t 12t 22
= ( -lt 12 ) ( -lt 22 )
[from Eqs. (iv) and (v)]
= PB × PC
Y 2 = ( 4a cosec 2 q ) X ,
\
which is the required parabola referred to diameter and
tangent at the extremity of the diameter as axes.
Remark
The quantity 4acosec 2q is called the parameter of the diameter
AQ. It is equal to length of the chord which is parallel to AT and
passes through the focus.
i.e.
a cosec 2q = a( 1 + cot 2 q) = a + a cot 2 q
a
= a+ 2
m
= a + VL = SP
Lengths of Tangent, Subtangent,
Normal and Subnormal
Let the parabola y 2 = 4ax . Let the tangent and normal at
P ( x , y ) meet the axis of parabola at T and G respectively
and let tangent at P ( x , y ) makes angle y with the positive
direction of X-axis.
Y
2
)
x, y
1
But length of focal chord if P ( at 2, 2at ) is [QS ( a, 0 )] a æçt + ö÷ .
è tø
tanq =
\
2
\
2at - 0
at 2 - a
=
2t
t2 - 1
P(
1
or t - = 2 cot q
t
X′
90°– ψ
N
G
X
2
1
1
ïì
ïü
a æçt + ö÷ = a íæçt - ö÷ + 4 ý
è
ø
è tø
t
ïþ
îï
Y′
= a{( 2 cot q) 2 + 4}
Then,
= 4 acosec 2q = 4 × SP
y Example 52 If the diameter through any point P of a
parabola meets any chord in A and the tangent at the
end of the chord meets the diameter in B and C, then
prove that PA 2 = PB ×PC .
Sol. The equation of the parabola referred to the diameter
through P and tangent at P as axes is
…(i)
y 2 = 4 lx
where,
T
ψ
A
l = a cosec 2q
[from previous corollary]
Let QR be any chord of the parabola Eq. (i). Let the
extremities Q and R be ( lt 12 , 2lt 1 ) and ( lt 22 , 2lt 2 ).
Then, the equation of QR is
… (ii)
y (t 1 + t 2 ) - 2x - 2lt 1t 2 = 0
It meets the diameter through P i.e. X -axis or y = 0, then
Eq. (ii) reduces
0 - 2x - 2l t 1t 2 = 0
Þ
x = - lt 1t 2 = PA
…(iii)
Now, tangent at Q is
t 1y = x + lt 12
PT = Length of Tangent
PG = Length of Normal
TN = Length of Subtangent
and
NG = Length of Subnormal
If A (0, 0 ) is the vertex of the parabola.
Q
PN = y
\
PT = PN cosec y = y cosec y
PG = PN cosec (90° - y ) = y sec y
TN = PN cot y = y cot y
and
NG = PN cot(90° - y ) = y tan y
2a
[slope of tangent at P ( x , y )]
where, tan y =
=m
y
y Example 53 Find the length of tangent, subtangent,
normal and subnormal to y 2 = 4ax at (at 2 , 2at ) .
Sol. Q Equation of tangent of (at 2 , 2at ) of parabola y 2 = 4ax is
ty = x + at 2
Slope of this tangent m =
1
t
Chap 05 Parabola
Let tangent makes angle y with positive direction of X -axis
1
tan y =
t
then
t = cot y
\ Length of tangent at
(at 2 , 2at ) = 2at cosecy
= 2at
(1 + cot 2 y ) = 2at
(1 + t 2 )
and adding Eqs. (i) and (ii), we get
( x + a ) (1 + t 2 ) = 0 Þ 1 + t 2 ¹ 0
\ x + a = 0, which is directrix.
(2) The portion of a tangent to a parabola intercepted
between the directrix and the curve subtends a right
angle at the focus.
The equation of the tangent to the parabola y 2 = 4ax at
Length of normals at
(at 2 , 2at ) = 2at sec y = 2at
P (at 2 , 2at ) is
(1 + tan 2 y )
ty = x + at 2
= 2a (t 2 + t 2 tan 2 y ) = 2a (t 2 + 1)
…(i)
Let Eq. (i) meet the directrix x + a = 0 at Q,
æ
at 2 - a ö
then coordinates of Q are ç - a,
÷ , also focus S
t ø
è
is (a , 0 ).
Length of subtangent at (at 2 , 2at ) = 2at cot y
= 2at 2
Length of subnormal at (at 2 , 2at ) = 2at tan y
= 2a
Y
P
Some Standard Properties
of the Parabola
Q
\
and
Q
Let parabola is y 2 = 4ax
\ Tangent at P (at 2 , 2at ) is
ty = x + at 2
1
Its slope = m 1
t
(say)
2a ö
æa
and tangent at Q ç , - ÷ is
2
èt
t ø
y
a
- =x +
t
t2
Þ
…(i)
at 2 - a
2t
2
t -1
(say)
= m1
at 2 - a
-0
t2 - 1
t
slope of SQ =
=
= m2
-a -a
-2 t
Let the tangent at P (at 2 , 2at ) to the parabola y 2 = 4ax
meets the axis of the parabola i.e. X-axis or y = 0 at T. The
equation of tangent to the parabola y 2 = 4ax at P (at 2 , 2at )
is
ty = x + at 2
…(ii)
Y
M
- ty = xt + a
(say)
(say)
\
m 1m 2 = - 1
i.e. SP is perpendicular to SQ i.e. Ð PSQ = 90°
(3) The tangent at any point P of a parabola bisects
the angle between the focal chord through P and the
perpendicular from P to the directrix.
2
Its slope
- t = m2
Q
m 1m 2 = - 1
\
Ð PTQ = 90°
2at - 0
Slope of SP =
=
P
X
X
S (a, 0)
Directrix
90°
S (a, 0)
(at,2 2at)
90°
V
Z
x+a=0
(1) The tangents at the extremities of a focal chord
intersect at right angles on the directrix.
The extremities of a focal chord PQ are
2a ö
æa
P º (at 2 , 2at ) and Q º ç , - ÷
2
èt
t ø
T
403
θ
P
(at2, 2at)
θ
θ
T
Z
x+a=0
V
Directrix
S (a, 0)
X
404
Textbook of Coordinate Geometry
For coordinates of T solve it with y = 0.
SP = PM = a + at 2
\
T ( -at 2 , 0 )
\
SG = VG - VS = 2a + at 2 - a
\
ST = SV + VT = a + at 2 = a (1 + t 2 )
Also,
SP = PM = a + at 2 = a (1 + t 2 )
= a + at 2
ST = VS + VT = a + at 2
and
\
But
SP = ST , i.e. ÐSTP = ÐSPT
[alternate angles]
ÐSTP = Ð MPT
ÐSPT = Ð MPT
(4) The foot of the perpendicular from the focus on any
tangent to a parabola lies on the tangent at the vertex.
Hence, SP = SG = ST
(6) If S be the focus and SH be perpendicular to the
tangent at P, then H lies on the tangent at the vertex
and SH 2 = OS × SP , where O is the vertex of the
parabola.
Equation of tangent at P (at 2 , 2at )
Let P (at 2 , 2at ) be any point on the parabola
On the parabola y 2 = 4ax
is
Þ
ty = x + at
y 2 = 4ax
2
then, tangent at P (at , 2at ) to the parabola Eq. (i) is
2
ty = x + at 2
…(i)
x - ty + at = 0
Now, the equation of line through S (a, 0 ) and
perpendicular to Eq. (i) is
tx + y = l
Since, it passes through (a, 0 ).
\
ta + 0 = l
\
Equation tx + y = ta
It meets the tangent at the vertex i.e. x = 0.
P (at2, 2at)
M
H
…(ii)
Directrix
x (1 + t 2 ) = 0
\ Coordinates of H is (0, at )
[Q1 + t 2 ¹ 0 ]
x =0
Hence, the point of intersection of Eq. (i) and (ii) lies on
x = 0 i.e. on Y-axis (which is tangent at the vertex of a
parabola).
(5) If S be the focus of the parabola and tangent and
normal at any point P meet its axis in T and G
respectively then ST = SG = SP .
Let P (at 2 , 2at ) be any point on the parabola y 2 = 4ax ,
then equation of tangent and normal at P (at 2 , 2at ) are
ty = x + at 2 and y = - tx + 2at + at 2 , respectively.
and SP = PM = a + at 2 Þ
and SH = (a - 0 ) 2 + (0 - at ) 2 = a 2 + a 2 t 2
or (SH ) 2 = a {a (1 + t 2 )} = OS × SP.
Reflection Property of Parabola
The tangent ( PT ) and normal ( PN ) of the parabola
y 2 = 4ax
at P are the internal and external bisectors of ÐSPM and
BP is parallel to the axis of the parabola and
Ð BPN = ÐSPN .
Y
M
P (at , 2at)
T
A
N
M
90°
Z
V
Directrix
Y′
G
S
(a, 0)
X
β α
S
Light ray
B
α
Reflected ray
al
X′
2
P
β
nt
nge
rm
Ta
Y
x+a=0
OS = a
No
Since, tangent and normal meet its axis in T and G.
\ Coordinates of T ¢ and G are ( -at 2 , 0 ) and (2a + at 2 , 0 )
respectively.
T
X
S (a, 0)
O
or t 2 x + ty - at 2 = 0
Þ
Y
x+a=0
By adding Eqs. (i) and (ii), we get
X′
… (i)
2
X
Light ray
Light ray
Light ray
Y′
All rays of light coming from the positive direction of
X-axis and parallel to the axis of the parabola after
reflection pass through the focus of the parabola.
405
Chap 05 Parabola
y Example 54 A ray of light is coming along the line
y = b from the positive direction of X-axis and strikes
a concave mirror whose intersection with the
xy-plane is a parabola y 2 = 4ax . Find the equation
of the reflected ray and show that it passes through
the focus of the parabola. Both a and b are positive.
Sol. Given parabola is y 2 = 4ax
P
X′
A
2α
S
(a, 0)
α
y=b
α
α
i.e.
Normal
180°–α
X
180°–2α
i.e.
æ
æ b2 ö
b2 ö
Equation of tangent at P ç , b ÷ is yb = 2a ç x +
÷,
4a ø
è
è 4a ø
2a
Slope of tangent is .
b
b
Hence, slope of normal = = tan (180° - a ).
