Carto Wong
Notes for HKCEE Mathematics
Coordinate Geometry of Circles
Distance Formula.
Given two points A( x1 , y1 ) , B( x2 , y2 ) in the rectangular coordinate plane. The distance
between A and B is defined to be
AB  ( x1  x2 ) 2  ( y1  y2 ) 2 .
P

The distance from a point P ( x, y ) to the origin is
x y .
2
2
2
x +y2
y
O
x
Section Formula.
Suppose the point P ( x, y ) divides the segment joining A( x1 , y1 ) , B( x2 , y2 ) internally in
the ratio AP : PB  h : k , then
k
h

 x  h  k x1  h  k x2

 y k y  h y
1
2
hk
hk

k
h
B (x 2 , y 2 )
P (x, y)
A (x 1 , y 1 )
x1  x2 y1  y2
,
).
2
2

In particular, the midpoint of AB is (

Sometimes it is more convenient to denote  
AP
and the formula written as
AB
 x  (1   ) x1   x2

 y  (1   ) y1   y2
Definition. (Slope)
Given two distinct points P( x1 , y1 ) , Q( x2 , y2 ) . The slope of the segment/ray/line PQ is
m
y2  y1
x2  x1
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When x1  x2 , the slope is  . Some authors said the slope is undefined in this case.

Line with slope 0 is called horizontal line. Line with slope  is called vertical line.

A straight line which makes a positive angle  with x-axis has slope m  tan .
y


L
x
If 0    90 , then tan  0 and the line has positive slope. If 90    180 , then
the line has negative slope.

Two lines L1 , L2 are said to be parallel （平行） if they have the same slope. Parallel
lines do NOT intersect with each others (except the trivial case L1  L2 ).

Two lines L1 , L2 are said to be perpendicular （垂直） if m1m2  1, where m1 ,
m2 are their slopes. Perpendicular lines make a right angle with each others.
L1
L1
L2
L2
Parallel Lines
Pe rp e n d ic u la r Lin e s
Equation of Straight Line.
(Point-slope Form)
Given a point ( x1 , y1 ) on the line and slope m, the equation is
y  y1  m( x  x1 ) .
(Two-point Form)
Given two points ( x1 , y1 ) , ( x2 , y2 ) on the line, the equation is
y  y1 y2  y1
.

x  x1 x2  x1
(Intercept Form)
Given x-intercept a and y-intercept b, the equation is
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x y
 1.
a b
(Slope-intercept Form)
(General Form)
Given slope m and y-intercept c, the equation is
y  mx  c .
The equation of any straight line can be written as
Ax  By  C  0 .
In this case, the slope and y-intercept are given by
slope  
A
B
y -intercept  
and
C
.
B
When A  0 and B  0 , the line is horizontal. When A  0 and B  0 , the line is
vertical. When C  0 , the line passes through origin.
Recall that a circle is a set of points for which they are equidistant （等距） from center.
P(x, y)
r
C(a, b)
Equation of Circle.
(Standard Form) The equation of circle centered at C (a, b) with radius r is
( x  a ) 2  ( y  b) 2  r 2 .
In particular, when C  (0, 0) , the equation of circle centered at origin with radius r is
x2  y 2  r 2 .
(General Form)
The equation of any circle can be written as
x 2  y 2  Dx  Ey  F  0 .
In this case,
center  ( 

D E
, )
2
2
and
radius 
1
D2  E 2  4F .
2
Given the values of D, E and F, the equation x 2  y 2  Dx  Ey  F  0 may NOT
represent a circle. It represents a circle if and only if D 2  E 2  4 F  0 .

If D 2  E 2  4 F  0 , we say the equation x 2  y 2  Dx  Ey  F  0 represents a
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point circle （點圓）. If D 2  E 2  4 F  0 , it represents an imaginary circle （虛圓）.
Given a line and a circle in the same plane. One and only one of the following cases occur.
Case 1
Case 2
y
y
x
y
x
Intersect at 2 distinct points

