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

2. From the circle: x 2 + y 2 = 1 to the circle: (x-1) 2 + (y+3) 2 = 9 What transformations have occurred? x 2 + y 2 = 1 Centre (0,0) Radius 1 Centre (0,0) Radius 3 x 2 + y 2 = 3 2 (x-1) 2 + (y+3) 2 = 9 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 … 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 (x 2 – 2x +1) + (x 2 + 14x + 49) = 9 2x 2 + 12x + 41 = 0 Solve equation to find intersect

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

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

6. “b 2 -4ac = 0” - one solution - tangent “b 2 -4ac > 0” - two solutions - crosses “b 2 -4ac < 0” - no solutions - misses What the discriminent tells us ……

7. 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

8. 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

9. 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 2 M Gradient of AB is : 14 - 10 11 - 5 = 4/6 = 2/3 y - 12 = -3/2 (x - 8) y - 12 = -3/2 x + 12 y = -3/2 x + 24 Equations of form y-y 1 =m(x-x 1 ) Line goes through (x 1,y 1 ) with gradient m Equation of perpendicular bisector of AB is: C Gradient MC x 2/3 = -1 Gradient MC = -3/2

10. 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 - 6 y = x - 1 Equations of form y-y 1 =m(x-x 1 ) Line goes through (x 1,y 1 ) with gradient m Equation of perpendicular bisector of RS is: C Gradient NC x -1 = -1 Gradient NC = 1

11. 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 = x - 1 subtract - 0 = x - -3/2x -1 - 24 5/2 x -25 = 0 5/2 x = 25 5x = 50 x = 10 y = x - 1 y = 10 -1 = 9 Centre is at (10,9)