2a
b
\
tana =
2a
\ Slope of reflected ray = tan (180° - 2a )
= - tan2a
ì
b ü
ïï 2 ×
ï
ì 2 tan a ü
4ab
2a ï = = -í
=
ý
í
2
2 ý
( 4a 2 - b 2 )
î 1 - tan a þ
ï1 - b ï
2
îï
4a þï
Hence, equation of reflected ray is
æ
4ab
b2 ö
x
y -b= ÷
ç
4a ø
( 4a 2 - b 2 ) è
(y - b ) ( 4a 2 - b 2 ) = - ( 4ax - b 2 )
which obviously passes through the focus S (a , 0).
Study of Parabola of the Form
(ax + by ) 2 +2 gx + 2 fy + c = 0
Given equation can be written as
(ax + by ) 2 = - 2 gx - 2 fy - c
Now, add an arbitrary constant l in the square root of the
second degree terms. Then the equation will be of the form
i.e.
(ax + by + l) 2 = xf 1 ( l) + yf 2 ( l) + f 3 ( l)
Now, choose l such that the lines
ax + by + l = 0 and
are perpendicular
æ ax + by + l ö
÷
(ax + by + l) = (a + b ) ç
ç a 2 + b2 ÷
ø
è
2
xf 1 ( l) + yf 2 ( l) + f 3 ( l) = 0
2
2
2
and RHS of Eq. (i) by (a 2 + b2 )
Y′
Þ
…(ii)
Þ
af 1 ( l) + bf 2 ( l) = 0
Now, substitute the value of l in Eq. (i) from Eq. (ii).
Multiply and divide (a 2 + b2 ) in LHS of Eq. (i)
) b4a , b )
2
Y
i.e. (slope of ax + by + l = 0) ´ (slope of
xf 1 ( l) + yf 2 ( l) + f 3 ( l) = 0 ) = -1
f ( l)
a
Þ
= -1
- ´- 1
b
f 2 ( l)
…(i)
xf 1 ( l) + yf 2 ( l) + f 3 ( l)
æ xf ( l) + yf ( l) + f ( l) ö
2
3
÷
= (a 2 + b2 ) ç 1
ç
÷
2
2
(a + b )
è
ø
Then, Eq. (i) reduce in the form
2
ö
æ
æ ax + by + l ö
÷ = 4r ç bx - ay + m ÷
ç
ç (a 2 + b2 ) ÷
ç a 2 + b2 ÷
ø
è
ø
è
which is of the form Y 2 = 4rX
Y=
ax + by + l
2
2
,X =
(a + b )
Latusrectum is 4r =
bx - ay + m
2
2
(a + b )
1
2
(a + b2 )
and 4r =
1
2
(a + b2 )
.
.
Axis is Y = 0 or ax + by + l = 0.
Equation of tangent at vertex is
X = 0 or bx - ay + m = 0.
Vertex is the point of intersection of
X = 0 and Y = 0
i.e. bx - ay + m = 0 and ax + by + l = 0.
Equation of directrix is
X + r = 0.
Equation of latusrectum is X - r = 0.
Focus Since, axis and latusrectum intersect at the focus S
its coordinates are detained by solving
X - r = 0 and Y = 0.
y Example 55. Find the length of latusrectum of the
parabola (a 2 + b 2 ) ( x 2 + y 2 ) = (bx + ay - ab ) 2 .
Sol. Given equation may be written as
a 2 x 2 + a 2y 2 + b 2 x 2 + b 2y 2 = b 2 x 2 + a 2y 2 + a 2b 2
+ 2abxy - 2a 2by - 2ab 2 x
406
Textbook of Coordinate Geometry
a 2 x 2 - 2abxy + b 2y 2 = - 2ab 2 x - 2a 2by + a 2b 2
Þ (x 2 + y 2 ) =
ab ö
æ
(ax - by ) = - 2ab çbx + ay - ÷
Þ
è
2ø
ab
Since, ax - by = 0 and bx + ay = 0 are perpendicular.
2
ab ö
æ
2
æ
ö
÷
ç bx + ay 2
2 ç ax - by ÷
2
2
2 ÷
\ (a + b )
= - 2ab (a + b ) ç
ç a2 + b2 ÷
2
2
ç (a + b ) ÷
è
ø
ø
è
| bx + ay - ab |
2
Þ
æ ax - by
ç
ç a2 + b2
è
2
ö
-2ab
÷ =
÷
(a 2 + b 2 )
ø
( x - 0) 2 + ( y - 0) 2
Þ
=
| bx + ay - ab |
(a 2 + b 2 )
\
4r =
which is of the form Y 2 = - 4 rX .
2ab
(a 2 + b 2 )
.
Aliter :
Given equation may be written as
(bx + ay - ab )2
x2 + y2 =
(a 2 + b 2 )
P (x, y)
M
N
S (0, 0)
which is of the form SP = PM .
1
Since, distance from focus S to ( bx + ay - ab = 0) = ( 4r )
2
Y
1
ab
, y)
Þ
( 4r ) =
P (x
2
2
N
2
a +b
ab ö
æ
÷
ç bx + ay 2 ÷
ç
2
2
ç (a + b ) ÷
ø
è
Therefore, the latusrectum = 4r =
(a 2 + b 2 )
bx+ay–ab = 0
Þ
2ab
O
(a 2 + b 2 )
M
Remark
Consider the equation of parabola is y 2 = 4 ax.
i.e.
( MP ) 2 = (Latusrectum) NP.
i.e. if P is any point on the given parabola, then
(the distance of P from its axis) 2 = (Latusrectum)
(The distance of P from the tangent at its vertex).
Exercise for Session 3
1.
If m1, m 2 are slopes of the two tangents that are drawn from (2, 3) to the parabola y 2 = 4x , then the value of
1
1
is
+
m1 m 2
(a) –3
2.
(b) 30°
1
(c) tan-1 æç ö÷
è 2ø
(d) 45°
(b) 2a = b
(c) a 2 = 2b
(d) 2a = b 2
(b) y - 2 = 0
(c) y + 1 = 0
(d) y + 2 = 0
(b) 8 3
(c) 8 2
(d) None of these
For parabola x 2 + y 2 + 2xy - 6x - 2y + 3 = 0, the focus is
(a) (1, - 1)
7.
3
2
From the point ( -1, 2) tangent lines are drawn to the parabola y 2 = 4x , the area of triangle formed by chord of
contact and the tangents is given by
(a) 8
6.
(d)
The diameter of the parabola y 2 = 6x corresponding to the system of parallel chords 3x - y + c = 0 is
(a) y - 1 = 0
5.
2
3
If (a, b) is the mid-point of chord passing through the vertex of the parabola y 2 = 4x , then
(a) a = 2b
4.
(c)
The angle between the tangents drawn from the origin to the parabola y 2 = 4a( x – a ) is
(a) 90°
3.
(b) 3
(b) (–1, 1)
(c) (3, 1)
(d) None of these
The locus of the mid-point of that chord of parabola which subtends right angle on the vertex will be
(a) y 2 - 2ax + 8a 2 = 0
(b) y 2 = a (x - 4a )
(c) y 2 = 4a (x - 4a )
(d) y 2 + 3ax + 4a 2 = 0
X
Chap 05 Parabola
8.
A ray of light moving parallel to the X-axis gets reflected from a parabolic mirror whose equation is
( y - 2)2 = 4( x + 1). After reflection, the ray must pass through the point
(a) (–2, 0)
9.
407
(b) (–1, 2)
(c) (0, 2)
(d) (2, 0)
2
Prove that the locus of the point of intersection of tangents to the parabola y = 4ax which meet at an angle a is
( x + a )2 tan2 a = y 2 - 4ax .
10.
Find the locus of the middle points of the chords of the parabola y 2 = 4ax which pass through the focus.
11.
From the point P( -1, 2) tangents are drawn to the parabola y 2 = 4x . Find the equation of the chord of contact.
Also, find the area of the triangle formed by the chord of contact and the tangents.
Shortcuts and Important Results to Remember
1 Second degree terms in the equation of a parabola
should make perfect squares.
2 If l1 and l 2 are the lengths of segments of a focal chord
4l l
then the latusrectum of the parabola is 1 2 .
l1 + l 2
3 If a be the inclination of a focal chord with axis of the
parabola then its length is 4acosec 2 a.
4 If tangents of y 2 = 4ax at P(t1 ) and Q(t 2 ) meets at R, then
1
area of DPQR is a2 (t1 - t 2 )3 .
2
5 The foot of the perpendicular from the focus on any
tangent to a parabola lies on the tangent at the vertex.
6 The area of the triangle formed by three points on a
parabola is twice the area of the triangle formed by the
tangents, at these points.
7 The equation of the common tangent to the parabolas
x 2 = 4ay and y 2 = 4ax is x + y + a = 0.
8 If the chord joining t1, t 2 and t 3 , t 4 pass through a point
(c, 0) on the axis, then
t1t 2 = t 3t 4 = - c / a.
9 If the normals to the parabola y 2 = 4ax at the points t1 and
t 2 intersect again on the parabola at the point
‘t 3 ’, then t1t 2 = 2; t 3 = -(t1 + t 2 ) and the line joining t1 and t 2
passes through a fixed point (-2 a, 0).
10 If the length of a focal chord of y 2 = 4ax at a distance b
from the vertex is c then b2c = 4a3 .
11 From an external point only one normal can be drawn.
12 If a normal to y 2 = 4ax makes an angle q with the axis of
y 2 = 4ax then it will cut the curve again at an angle of
æ tan q ö
tan -1 ç
÷.
è 2 ø
13 The orthocentre of any triangle formed by the three
tangents to a parabola y 2 = 4ax ‘t1’, ‘t 2 ’ and ‘t ’3 lies on the
directrix and has the coordinates
(- a, a(t1 + t 2 + t 3 + t1t 2t 3 )).
14 Normals at the end points of the latusrectum of a parabola
y 2 = 4ax intersect at right angle on the axis of the parabola
and their point of intersection is (3a, 0 ).
15 A line ray parallel to axis of the parabola after reflection
passes through the focus.
