1 / 11

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
Télécharger la présentation

Circles Revision

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related