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Circles Revision

Explore the transformations and intersections of circles and their related equations, including translations, enlargements, and the use of the discriminant. Learn how to find the centre of a circle using perpendicular bisectors of chords.

donaldwing
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Circles Revision

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  1. Circles Revision Transformations Intercepts Using the discriminant Chords

  2. x2 + y2= 1 x2 + y2 = 32 (x-1)2 + (y+3)2 = 9 Centre (1,-3) Radius 3 (1,-3) TRANSLATED BY [ ] 1 -3 From the circle: x2 + y2= 1 to the circle:(x-1)2 + (y+3)2 = 9 What transformations have occurred? Centre (0,0) Radius 1 Centre (0,0) Radius 3 ENLARGED BY SCALE FACTOR 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 +10-3)2= 9 Solve equation to find intersect (x-1)2 + (x +7)2= 9

  4. Use discriminent “b2-4ac” To find out about roots Circle Intersect 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)

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

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

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

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