JEE Type Solved Examples :
Single Option Correct Type Questions
n
This section contains 10 multiple choice examples.
Each example has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
Ex. 1 A ray of light travels along a line y = 4 and strikes
the surface of curves y 2 = 4( x +y ), then the equation of the
line along which the reflected ray travels is
Ex. 3 Two parabolas have the same focus. If their
directrices are the X-axis and the Y -axis, respectively, then
the slope of their common chord is
l
(a) ± 1
l
(a) x = 0
(c) x + y = 4
(b)
4
3
(c)
3
4
(d) None of these
Sol. (a) Let the focus be (a, b )
Then, equations are ( x - a ) 2 + (y - b ) 2 = y 2 and
( x - a ) 2 + (y - b ) 2 = x 2
(b) x = 2
(d) 2x + y = 4
Sol. (a) The given curve is (y - 2 ) 2 = 4( x + 1 )
Y
If
S1 º ( x - a ) 2 + (y - b ) 2 - y 2 = 0
and
S 2 º ( x - a ) 2 + (y - b ) 2 - x 2 = 0
\Equation of common chord S1 - S 2 = 0 gives
y=4
(0,4)
x2 -y 2 = 0
or
y =± x
Hence, slope of common chord is ± 1.
(–1,2)
(0,2)
Ex. 4 Let us define a region R in xy-plane as a set of
points ( x , y ) satisfying [ x 2 ] = [y ] (where [ x ] denotes greatest
integer £ x), then the region R defines
l
X
(0,0)
The focus is ( 0, 2 ).
The point of intersection of the curve and y = 4 is ( 0, 4 ). From
the reflection property of parabola the reflected ray passes
through the focus.
Therefore, x = 0 is the reflected ray.
Ex. 2 A parabola is drawn with focus at (3, 4 ) and vertex
at the focus of the parabola y 2 - 12 x - 4y + 4 = 0. The
equation of the parabola is
l
(a) x 2 - 6 x - 8y + 25 = 0 (b) y 2 - 8 x - 6y + 25 = 0
(c) x 2 - 6 x + 8y - 25 = 0 (d) x 2 + 6 x - 8y - 25 = 0
2
2
Sol. (a) y - 12 x - 4y + 4 = 0 Þ (y - 2 ) = 12 x
S (3,4)
A (3,2)
Its vertex is ( 0, 2 ) and a = 3,
its focus = (3, 2 ).
Hence, for the required parabola; focus is (3, 4 ), vertex = (3, 2 )
and a = 2,
Hence, the equation of the parabola is
( x - 3 ) 2 = 4(2 ) (y - 2 )
or
x 2 - 6 x - 8y + 25 = 0
(a) a parabola whose axis is horizontal
(b) a parabola whose axis is vertical
(c) integer point of the parabola y = x 2
(d) None of the above
Sol. (d)Q[ x 2 ] = [y ]
Y
If
0 £ y < 1,
then, [y ] = 0
\
3
[x 2 ] = 0
0 £ x2 <1
Þ
x Î( - 1, 1 )
for
1 £ y < 2,
then [y ] = 1
\
[x 2 ] = 1
2
1
X′
–2 – √3 – √2 –1 O
Þ
1 £ x2 <2
Þ
x Î( - 2, - 1 ] È [1, 2 )
for
2 £ y < 3, then [y ] = 2
1 √2 √3
2
X
Y′
then, [ x 2 ] = 2 Þ2 £ x 2 < 3
\
x Î( - 3, - 2 ] È [ 2, 3 )
The graph of the region will not only contain of the parabola
y = x 2 but [ x 2 ] = [y ] contain points within the rectangles of side
1, 2, ; 1, 2 - 1; 1, 3 - 2 etc.
Hence, a, b, c are incorrects.
409
Chap 05 Parabola
Ex. 5 The minimum area of circle which touches the
parabolas y = x 2 +1 and x = y 2 + 1 is
l
9p
(a)
sq units
16
9p
(c)
sq units
8
\
9p
(b)
sq units
32
9p
(d)
sq units
4
Sol. (b) The parabolas y = x 2 + 1 and x = y 2 + 1 are symmetrical
about y = x.
Therefore, the tangent at point A is parallel to y = x
dy
Y
then
= 2x =1
y=x
dx
1
5
or
x = ,y =
2
4
A
1 5ö
æ
\
Aºç , ÷
B
è2 4ø
X′
O
æ 5 1ö
B ºç . ÷
è 4 2ø
and
\
æ - 2m ö 2
æ - 2m ö ö÷
D º ça æç
÷ , - 2a ç
÷
è
ø
è l øø
l
è
Þ
æ 4am 2 4am ö
Dºç 2 ,
÷
l ø
è l
Ex. 7 If d is the distance between the parallel tangents
with positive slope to y 2 = 4 x and
x 2 + y 2 - 2 x + 4y - 11 = 0, then
(a) 10 < d < 20
(c) d < 4
(b) 4 < d < 6
(d) None of these
Sol. (c) Tangent to the parabola y 2 = 4 x having slope m is
X
y = mx +
1
m
and tangent to the circle ( x - 1 ) 2 + (y + 2 ) 2 = 4 2 having slope m is
2
y + 2 = m( x - 1 ) + 4 (1 + m 2 )
1
1 æ1 5 ö æ 5 1ö
3 2
Hence, Radius (r ) = AB =
ç - ÷ +ç - ÷ =
2
2 è2 4ø è 4 2ø
8
4 (1 + m 2 ) - m - 2 -
9p
Area = pr =
sq units
32
2
\
2m
l
l
Y′
2
m3 = -
\Distance between tangents (d ) =
Ex. 6 Let the line lx + my =1 cut the parabola y 2 = 4ax
in the points A and B. Normals at A and B meet at point C.
Normal from C other than these two meet the parabola at D,
then the coordinate of D are
(1 + m 2 )
l
(a) (a, 2a )
æ 4am 4a ö
(b) ç 2 , ÷
è l
l ø
æ 2am 2 2a ö
(c) ç 2 , ÷
l ø
è l
æ 4am 2 4am ö
÷
(d) ç 2 ,
l ø
è l
A and B lie on lx + my = 1
Þ
l (am12 ) + m( - 2am1 ) = 1
...(i)
and
l (am22 ) + m( - 2am2 ) = 1
...(ii)
Subtracting Eq. (ii) from Eq. (i), then
a(m1 - m2 ) ¹ 0
l (m1 + m2 ) - 2m = 0
2m
m1 + m2 =
l
Þ
Let
D
º (am32, - 2am3 ) and C
y = Mx - 2aM - aM
then
\
Þ
3
m1 + m2 + m3 = 0
2m
+ m3 = 0
l
(given)
m>0
d<4
(a) 1
(b) 2
(c) 3
(d) 4
Sol. (c) C : y = x 2 - 3 and D : y = kx 2
kx 2 = x 2 - 3
or
...(iii)
then,
...(i)
3
,
x2 =
1 -k
y=
3k
1 -k
\
æ 3
3k ö
Aºç
,
÷ (given x-value of A is positive)
è 1 -k 1 -k ø
and
æ 3 ö
a= ç
÷
è 1 -k ø
then
A º (a, ka 2 ) º (a, a 2 - 3 )
3
aM - (h - 2a ) M + k = 0
(1 + m 2 )
m
Ex. 8 Two parabolas C and D intersect at two different
points, where C is y = x 2 - 3 and D is y = kx 2 . The
intersection at which the x value is positive is designated
point A , and x = a at this intersection the tangent line l at A
to the curve D intersects curve C at point B, other than A. If
x-value of point B is 1, then a is equal to
º (h, k )
\ Equation of normal in terms of slope
(1 + m 2 )
-
Solving C and D, then
la(m12 - m22 ) - 2am(m1 - m2 ) = 0
Þ
\
As
we get
2
l
Sol. (d) Let A º (am12, - 2am1 ) and B º (am22, - 2am2 )
Now,
= 4-
1
m
[from Eq. (i)]
410
Textbook of Coordinate Geometry
tangent ‘l’ at A to the curve D is
Hence, the corresponding point on the parabola is (1, 2 )
y + a2 -3
= kx × a
2
Þ
\ ( x1 - x 2 ) 2 + (3 + (1 - x12 ) - 4 x 2 ) 2 = ( x1 - x 2 ) 2 + (y1 - y 2 ) 2
= Distance between (1, 2 ) and ( 0, 3 ) – radius
3ö
æ
y + a 2 - 3 = 2ax ç1 - 2 ÷
è a ø
[from Eq. (i)] …(ii)
= (1 + 1 ) - 1 = ( 2 - 1 )
\
B º (1, - 2 ) (a ¹ 1)
or
( x1 - x 2 ) 2 + (3 + (1 - x12 ) - 4 x 2 ) 2 = ( 2 - 1 ) 2
From Eq. (ii),
3ö
æ
- 2 + a 2 - 3 = 2a ç1 - 2 ÷
è a ø
\
min× [( x1 - x 2 ) 2 + (3 + (1 - x12 ) - 4 x 2 ) 2 ] = 3 - 2 2
a 3 - 5a = 2a 2 - 6
Þ
3
Þ
a - 2a - 5a + 6 = 0
Þ
\
(a - 1 ) (a + 2 ) (a - 3 ) = 0
a =3
(Qa ¹ 1, a ¹ - 2)
Ex. 9 min [( x 1 - x 2 ) 2 + (3 + (1 - x 12 ) - 4 x 2 ) 2 ], "
x 1 , x 2 Î R, is
l
(a) 4 5 + 1 (b) 3 - 2 2
Sol. (b) Let y1 = 3 +
or
x12
(1 - x12 )
2
+ (y1 - 3 ) = 1
Ex. 10 The condition that the parabolas y 2 = 4c ( x - d )
and y 2 = 4ax have a common normal other than X-axis
(a > 0, c > 0 ) is
l
2
(c) 5 + 1
(a) 2a < 2c + d
(c) 2d < 2a + c
(b) 2c < 2a + d
(d) 2d < 2c + a
Sol. (a) Normals of parabolas y 2 = 4ax and y 2 = 4c ( x - d ) in terms
of slope are
y = mx - 2am - am 3
(d) 5 - 1
and
y 2 = 4x 2
and
and
y 22
Subtracting Eqs. (ii) from (i), then
2
= 4x 2
Thus, ( x1, y1 ) lies on the circle x + (y - 3 ) = 1 and ( x 2, y 2 ) lies
on the parabola y 2 = 4 x.