Case 3
x
Intersect at exactly 1 point.
No intersection.
A line for which intersects a circle C at exactly 1 point is called a tangent to the circle C
at that point.
Intersection of Line and Circle.
Given a line L : y  mx  c and a circle C : x 2  y 2  Dx  Ey  F  0 . To find their
intersection, we consider the system
 y  mx  c
 2
2
 x  y  Dx  Ey  F  0
Substitute the first equation into the second one, we obtain
x 2  (mx  c)2  Dx  E (mx  c)  F  0 .
After simplification, it reduces to a quadratic equation
x 2  px  q  0
----------------------------------
(*)
The number of intersections of L and C is equal to the number of (real) roots of (*). So,
 p 2  4q  0  2 intersections
 2
 p  4q  0  1 intersection
 p 2  4q  0  no intersection

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Basic Techniques and Typical Examples
Example 1.
Find the equation of circle with diameter AB, where A  (1, 2) and B  (3, 4) .
Solution.
1 3 2  4
1
1
,
)  (2, 3) . Radius is
AB 
(3  1) 2  (4  2) 2  2 .
The center is M  (
2
2
2
2
The equation of circle is given by standard form:
( x  2)2  ( y  3)2  2 .
After simplification,
x 2  y 2  4 x  6 y  11  0 .
Tip.
In the examination, you are suggested to substitute the points A, B into the
equation you obtained to see whether they satisfy the equation. Of course, even if
they satisfy the equation, it does NOT mean your answer must be correct (the
circle passes through A, B does not mean it has AB as diameter).
In general, the circle with diameter PQ, where P( x1 , y1 ) and Q( x2 , y2 ) , is given by
( x  x1 )( x  x2 )  ( y  y1 )( y  y2 )  0 .
It is easy to see why this circle passes through P, Q. To prove it has PQ as diameter, we write
it into general form:
x 2  y 2  ( x1  x2 ) x  ( y1  y2 ) y  x1 x2  y1 y2  0
The center is (
x1  x2 y1  y2
,
) , which is the midpoint of PQ. Hence, PQ is the diameter of
2
2
this circle.
Example 2.
Find the equation of circle passes through A(0, 5) , B(3, 4) , C (3, 4) .
Solution 1. (Algebraic Method)
Let the required equation be
x 2  y 2  Dx  Ey  F  0
---------------------------
(*)
Substitute the point A(0, 5) into (*), 02  52  0 D  5E  F  0 , i.e.
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5E  F  25
------------------------------------
(1)
Substitute the point B(3, 4) into (*), 32  42  3D  4 E  F  0 , i.e.
3D  4E  F  25
--------------------------------
(2)
Substitute the point C (3, 4) into (*), (3)2  42  3D  4 E  F  0 , i.e.
3D  4E  F  25
-------------------------------
(3)
(2)  (3), gives 6D  0  D  0 . Put D  0 into (2),
----------------------------------4E  F  25
(4)
(1)  (4), gives E  0 . Put E  0 into (4), we have F  25 .
A(0, 5)
C(-3, 4)
B(3, 4)
Hence, the required equation is
x 2  y 2  25  0 .
Solution 2. (Geometric Method)
In the view of Euclidean geometry, the center of circle is the intersection of perpendicular
bisectors of two chords.
Step I
(Find the equation of perpendicular bisector of AB)
3 9
Midpoint of AB is ( , ) .
2 2
Slope of AB is mAB 
A(0, 5)
54
1
 .
03
3
C(-3, 4)
Slope of the perpendicular bisector is 
B(3, 4)
1
 3.
mAB
Equation of perpendicular bisector is
center
9
3
y   3( x  )
2
2
i.e.
y  3x
Step II
----------------------------------------
(1)
(Find the equation of perpendicular bisector of AC)
3 9
Midpoint of AC is ( , ) .
2 2
Slope of AC is mAC 
54
1
 .
0  (3) 3
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Slope of the perpendicular bisector is 
1
 3 .
mAC
Equation of perpendicular bisector is
y
9
3
 3( x  )
2
2
i.e.
y  3 x
--------------------------------------
(2)
Step III (Find the coordinates of the center)
By solving (1) and (2),
center of the circle  (0, 0) .
Step IV (Find the equation of the circle)
The radius of circle is given by the distance from the center (0, 0) to A(0, 5) , which is
r  02  52  5 .
The equation of circle is therefore
( x  0)2  ( y  0)2  r 2 ,
i.e.
x 2  y 2  25  0 .
Example 3.
Find the coordinates of the intersection point(s) of C : x 2  y 2  25 and L : x  2 y  5  0 .
Solution.
Substitute x  2 y  5 into the circle equation,
(2 y  5)2  y 2  25 .
This is a quadratic equation in y, after simplification becomes
5 y 2  20 y  0
5 y ( y  4)  0
The roots are y  0, 4 (2 intersections!).
L
(3, 4)
When y  0 , x  2 y  5  5 .
When y  4 , x  2 y  5  3 .
(-5, 0)
C
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Hence, the intersection points are (5, 0) and (3, 4) .
Example 4.
Given a circle x 2  y 2  2 x  2 y  23  0 and a point P(4, 5) on this circle. Find the
equation of the tangent at P.
Solution.
The center of the circle is C  (1, 1) .
P(4, 5)
5 1 4
 .
Slope of CP is mCP 
4 1 3
1
3
 .
Slope of the tangent l is 
mCP
4
3
The equation of l is y  5   ( x  4) , i.e.
4
3x  4 y  32  0 .
TIP.
Given a circle x 2  y 2  Dx  Ey  F  0 and a point P( x1 , y1 ) on it. The
equation of tangent at P is
 x  x1 
 y  y1 
xx1  yy1  D 
 E

