Circles RevisionCircles Revision• Transformations Transformations • InterceptsIntercepts• Using the discriminantUsing the discriminant• ChordsChords

From the circle: x2 + y2 = 1

to the circle: (x-1)2 + (y+3)2 = 9

What transformations have occurred?

x2 + y2 = 1Centre (0,0)Radius 1

Centre (0,0)Radius 3

x2 + y2 = 32 (x-1)2 + (y+3)2 = 9 Centre (1,-3)

Radius 3(1,-3)ENLARGED BY

SCALE FACTOR 3

TRANSLATED BY[ ]1-3

Where do they intersect?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 …

y = x +10(x-1)2 + (y-3)2 = 9

Substitute y:

(x-1)2 + (x +10 -3)2 = 9

(x-1)2 + (x +7)2 = 9

(x2 – 2x +1) + (x2 + 14x + 49) = 9

2x2 + 12x + 41 = 0Solve equation to find intersect

Circle IntersectCircle Intersect

Use discriminent “b2-4ac”

To find out about roots

Does 2x2 + 12x+41=0 have real 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)

Circle IntersectCircle Intersect

2x2 + 12x+41=0 has no real roots

-> no solutions; so lines do not cross

y = x +10(x-1)2 + (y-3)2 = 9

“b2-4ac = 0”- one solution- tangent

“b2-4ac > 0”- two solutions- crosses

“b2-4ac < 0”- no solutions- misses

What the discriminen

t tells us ……

The really important stuff you need to know about chords - but were afraid to ask 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

centre

Using these facts we can solve circle problems

Given these 2 chords … find the centre of the circle

(5,10)

(11,14)

(4,7)

(8,3)

The perpendicular bisector of a chord passes through the centre of the circle

If you find the equation of the two perpendicular bisectors, where they cross is the centre

Given these 2 chords … find the centre of the circle

(5,10)

(11,14)

(4,7)

(8,3)

A

B

R

S

Midpoint (M) of AB is …(5 + 11 , 10 + 14) = (8, 12) 2 2M

Gradient of AB is : 14 - 10 11 - 5= 4/6 = 2/3

y - 12 = -3/2 (x - 8)y - 12 = -3/2 x + 12y = -3/2 x + 24

Equations of form y-yEquations of form y-y11=m(x-x=m(x-x11) ) Line goes through (x1,y1) with gradient m

Equation of perpendicular bisector of

AB is:

CGradient MC x 2/3 = -1 Gradient MC = -3/2

Given these 2 chords … find the centre of the circle

(5,10)

(11,14)

(4,7)

(8,3)

A

B

R

S

Midpoint (N) of RS is …(4 + 8 , 7 + 3 ) = (6, 5) 2 2

N

Gradient of RS is : 7 - 3 4 - 8= 4/-4 = -1

y - 5 = 1 (x - 6)y - 5 = x - 6y = x - 1

Equations of form y-yEquations of form y-y11=m(x-x=m(x-x11) ) Line goes through (x1,y1) with gradient m

Equation of perpendicular bisector of

RS is:

CGradient NC x -1 = -1 Gradient NC = 1

Finding the centre ….

(5,10)

(11,14)

(4,7)

(8,3)

If you find the equation of the two perpendicular bisectors, where they cross is the centre

y = x - 1

y = -3/2 x + 24

C

y = -3/2 x + 24

y = x - 1subtract-

0 = x - -3/2x -1 - 24

5/2 x -25 = 05/2 x = 25 5x = 50 x = 10

y = x - 1

y = 10 -1 = 9

Centre is at (10,9)