Now, the shortest distance always occurs along the common
normal to the curves and normal to the circle passes through
the centre of the circle.
Normal to parabola y 2 = 4 x is y = mx - 2m - m 3. It passes
through centre of circle ( 0, 3 ).
Therefore, 3 = - 2m - m 3 Þm 3 + 2m + 3 = 0 which has only one
real value m = - 1.
…(ii)
md - 2am - am 3 + 2cm + cm 3 = 0
2
Thus, the given expression is the shortest distance between
the curve x 2 + (y - 3 ) 2 = 1 and y 2 = 4 x.
y = m ( x - d ) - 2cm - cm
…(i)
3
m¹0
\
Þ
d - 2a - am 2 + 2c + cm 2 = 0
(a - c )m 2 = d - 2a + 2c
d - 2a + 2c
(a - c )
Þ
m2 =
Þ
d
-2 > 0
a -c
Þ
Þ
d > 2a - 2c
2a < 2c + d
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
Ex. 11 The locus of the mid-point of the focal radii of a
variable point moving on the parabola y 2 = 4ax is a
l
parabola whose
(a) latusrectum is half the latusrectum of the original
parabola
æa ö
(b) vertex is ç , 0÷
è2 ø
(c) directrix is Y -axis
(d) focus has the coordinates (a, 0)
Sol. (a, b, c, d) Let P (at 2, 2 at ) be a point on the parabola y 2 = 4ax
with focus S (a, 0 )
a
Now, mid-point of focal radii SP is M æç (t 2 + 1 ), at ö÷.
è2
ø
a 2
Let
x = (t + 1 ) and y = at
2
ö
a æy 2
or
x = ç 2 + 1÷
2 èa
ø
a
or y 2 = 2a æç x - ö÷
Þ
y 2 = 2ax - a 2
è
2ø
a
which is a parabola with vertex æç , 0 ö÷ and latusrectum is 2a,
è2 ø
a a
a a
directrix is x - = i.e. x = 0 (Y ) and focus æç + , 0 ö÷ i.e. (a, 0 ).
è2 2 ø
2 2
Chap 05 Parabola
Also, line (i) touches the circle x 2 + (y - 2 ) 2 = 4, then
Ex. 12 If P1P2 and Q 1Q 2 , two focal chords of a parabola
2
y = 4ax at right angles, then
l
| 2 + am 2 |
(1 + m 2 )
(a) area of the quadrilateral P1Q 1P2Q 2 is minimum when
the chords are inclined at an angle p / 4 to the axis of
the parabola.
(b) minimum area is twice the area of the square on the
latusrectum of the parabola.
(c) minimum area of quadrilateral P1Q 1P2Q 2 cannot be
found.
(d) minimum area is thrice the area of the square on the
latusrectum of the parabola.
Sol. (a, b) Let coordinates of P1 are (at 2, 2 at ), then coordinates of P2
a 2a
are æç 2 , - ö÷.
èt
t ø
Y
P1
P
Q1
X′
α
A
\
m2 =
4 - 4a
and m 2 = 0
a2
for y = x 2, x ³ 0, then m 2 = 48
4a = 1
and put 4a = - 1 for y = - x 2, x < 0, then m 2 = 80
\Common tangents are
y = 0, y = 4 3 x - 12 and y = 4 5 x + 20
Ex. 14 Let V be the vertex and L be the latusrectum of
the parabola x 2 = 2y + 4 x - 4. Then, the equation of the
parabola whose vertex is at V , latusrectum L / 2 and axis is
perpendicular to the axis of the given parabola.
l
(a) y 2 = x - 2
(b) y 2 = x - 4
(c) y 2 = 2 - x
(d) y 2 = 4 - x
Let focal chord P1P2 makes an angle a with X -axis, then
æ 2a ö
2at - ç - ÷
è t ø
2
tan a =
=
1
a
tat 2 - 2
t
t
1
t - = 2 cot a
Þ
t
2
ü
ìæ 1ö 2
æ 1ö
Now,
P1P2 = a çt + ÷ = aí çt - ÷ + 4ý
è tø
è tø
þ
î
Vertex of the parabola is (2, 0) and length of latusrectum = 2
\
V (2, 0 ) and L = 2
Y
X′
… (i)
Þ
(Q cosec 2 a = 1)
Ex. 13 The equation of the line that touches the curves
y = x | x | and x 2 + (y - 2 ) 2 = 4, where x ¹ 0 , is
l
(d) y = - 4 5x - 20
y 2 = x - 2 or y 2 = - x + 2
Ex. 15 Consider a circle with its centre lying on the
focus of the parabola y 2 = 2ax such that it touches the
directrix of the parabola. Then a point of intersection of the
circle and the parabola is
l
\Minimum area = 32a 2 = 2 (latusrectum) 2 and is inclined at
(c) y = 0
X
Length of latusrectum of required parabola = L / 2 = 1
Q1Q2 = 4a cosec 2(90 - a ) = 4a sec 2a
1
\Area of quadrilateral P1Q1P2Q2 = ( P1P2 ) (Q1Q2 )
2
2
2
2
= 8a sec a cosec a = 32a 2 cosec 2a
(b) y = 4 3 x - 12
(2,0)
\Equation of the required parabola is (y - 0 ) 2 = ± 1( x - 2 )
Similarly,
(a) y = 4 5x + 20
axis of
parabola
Y′
= a { 4 cot 2 a + 4 } = 4a cosec 2a [from Eq. (i)]
a=p /4
( x - 2 ) 2 = 2y
Þ
Q2
æa ö
(a) ç , a ÷
è2 ø
ö
æa
(b) ç , - a ÷
ø
è2
æ a ö
(c) ç - , a ÷
è 2 ø
ö
æ a
(d) ç - , - a ÷
ø
è 2
Sol. (a, b) Given parabola is y 2 = 2ax
…(i)
a
a
\ Focus and equation of directrix are S æç , 0 ö÷ and x = è
ø
2
2
respectively.
ì x 2, x ³ 0
Sol. (a, b, c)Qy = x | x | = í 2
î- x , x < 0
2
a
æ
\Equation of circle is ç x - ö÷ + (y - 0 ) 2 = a 2
è
2ø
Q Equation of tangent in terms of slope (m ) of x 2 = 4ay is
y = mx - am 2
4 + a 2m 4 + 4am 2 = 4 + 4m 2
Sol. (a, c) Given parabola is x 2 = 2y + 4 x - 4
P2
Y′
=2
Þ
Put
X
S
411
… (i)
… (ii)
(Q radius = distance between focus and directrix)
412
Textbook of Coordinate Geometry
From Eq. (i),
From Eqs. (i) and (ii),
y 2 =a2
2
aö
3a 2
æ
=0
ç x - ÷ + 2ax = a 2 Þ x 2 + ax è
4
2ø
or
a
3a
x = ,x ¹2
2
3a
(Q for x = - , y = imaginary)
2
\
y =±a
a
Hence, required point of intersection are æç , ± a ö÷.
è2
ø
JEE Type Solved Examples :
Paragraph Based Questions
n
This section contains 2 Solved Paragraphs based upon
each of the paragraph 3 multiple choice question. Each
of these questions has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
Therefore,
4 + 2 l = 0 or l = - 2
Thus, the required circle is
( x - 1 ) 2 + (y - 2 ) 2 - 2( x - y + 1 ) = 0
or
x 2 + y 2 - 4 x - 2y + 3 = 0
Paragraph I
Y
(Q. Nos. 16 to 18)
Tangents are drawn to the parabola y 2 = 4 x at the point P
which is the upper end of latusrectum.
16. Image of the parabola y 2 = 4 x in the tangent line at
the point P is
(a) ( x + 4 )2 = 16y
(b) ( x + 2)2 = 8 (y - 2)
(c) ( x + 1)2 = 4 (y - 1)
(d) ( x - 2)2 = 2(y - 2)
17. Radius of the circle touching the parabola y 2 = 4 x at
the point P and passing through its focus is
(a) 1
(b) 2
(c) 3
(1,2) P
2
sq units
3
14
sq units
(c)
3
O
A
18. (a) Area bounded by AOPA = Area of DAOB + Area of OPBO
Y
(d) 2
4
sq units
3
16
(d)
sq units
3
X
S (1, 0)
Its radius is ( 4 + 1 - 3 ) = 2.
18. Area enclosed by the tangent line at P, X-axis and the
parabola is
(a)
0
1=
y+
x–
M
P(1, 2)
B
(b)
A
X
O
Sol. Upper end of latusrectum is P (1, 2 )
\The equation of tangent at P (1, 2 ) is
y × 2 = 2 (x + 1) Þ x - y + 1 = 0
16. (c) Any point on the given parabola is (t 2, 2t ). The image of
(h, k ) of the point (t 2, 2t ) on x - y + 1 = 0 is given by
Þ
or
h - t 2 k - 2 t - 2(t 2 - 2 t + 1 )
=
=
-1
1
1+1
h = 2 t - 1 and k = t 2 + 1
2
æh + 1ö
k =ç
÷ + 1 Þ (h + 1 ) 2 = 4(k - 1 )
è 2 ø
The required equation of image is ( x + 1 ) 2= 4(y - 1 ).
17. (c)Q Focus is S(1, 0) and P is (1, 2).
Equation of circle touching the parabola at (1, 2) is
( x - 1 ) 2 + (y - 2 ) 2 + l( x - y + 1 ) = 0 it passes through (1, 0 ).
2
2y 2
éy 3 ù
8 2
1
1
= ´1 ´1 + ò
dy - ´ 1 ´ 1 = ê ú = = sq unit
0 4
12
12
3
2
2
ë û0
Paragraph II
(Q. Nos. 19 to 21)
Let C 1 and C 2 be respectively, the parabolas x 2 = y - 1 and
y 2 = x - 1. Let P be any point on C 1 and Q be any point on
C 2 . Let P1 and Q 1 be the reflections of P and Q , respectively
with respect to the line y = x .
