1. Lecture 2 of 9
SECTIONS
1.2 CIRCLES
CONIC

2. LEARNING OBJECTIVES Find the equation of a circle
i. passing through three points
ii. passing through two points where the equation of the diameter is given
b. Find the points of intersection of two circles.

3. Circle passing through three given pointsB C
Example 1
Find the equation of the circle passing
through the points (-2,-1) , (-6,1) and (1,8).

4. perpendicular bisector
Solution
perpendicular bisector
R(1,8)
chord C
Q(-6,1)
P(-2,-1)
The perpendicular bisectors of the chords
PQ and QR meet at the centre of the circle, C.

5. L1 C Find the equation of L1: R(1,8) Q(-6,1) P(-2,-1)
Thus, the equation of L1: y – 0 = 2 (x – (– 4))
y – 2x = (1)

6. L2 C Thus, the equation of L2: Find the equation of L2: R(1,8) Q(-6,1)
P(-2,-1)
Q(-6,1) C
R(1,8) L2
Thus, the equation of L2:

7. The centre of the circle is (–2,4)
By solving these equations simultaneously ,
we will get the centre of the circle.
P(-2,-1)
Q(-6,1) C
R(1,8) L1 L2
y – 2x = (1)
y + x = (2)
(2)-(1): x = -6
x = -2
y = 4
The centre of the circle is (–2,4)

8. CP is the radius of the circle,
Thus, the equation of the circle is or
x2 + y2 + 4x - 8y - 5 = 0

9. An Alternative Method
The general equation of the circle is
x2+ y2 + 2gx + 2fy + c = 0
Substitute each point into the equation.

10. At point (-2,-1),
(-2)2 + (-1)2 + 2g(-2) + 2f(-1) + c = 0
5 - 4g -2f + c = (1)
At point (-6,1),
(-6)2 + (1)2 + 2g(-6) + 2f(1) + c = 0
g + 2f + c = (2)
At point (1,8),
(1)2 + (8)2 + 2g(1) + 2f(8) + c = 0
65 + 2g + 16f + c = (3)

11. Then, solve these equations simultaneously to get the values of g, f and c.
(2)-(1)
g + 2f + c = (2)
5 - 4g -2f + c = (1)
32 – 8g + 4f = 0
8 – 2g + f = 0 ……………….. (4)
(3)-(2)
65 + 2g + 16f + c = (3)
g + 14g = 0
2 + g + f = 0 …………………. (5)

12. x2 + y2 + 4x - 8y - 5 = 0 (4)-(5) 8 – 2g + f = 0 ……………….. (4)
(1) g -2f + c = (1)
c = (2) + 2(-4)
c = - 5
x2 + y2 + 4x - 8y - 5 = 0

13. Circle passing through two points with the equation of the diameter given.
C(-g,-f) Q
Example 2
Find the equation of the circle passing
through the points (0,0) and (4,2) with its
diameter is x + y = 1.

14. Solution P(4,2) Q(0,0) C Find the equation of L1: L1 Diameter,
x + y = 1 L1 y x
Find the equation of L1:
Thus, the equation of L1: y - 1 = -2 (x - 2)
y + 2x = (1)

15. equation of the diameter The centre of the circle is (4, - 3)
Then, solve these equations simultaneously to get the centre of the circle.
y + 2x = (1)
equation of the diameter
y + x = (2)
(1)-(2):
x = 4,
y = - 3
The centre of the circle is (4, - 3)
Thus, the equation of the circle is
(x - 4)2 + ( y + 3)2 = 25 or
x2 + y2 – 8x + 6y = 0

16. Example 3
Find the equation of the circle having AB as
its diameter where A (1,5) and B (-2,3).
Solution
Centre of the circle is
the mid point of AB and
• B
• C A•
radius is ½ AB.

18. An Alternative Method
If P is any point on the circle and AB is the diameter, then
P(x,y)
mAP · mPB = -1
B(-2,3)
A(1,5)
(y - 5)(y - 3)= - (x - 1)(x +2)
y2 - 8y + 15 = - ( x2 + x - 2)
x2+y2+ x – 8y +13=0

19. Intersection of two circles
Consider two circles C1 and C2 whose equations are ;
…….(1)
…….(2) C1 C2 P Q

20. Find the points of intersection of the circles with equations ;
Example 4

21. Solution
The two circles will meet at the line: y = x – 1.
Substitute y = x – 1 into the equation of the first circle to find the points of intersection.
(i)
(ii)
(i) -
(ii)
y = x – 1

22. So the intersection points of the two circles are (2,1) and (4,3).

23. CONCLUSION through i. three points ~ find the equations of the
To find the equation of a circle passing
through
i. three points
~ find the equations of the
perpendicular bisectors (L1 and L2)
~ find the intersection of L1 and L2
to get the Centre
~ find the radius
~ form the equation of circle.

24. CONCLUSION To find the equation of a circle passing through
ii. two points and whose equation of
the diameter is given,
~ find the equation of the perpendicular
bisector L1
~ find the intersection of L1 and equation
of the diameter to get the Centre
~ find the radius
~ form the equation of circle.