 F  0.
 2 
 2 
Example 5.
Given a line L : 3x  4 y  32  0 and a point P(4, 5) on it. If a circle C touches the line at
P and the circle passes through Q (5,  2) . Find the equation of C.
Solution.
Step I
(Find the equation of perpendicular bisector of PQ)
9 3
Midpoint of PQ is ( , ) .
2 2
Slope of PQ is mPQ 
P(4, 5)
5  ( 2)
 7 .
45
Slope of the perpendicular bisector is 
1
1
 .
mPQ 7
center
Q(5, -2)
The equation of perpendicular bisector is
y
3 1
9
 (x  )
2 7
2
i.e.
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x  7y  6  0
Step II
-----------------------------------
(1)
(Find the equation of the line through P and perpendicular to L)
3
Slope of L is mL   .
4
Slope of the line perpendicular to L is 
1
4
 .
mL 3
The equation of the line through P and perpendicular to L is
y 5 
4
( x  4) .
3
After simplification,
4x  3 y  1  0
-----------------------------------
(2)
Step III (Find the coordinates of the center)
By solving (1) and (2),
center of the circle  (1, 1) .
Step IV (Find the equation of the circle)
Radius is r  (4  1) 2  (5  1) 2  5 . The equation of the circle is
( x  1)2  ( y  1)2  25 .
Example 6.
Given a circle x 2  y 2  2 x  2 y  23  0 and a point R(8, 2) outside the circle. Suppose
the tangents through R touch the circle at M, N. Find the equation of MN.
Solution.
Let M  ( x1 , y1 ) and N  ( x2 , y2 ) .
M (x 1 , y 1 )
By the tip in example 4, the equation of MR is
xx1  yy1  ( x  x1 )  ( y  y1 )  23  0 .
R(8, 2)
N (x 2 , y 2 )
Put ( x, y )  (8, 2) ,
8x1  2 y1  (8  x1 )  (2  y1 )  23  0
Replace ( x1 , y1 ) by ( x2 , y2 ) in (1), we get a similar equality
8x2  2 y2  (8  x2 )  (2  y2 )  23  0
---------------------
(1)
---------------------
(2)
By (1) and (2), we see that M ( x1 , y1 ) and N ( x2 , y2 ) satisfy the equation
8 x  2 y  (8  x)  (2  y )  23  0 .
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The equation of MN is
7 x  y  33  0 .
Example 7.
In the situation of example 6, find the equations of the tangents MR and NR.
Solution.
Step I
(Find the coordinates of M, N)
The intersections of MN and the circle are given by
 7 x  y  33  0
 2
2
 x  y  2 x  2 y  23  0
By solving this system, the intersections are
(4, 5)
and

(5,  2) .
We don’t need to know which one is M and which one is N.
(4, 5)
R(8, 2)
(5, -2)
Step II
(Find the equations)
The equation of tangent passing through (4, 5) and (8, 2) is
3x  4 y  32  0 .
The equation of tangent passing through (5,  2) and (8, 2) is
4 x  3 y  26  0 .
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