19. P1 and Q 1 lie on
(a) C1 and C 2 respectively (b) C 2 and C1 respectively
(c) Cannot be determined (d) None of these
Chap 05 Parabola
\ P1 and Q1 are ( l2 + 1, l)
20. If the point P( l, l2 + 1) and Q(m 2 + 1, m), then P1 and
Q 1 are
2
2
2
(a) ( l + 1, l) and (m + 1, m) (b) ( l + 1, l ) and (m, m + 1)
(c) ( l, l2 + 1) and (m, m 2 + 1) (d) ( l, l2 +1) and (m 2 + 1, m)
(a) PQ
1
(b) PQ
2
(c) 2PQ
(d)
3
PQ
2
Sol. Since, the reflection of a point ( p, q ) with respect to line y = x is
(q, p ).
Let P( l, l2 + 1 ) and Q(m 2 + 1, m) be points on C1 and C 2,
respectively.
\ Reflection of P( l, l2 + 1 ) with respect to line y = x is
P1( l2 + 1, l) and reflection of Q(m 2 + 1, m) with respect to line
y = x is Q1(m, m 2 + 1 ).
x2 =y –1 Y
C1
Q1
y=x
P
Also, P1 and Q1 lie on y = x - 1
Q
P1
O
x 2 = y - 1.
and
(Ans. 19(b))
Hence, P1 and Q1 lie on C 2 and C1 , respectively.
21. (a)Q A is mid-point of PP1 and B is mid-point of QQ1.
1
PA = PP1
\
2
1
and
...(i)
QB = QQ1
2
...(ii)
Þ
PC ³ PA
and
...(iii)
QC ³ QB
On adding Eqs. (ii) and (iii), then
PC + QC ³ PA + QB
1
1
[from Eq. (i)]
= PP1 + QQ1
2
2
æ PP + QQ1 ö
=ç 1
÷
è
ø
2
\
C
A
(Ans. 20(b))
2
B
(0, 1)
(m, m 2 + 1 )
and
2
21. Arithmetic mean of PP1 and QQ 1 is always
less than
413
C2
Þ
æ PP + QQ1 ö
PC + QC ³ ç 1
÷
è
ø
2
PQ ³ (AM of PP1 and QQ1 )
X
(1, 0)
y2=x–1
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
at 2t 3 a(t 2 + t 3 ) 1
t t t +t 1
1
a2 2 3 2 3
Area of DPQR = | at 3t1 a(t 3 + t1 ) 1 | = | t 3t1 t 3 + t1 1 |
2 tt t +t 1
2 at t a(t + t ) 1
12
1
2
12
1
2
This section contains 2 solved examples. The answer to
each example is a single digit integer, ranging from 0 to
9 (both inclusive).
Ex. 22 Points A, B, C lie on the parabola y 2 = 4ax . The
tangents to the parabola at A, B, C taken in pairs intersect at
points P , Q , R, then, the ratio of the areas of the DABC and
DPQR is
=
l
and C respectively.
= a 2 | (t1 - t 2 ) (t 2 - t 3 ) (t 3 - t1 ) |
Coordinates of P , Q and R are (at 2t 3, a(t 2 + t 3 )), (at 3t1, a(t 3 + t1 )),
(at1t 2, a(t1 + t 2 )) respectively, then
a2
| (t1 - t 2 ) (t 2 - t 3 ) (t 3 - t1 ) |
2
1
= ´ Area of DABC
2
Area of DABC
=2
Area of DPQR
=
Sol. (2) Let (at12, 2at1 ), (at 22, 2at 2 ) and (at 32, 2at 3 ) be the points A, B
at12 2at1 1
1 t1 t12
1
2
2
\Area of DABC =
at 2 2at 2 1 | = a | 1 t 2 t 22
2
1 t 3 t 32
at 32 2at 3 1
a2
| St 2t 3(t 3 - t 2 ) |
2
\
Ex. 23 If the orthocentre of the triangle formed by the
points t 1 , t 2 , t 3 on the parabola y 2 = 4ax is the focus, the
value of | t 1t 2 + t 2 t 3 + t 3 t 1 | is
l
414
Textbook of Coordinate Geometry
A(at12, 2at1)
Sol. (5) Q SA is perpendicular to BC[S is focus (a, 0)]
F
Þ
æ 2at1 - 0 ö æ 2at 3 - 2at 2 ö
÷ = -1
֍
ç 2
è at1 - a ø è at 32 - at 22 ø
Þ
æ 2t1 ö æ 2 ö
ç 2
֍
÷ = -1
è t1 - 1 ø è t 3 + t 2 ø
Þ
4t1 + t12(t 2 + t 3 ) = t 2 + t 3
…(i)
Similarly,
4t 2 + t 22(t 3 + t1 ) = t 3 + t1
…(ii)
E
S
(a,0)
B
(at22, 2at2)
C
(at32, 2at3)
D
On subtracting Eq. (ii) from Eq. (i), then
4 + (t1t 2 + t 2t 3 + t 3t1 ) = - 1 Þ t1t 2 + t 2t 3 + t 3t1 = - 5
\
| t1t 2 + t 2t 3 + t 3t1 | = 5
JEE Type Solved Examples :
Matching Type Questions
n
l
This section contains one solved example. Example 24
has three statements (A, B and C) given in Column I and
five statements (p, q, r, s and t) in Column II. Any given
statement in Column I can have correct matching with
one or more statement(s) given in Column II.
(B) Equation of tangent to parabola y 2 = 4 x at P (t 2, 2 t ) is
ty = x + t 2
t 2 Î[1, 4 ]
Given
1
\Area of DAPN = ( AN ) ( PN )
2
Y
Ex. 24 Match the following.
Column I
Column II
(A) If PQ is any focal chord of the parabola y2 = 32x
and length of PQ can never be less than l units,
then l is divisible by
(p)
2
X′
(B) A tangent is drawn to the parabola y2 = 4 x at the
(q)
point ‘P’ whose abscissa lies in the interval [1, 4 ] .
If maximum possible area of the triangle formed
by the tangent at ‘P’, ordinate of the point ‘P’ and
the X-axis is l sq units, then l is divisible by
3
(C) The normal at the ends of the latusrectum of the
(r)
parabola y2 = 4 x meet the parabola again at A and
A¢. If length AA¢ = l unit, then l is divisible by
4
(s)
(t)
or
\
8
Þ
D = 2t 3 = 2(t 2 ) 2
Q
t 2 Î[1, 4 ]
Þ
D max occurs, when t 2 = 4
Þ
D max = 2( 4 ) 2 = 16 sq units
3
\
2
1
æ 1ö
³ 2 Þ a çt + ÷ ³ 4a
è tø
t
PQ ³ 32
l = 32
Y′
6
2
t+
X
3
(A) Let P (at 2, 2at ) be any point on the parabola y 2 = 4ax and
S (a, 0 ) be the focus, then the other end of focal chord through
æ a 2a
P will be Q ç 2 , - ö÷.
èt
t ø
Q
N
O
1
= (2 t 2 ) (2 t )
2
Sol. (A) ® (p, r, t); (B) ® (p, r, t); (C) ® (p, q, r, s)
1
Then, length of focal chord PQ = a æçt + ö÷
è tø
A (–t2,0)
)
(t2 , 2t
P
[Q 4a = 32]
l = 16
(C) Given parabola y 2 = 4 x. Now, the ends of latusrectum are
P(1, 2 ) and Q(1, - 2 ) or P (t 2, 2t ) and Q (t12, 2t1 ), where t = 1, t1 = - 1.
2
We know that the other end of normal is given by t 2 = - t t
Þ
A(t 22, 2t 2 ) and A ¢(t 32, 2t 3 ),
where
or
\
\
t 2 = - 3, t 3 = 3
A(9, - 6 ) and A¢(9, 6 )
AA¢ = 12 units
l = 12
JEE Type Solved Examples :
Statement I and II Type Questions
n
On differentiating w.r.t. t, we get
Directions (Ex. Nos. 25 and 26) are Assertion-Reason
type questions. Each of these question contains two
statements :
Statement I (Assertion) and
Statement II (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer.
You have to select the correct choice
3t 2 + (2 - l ) > 0, for l < 2
The cubic Eq. (i) has only one real root.
\ Statement I is true.
Hence, both statements are true but Statement II is not correct
explanation for Statement I.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for statement I.
(b) Statement I is true, Statement II is true; Statement II is not a
correct explanation for Statement I.
(c) Statement I is true, Statement II is false.
(d) Statement I is false, Statement II is true.
Ex. 26 Statement I If there exist points on the circle
x + y 2 = l2 from which two perpendicular tangents can be
1
drawn to the parabola y 2 = 2 x , then l ³ .
2
Statement II Perpendicular tangents to the parabola meet
at the directrix.
Ex. 25 Statement I Through the point ( l, l +1), l < 2,
there cannot be more than one normal to the parabola y 2 = 4 x .
Statement II The point ( l, l +1) cannot lie inside the
parabola y 2 = 4 x .
Sol. (a) Statement II is true as it is property of parabola. Equation
1
of directrix of parabola y 2 = 2 x is x = - .
2
1 ö
æ
Any point on directrix is ç - , y ÷ , now this point exists on the
è 2 ø
circle, then
1
+ y 2 = l2
4
1
Þ
y 2 = l2 - ³ 0
4
1
l³
\
2
Hence, both statements are true and Statement II is correct
explanation for Statement I.
l
Sol. (b) Let S = y 2 - 4 x
\
S1 = ( l + 1 ) 2 - 4 l = ( l - 1 ) 2 ³ 0
\ Point ( l, l + 1 ) cannot lie inside the parabola y 2 = 4 x.
\ Statement II is true.
Now, equation of normal at (t 2, 2 t ) is y + tx = 2t + t 3 passes
through ( l, l + 1 ).
l + 1 + tl = 2t + t 3
Þ
Þ
t 3 + (2 - l )t - ( l + 1 ) = 0
…(i)
l
2
Subjective Type Examples
n
On comparing the coefficients of x and y in Eq. (i) and (ii),
2a k
we get
=
h 2a
In this section, there are 16 subjective solved
examples.
