Higher Mathematics Unit 2.4
The Circle
Contents
1.
1.
Circle, centre O, radius r.
Circle, centre (a, b) , radius r.
2.
3.
4.
5.
6.
7.
x 2  y 2  2 gx  2 fy  c  0.
Finding the equation of a tangent line.
Intersections and tangency.
Finding the equation of a circle.
General points.
References/Resources.
Circle, centre O, radius r
Circle, centre (a,b), radius r
Go over the derivations of both equations:
Y
r
O
P lies on circle  x 2  y 2  r 2 .
P(x,y)
y
x
X
P(x,y)
P lies on the circle  ( x  a) 2  ( y  b) 2  r 2 .
r
O(a,b)
y-b
x-a
(able pupils should be able to relate this to transformations from unit 1)
The above results are most commonly used when we are given the coords of the centre and the
length of the radius. We are required to find the equation of the circle.
On the other hand, if we are given the equation of the circle in this form, the coords of the centre
and the length of the radius can be written down by inspection.
Pupils must have practice at doing this “both ways”.
2.
x 2  y 2  2 gx  2 fy  c  0.
This result may be derived with the class as follows:
x 2  y 2  2 gx  2 fy  c
x 2  2 gx  g 2  y 2  2 fy  f 2  g 2  f 2  c
(x  g)2  ( y  f )2  g 2  f 2  c
( x  ( g )) 2  ( y  ( f )) 2  g 2  f 2  c
c.f.
( x  a) 2
 ( y  b) 2
 r2
Hence this equation represents a circle, centre ( g , f ) , radius
g 2  f 2  c , provided that
g 2  f 2  c  0. Although formulae are given, encourage pupils to learn these.
This result is used to obtain the coords of the centre and the length of the radius.
Make sure that the pupils have practice at questions in which the equation is not in “standard
form”, e.g.
2 x 2  2 y 2  18x  8 y  10  0
 x 2  y 2  9 x  4 y  5  0 , etc, etc.
3.
Finding the Equation of a Tangent Line
A rough sketch is useful if one is not given.
The procedure is simple:
Find coords of centre
Find mradius
Find mtgt for m1 m2 = -1 (Stated)
Use y  b  m( x  a) to obtain equation of tangent.
Working may be set out as follows:
“A(1,1) is the centre of a circle of radius 5 units. Find the equation of the tangent at the point
P(4,5) on its circumference.”
P(4,5)
A(1,1)
m AP  43  mtgt   34
(tgt  radius property; m1 m2  1 for perpendicu lar lines .)
y  b  m( x  a )
y  5   34 ( x  4)
(may be left in this form, if no further work required)
4( y  5)  4( 34 )( x  4) (get rid of the fraction first before multiplying out)
4 y  20  3( x  4)
4 y  20  3x  12
3x  4 y  32
(equation of tangent at P.)
4.
Intersections/Tangency
Solve the equations of the circle and the line simultaneously to obtain the coords of the
intersection points.
It is worthwhile to illustrate the 3 distinct situations in sketches:
[Note:
the idea “Exactly one solution  tangency” works here (for circles and straight lines)
because a circle is closed and convex.
The argument is not generally true, as the following sketches show.
[In the first diagram there is one point of contact, but no tangency; in the second there are
several points of contact and there is tangency.]
Pupils should be made aware of these ideas.]
Many candidates will need much practice at algebraically awkward examples such as
“Prove that the straight line 3x  2 y  6  0 is tangent
x 2  y 2  2 x  4 y  8  0 and find the coords of the point of contact.”
2 y  6  3x
y  3  32 x
to
the
(easier to rearrange in terms of y)
y 2  (3  32 x) 2
y 2  9  2(3)( 2 x)  ( 2 x) 2
3
y 2  9  9 x  94 x 2
3
(weak candidates need practice at this)
circle
Now substitute into the equation of the circle:
x 2  y 2  2x  4 y  8  0
x 2  9  9 x  94 x 2  2 x  4(3  32 x)  8  0
x 2  9  9 x  94 x 2  2 x  12  6 x  8  0
4( x 2  9  9 x  94 x 2  2 x  12  6 x  8)  4  0
(clear fractions)
4 x  36  36 x  9 x  8 x  48  24 x  32  0
13x 2  52 x  52  0
13( x 2  4 x  4)  0
2
2
13( x  2) 2  0
x2 0
x2
Exactly one solution, hence we have tangency here.
x  2  y  0.
Also, y  3  32 x.
 point of tangency is (2,0).
Example
a) Find the equation of AB, the perpendicular bisector
of the line joining the points P(-3, 1) and Q(1, 9).
(4)
b) C is the centre of a circle passing through P and Q.
Given that QC is parallel to the y-axis, determine
the equation of the circle.
(3)
c) The tangents at P and Q intersect at T.
Write down (i) the equation of the tangent at Q
(ii) the coordinates of T
(2)
Answer
 1

2







3
4
5
6
7
8
9
midpoint =(-1, 5)
9 1
mPQ =
1  (3)
mperp = - ½
y – 5 = - 1/2 (x – (-1))
yc = 4
radius = 5
(x – 1)2+ (y – 4)2= 25
y=9
T = (-9, 9)
(twice)
(twice)
5.
Finding the Equation of a Circle
Method 1
Substitute the coords of each point into the general equation
x 2  y 2  2 gx  2 fy  c  0 .
Solution of 3 linear equations in 3 unknowns is not in the syllabus, so the
given points will include (0,0) and at least one point on the coordinate axes
if this method is to be used.
Method 2
The idea here is that the perpendicular bisector of a chord is a diameter.
Therefore, given 3 points on the circumference, the equations of 2 diameters can be found.
Centre of circle is the point of intersection of the diameters, and the radius
can be found by calculating the distance from the centre to one of the 3
points.
Then use ( x  a) 2  ( y  b) 2  r 2 .
A compass and straight edge construction is a good introduction to this
method.
But a rough sketch is also useful:
Method 3
If we have a right angle on the circumference of a circle then it “sits” on a
diameter.
B
A
C
If angle ABC is a right angle then AC is a diameter of the circle.
The centre is the mid-point of AC and the diameter is the length of AC.
Then use ( x  a) 2  ( y  b) 2  r 2 .
It is worthwhile to do an example which can be done by all 3 methods, e.g.
“Find the equation of the circle passing through the points A(0,6), B(4,0)
and O(0,0).”
6.
General Points
(a)
Once all topics have been taught, give pupils practice at “miscellaneous questions.”
(b)
Past papers contain many excellent examples.
(c)
Practice must be given at:
(i)
(ii)
(iii)
(iv)
(v)
Finding the equation of the common tangent through the point of contact
for circles touching externally.
(vi)
Proving that two circles touch internally. (i.e. dist. between centres =
difference of radii.)
Practice should be given at considering r = 0 situation.
If asked for the equation of a circle, it can be left in the form
(x – a)2 + (y – b) 2 = r2 without giving the fully expanded form( unless
explicitly asked for).
(vii)
(viii)
7.
Finding (by P.T.) the length of a tangent from an external point.
Calculating equation of circle, given end points of diameter.
Finding value of an unknown constant from the equation of circle. e.g.
“(8,a) lies on the circumference of the circle with equation …………. Find
the value of a.”
Proving that two circles touch externally. (i.e. dist. between centres = sum
of radii.)
References/Resources
1.
2.
H.H.M. pp 205-228.
M.I.A. pp168 - 180