1. Transformations Intercepts Using the discriminant Chords
Circles Revision
Transformations
Intercepts
Using the discriminant
Chords

2. ENLARGED BY SCALE FACTOR 3
From the circle: x2 + y2 = 1
to the circle: (x-1)2 + (y+3)2 = 9
What transformations have occurred?
x2 + y2 = 1
x2 + y2 = 32
Centre (0,0)
Radius 1
(x-1)2 + (y+3)2 = 9
Centre (0,0)
Radius 3
Centre (1,-3)
Radius 3
(1,-3)
ENLARGED BY SCALE FACTOR 3
TRANSLATED BY
[ ] 1 -3

3. Where do they intersect?
For the circle: (x-1)2 + (y-3)2 = 9
.. and the line y = x +10
Where do they cross?
Solve simultaneously to find intersect …
(x2 – 2x +1) +
(x2 + 14x + 49) = 9
(x-1)2 + (y-3)2 = 9
y = x +10
Substitute y:
2x2 + 12x + 41 = 0
(x-1)2 + (x )2 = 9
Solve equation to find intersect
(x-1)2 + (x +7)2 = 9

4. Circle Intersect Does 2x2 + 12x+41=0 have real roots
Use discriminent “b2-4ac”
To find out about roots
a = [coefficient of x2]
b = [coefficient of x]
c= [constant]
= 2
= 12
= 41
b2 - 4ac
= 122 – (4 x 2 x 41)
= 144 – 328
= -184
“b2 – 4ac < 0”
No roots
(solutions)

5. Circle Intersect (x-1)2 + (y-3)2 = 9 y = x +10
2x2 + 12x+41=0 has no real roots
-> no solutions; so lines do not cross

6. What the discriminent tells us ……
“b2-4ac = 0”
one solution
tangent
“b2-4ac > 0”
- two solutions
- crosses
“b2-4ac < 0”
- no solutions
- misses

7. The really important stuff you need to know about chords - but were afraid to ask
centre
A chord joins any 2 points on a circle and creates a segment
The perpendicular bisector of a chord passes through the centre of the circle
… conversely a radius of the circle passing perpendicular to the chord will bisect it

8. Using these facts we can solve circle problems
Given these 2 chords …
find the centre of the circle
(11,14)
The perpendicular bisector of a chord passes through the centre of the circle
(5,10)
(4,7)
If you find the equation of the two perpendicular bisectors, where they cross is the centre
(8,3)

9. Equations of form y-y1=m(x-x1)
Given these 2 chords …
find the centre of the circle
Midpoint (M) of AB is …
( , ) = (8, 12) M B
(11,14)
Gradient of AB is :
11 - 5
= 4/6 = 2/3 A
(5,10) C
(4,7)
Gradient MC x 2/3 = -1
Gradient MC = -3/2 R
y - 12 = -3/2 (x - 8)
y - 12 = -3/2 x + 12
y = -3/2 x + 24
Equations of form y-y1=m(x-x1)
Line goes through (x1,y1) with gradient m
Equation of perpendicular bisector of AB is: S
(8,3)

10. Equations of form y-y1=m(x-x1)
Given these 2 chords …
find the centre of the circle
Midpoint (N) of RS is …
( , ) = (6, 5) N B
(11,14)
Gradient of RS is :
4 - 8
= 4/-4 = -1 A
(5,10) C
(4,7)
Gradient NC x -1 = -1
Gradient NC = 1 R
y - 5 = 1 (x - 6)
y - 5 = x - 6
y = x - 1
Equations of form y-y1=m(x-x1)
Line goes through (x1,y1) with gradient m
Equation of perpendicular bisector of RS is: S
(8,3)

11. - Finding the centre …. Centre is at (10,9)
If you find the equation of the two perpendicular bisectors, where they cross is the centre
y = -3/2 x + 24
(11,14)
subtract -
y = x - 1
y = -3/2 x + 24
(5,10)
0 = x - -3/2x C
5/2 x -25 = 0
5/2 x = 25
5x = 50
x = 10
(4,7)
y = x - 1
(8,3)
y = x - 1
y = = 9
Centre is at (10,9)