Ex. 27 The two parabolas y 2 = 4a ( x - l ) and
x 2 = 4a (y - l1 ) always touch one another, l and l1 being
variable parameters. Prove that the point of contact lies on
l
Þ
hk = 4a 2
Hence, the locus is xy = 4a 2,
which is independent of l and l1.
2
the curve xy = 4a .
Sol. Let (h, k ) be the common point and if the parabolas touch
each other, then the tangents at (h, k ) should be identical.
Their equations are
ky = 2a( x - l + h )
…(i)
Þ
2ax - ky = 2a(l - h )
and
hx = 2a(y - l1 + k )
…(ii)
Þ
hx - 2ay = 2a(k - l1 )
Ex. 28 Show that the area formed by the normals to
y = 4ax at the points t 1 , t 2 , t 3 is
1 2
a |(t 1 - t 2 ) (t 2 - t 3 )(t 3 - t 1 )| l2 , where l = (t 1 + t 2 + t 3 ).
2
l
2
Sol. Equation of normals at P , Q, R are
y = - t1x + 2at1 + at13
416
Textbook of Coordinate Geometry
y = - t 2x + 2at 2 + at 23, y = - t 3x + 2at 3 + at 33, respectively.
2
(at1 , 2at1)
P
L
N
Sol. Given parabolas y 2 = 4ax and y 2 = 4c( x - b ) have common
normals. Then equation of normals in terms of slopes are
M
Q
2
(at2 , 2at2)
Ex. 29 Prove that the two parabolas y 2 = 4ax and
y 2 = 4c ( x - b ) cannot have common normal, other than the
axis unless b / (a - c ) > 2.
l
y = mx - 2am - am 3
R (at 2,2 at )
3
3
y = m( x - b ) - 2cm - cm 3
and
Subtracting them two by two, coordinates of L, M and N are
L º {2a + a(t12 + t1t 2 + t 22 ), - at1t 2(t1 + t 2 )}
respectively then normals must be identical, compare the
coefficients
M º {2a + a(t12 + t1t 3 + t 32 ), - at1t 3(t1 + t 3 )}
1=
N º {2a + a(t 22 + t 2t 3 + t 32 ), - at 2t 3(t 2 + t 3 )}
m [(c - a )m 2 + (b + 2c - 2a )] = 0,
Þ
\ Area of DLMN
½2a + a(t12 + t1t 2 + t 22 ) -at1t 2(t1 + t 2 ) 1½
1½
= | 2a + a(t12 + t1t 3 + t 32 ) -at1t 3(t1 + t 3 ) 1½|
2½
2
2
-at 2t 3(t 2 + t 3 ) 1½
½2a + a(t 2 + t 2t 3 + t 3 )
½
2am + am 3
mb + 2cm + cm 3
[Qother than axis]
m¹0
m2 =
and
2a - 2c - b
,
c -a
On applying R2 ® R2 - R1 and R3 ® R3 - R1
m=±
½ 2 + (t12 + t1t 2 + t 22 )
1½
-t1t 2(t1 + t 2 )
a
|
(t 3 - t 2 )(t1 + t 2 + t 3 ) t1(t 3 - t 2 )(t1 + t 2 + t 3 ) 0½
= |½
½
2 (t - t )(t + t + t ) t (t - t )(t + t + t ) 0½
½ 3 1 1 2 3 2 3 1 1 2 3
½
Expanding by last column
a2
1 t
= |(t 3 - t1 )(t 3 - t 2 )|(t1 + t 2 + t 3 ) 2 |½ 1½|
2
½1 t 2½
1 2
= a |(t1 - t 2 )(t 2 - t 3 )(t 3 - t1 )|(t1 + t 2 + t 3 ) 2
2
1
= a 2 |(t1 - t 2 )(t 2 - t 3 )(t 3 - t1 )| l2
2
2
æ
b ö
m = ± ç-2 ÷
c -a ø
è
Þ
\
-2 -
b
>0
c -a
Þ
-2 +
b
>0
a -c
b
>2
a -c
Þ
Aliter
Equations of sides of DLMN formed by the normals are
y + t1x - (2at1 + at13 ) = 0
y + t 2x - (2at 2 + at 23 ) = 0
y + t 3x - (2at 3 + at 33 ) = 0
2(a - c ) - b
c -a
Ex. 30 If on a given base BC, a triangle is described such
that the sum of the tangents of the base angles is m, then
prove that the locus of the opposite vertex A is a parabola.
l
Sol. Let the given point B and C be (0, 0) and (a, 0 ), respectively.
2
½t1 1 -(2at1 + at13 ) ½
1
½t 1 -(2at + at 3 )½
\ Required area =
2
2
2
½
2 | C1C 2C 3 | ½t 1
-(2at 3 + at 33 )½
½3
Given, tan a + tan b = m (constant)
Y
A(x,y)
2
½t1 1 2 t1 + t13 ½
a2
½t 1 2 t + t 3½
=
2
2
2
| 2(t 2 - t 3 )(t 3 - t1 )(t1 - t 2 )| ½t 1
3½
t
t
2
+
3
3
3½
½
ì t 1 2 t ½t 1 t13½ü
1½
1
a2
ï½ 1
3 ï
=
í½t 2 1 2 t 2½+½t 2 1 t 2 ½ý
2|(t 2 - t 3 )(t 3 - t1 )(t1 - t 2 )| ï t 1 2 t ½t 1 3½ï
½
t3 ½
3½
½3
þ
î 3
a2
=
2|(t 2 - t 3 )(t 3 - t1 )(t1 - t 2 )|
2
{ 0 - (t1 - t 2 )(t 2 - t 3 )(t 3 - t1 )(t1 + t 2 + t 3 )} 2
1
= a 2 |(t1 - t 2 )(t 2 - t 3 )(t 3 - t1 )|(t1 + t 2 + t 3 ) 2
2
1 2
= a |(t1 - t 2 )(t 2 - t 3 )(t 3 - t1 )| l2.
2
X′
α
B (0,0)
Y′
Þ
y
y
+
=m
x a -x
Þ
ay
=m
x(a - x )
\
β
D
ay = amx - mx 2
which is the equation of parabola.
X
C (a,0)
Chap 05 Parabola
Ex. 31 A parabolic mirror is kept along y 2 = 4 x and two
light rays parallel to its axis are reflected along one straight
line. If one of the incident rays is at 3 units distance from the
axis, then find the distance of the other incident ray from the
axis.
l
Ex. 33 A parabola of latusrectum 4a , touches a fixed
equal parabola , the axes of the two curves being parallel;
prove that the locus of the vertex of the moving curve is a
parabola of latusrectum 8a.
l
Sol. Let the given parabola is
y 2 = 4ax
Sol. Let the incident rays be PA and QB.
After reflection, both rays passes through the focus S(1, 0 ).
Y
A
417
…(i)
If the vertex of moving parabola is ( a, b ), then equation of
moving parabola is
P
(y - b ) 2 = - 4a ( x - a )
…(ii)
2
X′
O
On substituting the value of i.e. x =
X
S(1,0)
y
in Eq. (ii),
4a
Q
Y′
(0,0)
Therefore, AB is a focal chord.
1 -2ö
Let A be (t 2, 2t ), then B be æç 2 ,
÷.
èt t ø
Given, 2t = 3 \ t = 3 / 2
(α, β)
Hence, distance of B from the axis of parabola is -
2 4
= units.
t
3
æy 2
ö
(y - b ) 2 = -4a ç - a ÷
è 4a
ø
then
Ex. 32 Prove that the length of the intercept on the
normal at the point P (at 2 , 2at ) of a parabola y 2 = 4ax made
by the circle on the line joining the focus and point P as
l
y 2 - 2 by + b 2 = - y 2 + 4aa
Þ
2y 2 - 2 by + b 2 - 4aa = 0
diameter is a (1 + t 2 ) .
Since, two parabolas (i) and (ii) touch each other.
Hence, roots of Eq. (iii) are equal i.e. Discriminant = 0
Sol. Let the normal at P (at 2, 2at ) cut the circle in K and the axis
of parabola at G then PK is required intercept.
Þ
Y
M
T
φ
P (a
θ
A
2,
t
)
2at
φ θ
90° K
S (a,0)
G
“B 2 - 4 AC = 0”
( -2 b ) 2 = 4 × 2 × (b 2 - 4aa )
\
SP = PM = a + at 2
…(iii)
Þ
4b 2 = 32aa
or
b 2 = 8aa
the required locus is
y 2 = 8ax
which has latusrectum double that of given parabola.
X
Ex. 34 The normal at point P on a given parabola meet
the axis of parabola at Q. Then prove that a line through Q
and perpendicular to this normal always touches a fixed
parabola whose length of latusrectum is same as that of
given parabola.
l
Directrix (x+a=0)
Since angle in a semi-circle being right angle.
\
Ð SKP = 90°
Sol. Let the equation of parabola is
and normal at P (at 2, 2at ) is
y 2 = 4ax.
y = - tx + 2at + at 3
tx + y - 2at - at 3 = 0
Þ
…(i)
\ SK is the perpendicular distance from S (a, 0 ) to the normal (i),
then
SK =
| at + 0 - 2at - at 3|
t2 + 1
=a |t | 1 + t2
\ In DSPK , ( Pk ) 2 = (SP ) 2 - (SK ) 2
= a 2(1 + t 2 ) 2 - a 2t 2 (1 + t 2 ) = a 2(1 + t 2 )
\
PK = a (1 + t 2 )
…(i)
2
Let P (at , 2at )
Normal at P is
y + tx = 2at + at 3
Q Normal meet the axis of parabola (i.e. X-axis), then
Q(2a + at 2, 0 ).
Now, equation of the line through Q and perpendicular to the
normal is
1
y - 0 = ( x - 2a - at 2 )
t
418
Textbook of Coordinate Geometry
Þ
ty = x - 2a - at 2
Þ
ty = ( x - 2a ) - at 2
Points P and Q lie on y 2 = 4ax then
( KP ) 2 sin 2 q = 4a(d + KP cos q)
which is clearly tangent to the parabola
and
2
2
From Eq. (i),
KP =
From Eq. (ii),
KQ =
Sol. Let parabola be y 2 = 4ax and coordinates of P and Q on this
parabola are P º (at12, 2at1 ) and Q º (at 22, 2at 2 ) ; T is the point of
intersection of tangents at t1 and t 2.
P′
2 sin 2 q
R (t3)
Q ′ Q (t2)
\ Coordinates of T º {at1, t 2, a (t1 + t 2 )}
Similarly
P ¢ º {at 3t1, a (t 3 + t1 )}
Q ¢ º {at 2t 3, a (t 2 + t 3 )}
Let
TP ¢ : TP = l :1
t -t
TP ¢ t 3 - t 2
l= 3 2
Þ
=
\
t1 - t 2
TP t1 - t 2
\
-4a cos q + (16a 2 cos2 q + 16ad sin 2 q)
2 sin 2 q
1
1
2a cos2 q + d sin 2 q
+
=
2
2
( KP )
( KQ )
2ad 2
Ex. 37 If the distribution of weight is uniform, then the
rope of the suspended bridge takes the form of parabola. The
height of the supporting towers is 20 m, the distance between
these towers is 150 m and the height of the lowest point of
the rope from the road is 3 m. Find the equation of the
parabolic shape of the rope considering the floor of the
bridge as X-axis and the axis of the parabola as Y-axis. Find
the height of that tower which supports the rope and is at a
distance of 30 m from the centre of the road .
l
A
Sol. Here MZ is the road let V is lowest point of the rope given
VZ = 3 m.
S
TQ ¢ t1 - t 3
=
TQ t1 - t 2
Similarly,
2
For d = 2a
1
1
1
\
+
=
( KP ) 2 ( KQ ) 2 d 2
1
= 2 , which is independent of q.
4a
)
P (t 1
T
2
4a cos q + (16a cos q + 16ad sin q)
l
\
…(ii)
( KQ ) sin q = 4a(d - KQ cos q)
y 2 = 4a (2a - x ).
Ex. 35 TP and TQ are any two tangents to a parabola
and the tangent at a third point R cuts them in P ¢ and Q ¢,
TP ¢ TQ ¢
prove that
+
=1.
TP TQ
…(i)
2
150 m
P (75, 20)
TP ¢ TQ ¢
+
=1
TP TQ
20 m
Ex. 36 Prove that on the axis of any parabola there is a
certain point ‘K’ which has the property that, if a chord PQ
1
1
of parabola be drawn through it, then
is the
+
2
PK
QK 2
same for all positions of the chord .
l
Sol. Any line passing through K is
x -d y - 0
=
=r
cos q sin q
\ Coordinates of P and Q are
(d + KP cos q, KP sin q)
and
(d - KQ cos q, - KQ sin q).
3 V
M
Let MZ is X-axis and let MVS is Y-axis.
Taking X and Y -axes as shown in the given figure, the equation
of parabola is of the form
x 2 = 4ay
1
Coordinates of P is ìí (150 ), 20 - 3üý
î2
þ
Eq. (i), then
P
(75 ) 2 = 4a (17 )
θ
K (d, 0)
A
Q
Y′
Z
if V is origin
Y
X′
Road
X
or
From Eq. (i),
5625
17
2 5625
x =
y
17
4a =
i.e. (75 , 17 ) P lies on
Chap 05 Parabola
Now, shifting the origin (0, 0) to M ( 0, - 3 ),
then equation of parabola in new form is
5625
…(ii)
x2 =
(y - 3 )
17
Again, let the required height of supporting tower at a distance
of 30 m from the centre of road be h metres, then the coordinates
of the top of this tower are (30, h ) referred to given axes
5625
from Eq. (ii), (30 ) 2 =
(h - 3 )
17
900 ´ 17 ö
h = æç
\
÷ + 3 = 5.72 m
è 5625 ø
Ex. 39 A variable chord PQ of the parabola y = 4 x 2
subtends a right angle at the vertex. Find the locus of the
points of intersection of the normals at P and Q.
l
Sol. Parametric point on the parabola
y = 4 x 2 is (t, 4t 2 )
Let P (t1, 4t12 ) and Q (t 2, 4t 22 ),
if A is vertex of the parabola, then
Slope of AP ´ Slope of AQ = - 1
\
4t1 ´ 4t 2 = - 1
1
\
t1t 2 = 16
Equations of normals at P and Q are
Ex. 38 Tangent is drawn at any point ( x 1 , y 1 ) on the
parabola y 2 = 4ax . Now tangents are drawn from any point
on this tangent to the circle x 2 + y 2 = a 2 such that all the
chords of contact pass through a fixed point ( x 2 , y 2 ) . Prove
2
æ x1 ö æ y 1 ö
that 4 ç ÷ + ç ÷ = 0.
è x2 ø èy 2 ø
l
and
yy1 = 2a ( x + x1 )
Þ
æ 2a ( x ¢ + x1 ) ö
Any point on this tangent will be ç x ¢,
÷
y1
è
ø
Þ
(2ayx1 - a 2y1 ) + x ¢ ( xy1 + 2ay ) = 0
which is family of straight lines passing through point of
intersection of
2ayx1 - a 2y1 = 0 and xy1 + 2ay = 0
Þ
ay1
2 x1
y=
and
x =-
a
x1
and y 2 =
2
2
4 x12 4 x12
+ 2 =0
a2
a
æ x ö æy ö
Hence, 4 ç 1 ÷ + ç 1 ÷ = 0.
è x2 ø è y 2 ø
k=
1 + 16 (t12 + t 22 ) + 16t1t 2
4
2k = 8h 2 + 1
Sol. Let coordinates of P and Q are (at12, 2at1 ) and (at 22, 2at 2 ),
respectively.
ay1
2 x1
æ x
ö æ y1 ö
æ x ö æy ö
\ 4ç 1÷ + ç 1÷ = 4ç 21 ÷ + ç
÷
è x2 ø è y 2 ø
è -a / x1 ø è ay1 / 2 x1 ø
=-
1 + 16 (t12 + t 22 + t1t 2 )
4
Ex. 40 Equilateral triangles are circumscribed to the
parabola y 2 = 4ax . Prove that their angular points lie on the
conic (3 x + a ) ( x + 3a ) = y 2 .
a2
x1
2
x2 = -
k=
l
The fixed point is ( x 2, y 2 ).
\
…(ii)
x + 4t 2y
Hence, locus of (h, k ) is 2y = 8 x 2 + 1.
2ay æ 2a ö æ ay1 ö
= ç- ÷ ç
÷
y1 è y1 ø è 2 x1 ø
x =-
…(i)
= t 2 + 16t 23
1 + 16 ( (t1 + t 2 ) 2 - 2t1t 2 ) - 1
[Q16t1t 2 = - 1 ]
4
1ü
ì
= 4 í(t1 + t 2 ) 2 + ý
8þ
î
1
…(iii)
k = 4 (t1 + t 2 ) 2 +
2
Again
h = - 16t1t 2(t1 + t 2 )
…(iv)
Þ
h = t1 + t 2
On eliminating t1 and t 2 from Eqs. (iii) and (iv), we get
1
k = 4h 2 +
2
xx ¢y1 + 2ayx ¢ + 2ayx1 = a 2y1
\
x + 4t1y = t1 + 16t13
=
æ 2a ( x ¢ + x1 ) ö
Equation of chord of contact of the point ç x ¢,
÷ w.r.t.
y1
ø
è
a
x
x
2
¢
+
(
)
1
circle x 2 + y 2 = a 2 is xx ¢ + y ×
=a2
y1
Þ
[QAP^ AQ ]
Let the normals (i) and (ii) intersect at (h, k ), then
on solving Eqs. (i) and (ii), then we get
Sol. Tangent at ( x1, y1 ) on the parabola y 2 = 4ax
\
419
2
P
A
60°
Q
420
Textbook of Coordinate Geometry
Tangents at P and Q are t1y = x + at12 and t 2y = x + at 22,
respectively.
1
1
Slope of tangents at P and Q are and , respectively.
t1
t2
This is taken as directrix, then
SP = PM
Let one angular point of equilateral triangle is A.
\
A º (at1t 2, a(t1 + t 2 ))
1 1
t -t
t1 t 2
Given, tan60° =
= 2 1
1 1
1
+ t1t 2
1+ ×
t1 t 2
Þ
2
2
( 3 ) (1 + t1t 2 ) = (t 2 - t1 )
\
2
AS = AN
(h + a ) 2 + (k - 0 ) 2 = (a cos q + b ) 2
…(v)
a 2 cos2 q + b 2 = h 2 + k 2 + a 2
…(vi)
On subtracting Eq. (iv) from Eq. (v), then
h
4ab cos q = 4ah or cos q =
b
From Eqs. (vi) and (vii), we get
…(vii)
a 2h 2
+ b 2 = h2 + k 2 + a 2
b2
Þ
y 2 = 3 x 2 + 10ax + 3a 2
(3 x + a ) ( x + 3a ) = y
and
…(iv)
On adding Eqs. (iv) and (v), then
For locus of A, put at1t 2 = x and a(t1 + t 2 ) = y , then
2
2
x
y
x
3 æç1 + ö÷ = æç ö÷ - 4
è
a ø èa ø
a
Þ
BS = BM Þ(h - a ) 2 + (k - 0 ) 2 = (a cos q - b ) 2
Þ
3 (1 + t1t 2 ) 2 = (t1 + t 2 ) 2 - 4t1t 2
Þ
Þ
Þ
2
h2
(b 2 - a 2 )
+ k2 =b2 -a2
a2
h2
k2
+ 2
=1
2
a
b -a2
Hence, the locus of S (h, k ) is
Ex. 41 A parabola is drawn to pass through A and B, the
ends of diameter of a given circle of radius a and to have a
directrix a tangent to a concentric circle of radius b, the axes
being AB and the perpendicular diameter. Prove that the
y2
x2
locus of the focus of the parabola is
+
= 1.
b2 b2 -a2
l
Sol. Let AB be taken along X -axis, it mid-point O as origin and a
line through O perpendicular to AB as Y -axis.
Y
x2
y2
+
= 1.
a2 b2 -a2
Ex. 42 Two straight lines are at right angles to one
another and one of them touches y 2 = 4a ( x + a ) and the
other y 2 = 4b( x + b ) . Prove that the point of intersection of
the lines lies on the line x + a + b = 0.
l
Sol. We know that any tangent in terms of slope (m) of the
a
parabola y 2 = 4ax is y = mx + .
m
Replacing x by x + a, we get
a
which is tangent to
y = m( x + a ) +
m
y 2 = 4a( x + a )
X′
O
A
B
X
Similarly, tangent in terms of slope of y = 4b( x + b ) is
y = m1( x + b ) +
M
N
s
Y′
Equation of circle on AB as diameter is
x2 + y 2 =a2
…(i)
2
…(i)
The equation of a circle of radius b and concentric to circle (i) is
…(ii)
x2 + y 2 =b2
Let S (h, k ) be the focus of the parabola which passes through
A( - a, 0 ) and B(a, 0 ).
Let equation of tangent to circle (ii) be
…(iii)
x cos q + y sin q = 6.
b
m1
Given tangents are perpendicular, we have
1
mm1 = - 1 or m1 = m
(x + b )
then Eq. (ii) becomes y = - bm
m
On subtracting Eqs. (i) and (ii), then
1ö
æ
0 = (x + a + b ) çm + ÷
è
mø
1
m+ ¹0
Þ
m
Hence,
x +a +b =0
…(ii)
#L
Parabola Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. A common tangent is drawn to the circle
x 2 + y 2 = a 2 and the parabola y 2 = 4bx . If the angle
p
which this tangent makes with the axis of x is , then
the relationship between a and b is (a , b > 0 ) 4
(a) b = 2a (b) a = b 2 (c) c = 2a
(d) a = 2c
2. The equation of parabola whose vertex and focus lie
on the axis of x at distances a and a 1 from the origin
respectively, is
(a) y 2 = 4 (a1 - a ) x
(b) y 2 = 4 (a1 - a ) ( x - a )
2
(c) y = 4 (a1 - a ) ( x - a1 ) (d) y 2 = 4aa1 x
3. If parabolas y 2 = ax and
25 [( x - 3 ) 2 + (y + 2 ) 2 ] = (3 x - 4y - 2 ) 2 are equal,
then the value of a is
(a) 3
(b) 6
(c) 7
(d) 9
8. If a ¹ 0 and the line 2bx + 3cy + 4d = 0 passes through
the points of intersection of the parabolas y 2 = 4ax
and x 2 = 4ay , then
(a) d 2 + ( 2b + 3c )2 = 0
2
(c) d + ( 2b - 3c ) = 0
Y
(a)
bc
a
A
D
(a)
3 +1
4
(b)
3 +1
2
(b) cot 2 q
X
G
1
4a
(c)
5 +1
4
(c) tan 3 q
(b) (2, 2)
(c) (3, 3)
(d)
5 +1
2
(d) cot 3 q
1 1
(d) æç , ö÷
è 2 2ø
7. The circle x 2 + y 2 + 2 px = 0, p Î R, touches the
parabola y 2 = 4 x externally, then
(a) p > 0
(b) p < 0
(c)p > 1
b
a
(d)
c
a
(b)
1
a
(c)
2
a
(d)
4
a
11. If the normal to the parabola y 2 = 4ax at P meets the
curve again at Q and if PQ and the normal at Q make
angles a and b respectively with the X-axis, then
tan a (tan a + tan b) has the value equal to
(a) (SA ) 2
6. The vertex of the parabola whose parametric
equation is x = t 2 - t + 1, y = t 2 + t + 1, t Î R, is
(a) (1, 1)
(c)
(b) - 1
(c) -
1
2
(d) 0
12. If the normals to the parabola y 2 = 4ax at three
points P , Q and R meet at A and S is the focus, then
SP × SQ × SR is equal to
5. Let A and B be two points on a parabola y 2 = x with
vertex V such that VA is perpendicular to VB and q is
the angle between the chordVA and the axis of the
| VA |
is
parabola. The value of
| VB |
(a) tan q
(b) ac 2
F
B
C
(d) d 2 + ( 2b + 3c )2 = a 2
10. Two mutually perpendicular tangent of the parabola
y 2 = 4ax meet the axis at P1 and P2 . It S is the focus of
1
1
is equal to
the parabola, then
+
SP1 SP2
(a) - 2
E
(b) d 2 + (3b + 2c )2 = a 2
9. A parabola y = ax 2 + bx + c crosses the X-axis at (a, 0 )
and (b, 0 ) both to the right of the origin. A circle also
passes through these two points. The length of a
tangent from the origin to the circle is
(a)
4. ABCD and EFGC are squares and the curve y = l x
passes through the origin D and the points B and F.
FG
The ratio
is
BC
2
(d) p > 2
(b) (SA ) 3
(c) a (SA ) 2
(d) a (SA ) 3
13. The length of the shortest normal chord of the
parabola y 2 = 4ax is
(a) 2a 27
(b) 9a
(c) a 54
(d) 18a
14. The largest value of a for which the circle x 2 + y 2 = a 2
falls totally in the interior of the parabola
y 2 = 4 ( x + 4 ) is
(a) 4 3
(b) 4
(c)
4 6
7
(d) 2 3
15. From a point (sin q, cos q ), if three normals can be
drawn to the parabola y 2 = 4ax , then the value of a is
1
(a) æç , 1ö÷
è2 ø
1
(c) é , 1ù
êë 2 úû
1
(b) é - , 0ö÷
êë 2 ø
1
-1
(d) æç , 0ö÷ È æç0, ö÷
è 2 ø è 2ø
422
Textbook of Coordinate Geometry
16. If two different tangents of y 2 = 4 x are the normals
to x 2 = 4 by, then
(a) | b | <
(c) | b | >
1
(b) | b | <
2 2
1
(d) | b | >
2 2
1
2
1
2
17. The shortest distance between the parabolas
2y 2 = 2 x - 1 and 2 x 2 = 2y - 1 is
(a)
1
2 2
(b)
1
2
(d) 4
(c) 2 2
18. Normals at two points ( x 1 , y 1 ) and ( x 2 , y 2 ) of the
parabola y 2 = 4 x meet again on the parabola, where
x 1 + x 2 = 4, then | y 1 + y 2 | is equal to
(a) 2
(b) 2 2
(c)4 2
(d) 8 2
19. A line is drawn from A ( - 2, 0 ) to intersect the curve
y 2 = 4 x at P and Q in the first quadrant such that
1
1
1
+
< . Then the slope of the line is always
AP AQ 4
(a) <
1
3
(b) >
1
(c) > 2
3
(d) > 3
20. An equilateral triangle SAB is inscribed in the
parabola y 2 = 4ax having its focus at S. If chord AB
lies towards the left of S, then the side length of this
triangle is
(a) a ( 2 - 3 )
(b) 2a ( 2 - 3 )
(c) 4a ( 2 - 3 )
(d) 8a ( 2 - 3 )
21. Let C be a circle with centre (0, 1) and radius unity. P
is the parabola y = ax 2 . The set of values of a for
which they meet at a point other than origin is
(a) (0, ¥ )
1
(b) æç0, ö÷
è 2ø
1 1
(c) æç , ö÷
è 4 2ø
1
(d)æç , ¥ ö÷
è2 ø
2
22. Let S be the focus of y = 4 x and a point P be moving
on the curve such that its abscissa is increasing at the
rate of 4 units/s. Then the rate of increase of the
projection of SP on x + y =1 when p is at (4, 4) is
(a) - 2
(b) -
3
2
(c) - 1
(d) 2
23. If P is a point on the parabola y = 3 ( 2 x - 3 ) and M is
the foot of perpendicular drawn from P on the
directrix of the parabola, then the length of each side
of the equilateral triangle SMP, where S is the focus
of the parabola, is
(b) 4
(c) 6
1
(a) æç , 1ö÷
è4 ø
(b) (3, - 2 3 )
(c) (3, 2 3 )
(d) ( 4, 4 )
25. Through the vertex O of the parabola y 2 = 4ax , two
chords OP and OQ are drawn and the circles on OP
and OQ as diameters intersect at R. If q 1 , q 2 and f are
the angles made with the axis by the tangents at P
and Q on the parabola and by OR, then the value of
cot q 1 + cot q 2 is
(a) - 2 tan f
(b) 2 tan f
(c) 0
(d) 2 cot f
26. AB is a double ordinate of the parabola y 2 = 4ax .
Tangents drawn to the parabola at A and B meet the
Y -axis at A1 and B1 respectively. If the area of
trapezium AA1B1B is equal to 24a 2 , then the angle
subtended by A1B1 at the focus of the parabola is
equal to
(a) tan -1 2
(d) 8
(b) tan -1 3
(c) 2 tan -1 2 (d) 2 tan -1 3
n
1
27. If the 4th term in the expansion of æç px + ö÷ , n Î N
è
xø
5
2
is and three normals to the parabola y = x are
2
drawn through a point (q, 0 ), then
(a) q = p
(b) q > p
(c) q < p
(d) pq = 1
28. The set of points on the axis of the parabola
y 2 - 4 x - 2y + 5 = 0 from which all the three normals
to the parabola are real is
(a) (k , 0) ; k > 1
(c) (k , 2) ; k > 6
(b) (k , 1) ; k > 3
(d) (k , 3 ) ; k > 8
29. The triangle formed by the tangent to the parabola
y = x 2 at the point whose abscissa is x 0 ( x 0 Î[1, 2 ]),
the Y -axis and the straight line y = x 02 has the
greatest area if x 0 is equal to
(a) 0
2
(a) 2
24. Consider the parabola y 2 = 4 x . Let A º ( 4, - 4 ) and
B º ( 9, 6 ) be two fixed points on the parabola. Let C be
a moving point on the parabola between A and B
such that the area of the triangle ABC is maximum.
Then the coordinates of C are
(b) 1
(c) 2
(d) 3
30. The set of points ( x , y ) whose distance from the line
y = 2 x + 2 is the same as the distance from (2, 0) is a
parabola. This parabola is congruent to the parabola
in standard from y = kx 2 for some k which is equal to
(a)
4
5
(b)
12
5
(c)
5
4
(d)
5
12
Chap 05 Parabola
#L
423
Parabola Exercise 2 :
More than One Correct Option Type Questions
n
This section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which MORE TH