Unit 2 Outcome 4

1. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Lesson 6 Starter Q1 Solution Find radius (3,8) r2 = 42 + 42 4 r2 = 16 + 16 (-1,4) r2 = 32 4 r = √32 d = 2√16 √2 r = 2x4 √2 r = 8√2 Answer B Thursday, 18 September 2014

2. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Lesson 6 Starter Q2 Solution PQ is diam, = 12, radius = 6, r2 = 36. so must be A or C Centre co-ords (6, 2) (x – a)2 + (y – b)2 = 36 (x – 6)2 + (y – 2)2 = 36 Answer C Thursday, 18 September 2014

3. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Lesson 6 Starter Q3 Solution Find gradient radius Point C is (0,0) Point Q is (12, -5) m CQ = 0 - -5 0 - 12 m CQ = - 5 12 Tangent is perpendicular to radius 12 5 Answer C Thursday, 18 September 2014

4. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Lesson 6 Starter Q4 Solution 0 2 ∫ ∫ x dx + x dx -2 0 2 ∫ 2 x dx 0 Answer D Thursday, 18 September 2014

5. Circle Unit 2 Outcome 4 Intersection of a Line and a circle Reminder There are 3 possible scenarios 1 point of contact 0 points of contact 2 points of contact discriminant line is a tangent discriminant (b2- 4ac < 0) discriminant (b2- 4ac > 0) (b2- 4ac = 0) To determine where the line and circle meet we use simultaneous equations and the discriminant tells us how many solutions we have.

6. Circle Unit 2 Outcome 4 Intersection of a Line and a circle Find where the line y = 2x + 1 meets the circle (x – 4)2 + (y + 1)2 = 20 and comment on the answer Replace y by 2x + 1 in the circle equation (x – 4)2 + (y + 1)2 = 20 becomes (x – 4)2 + (2x + 1 + 1)2 = 20 (x – 4)2 + (2x + 2)2 = 20 x 2 – 8x + 16 + 4x 2 + 8x + 4 = 20 5x 2 = 0 x 2 = 0 x = 0 one solution tangent point Using y = 2x + 1, if x = 0 then y = 1 Point of contact is (0,1) Thursday, 18 September 2014

7. Circle Unit 2 Outcome 4 Intersection of a Line and a circle Find where the line y = 2x + 1 meets the circle (x – 4)2 + (y + 1)2 = 20 and comment on the answer Replace y by 2x + 1 in the circle equation (x – 4)2 + (y + 1)2 = 20 becomes (x – 4)2 + (2x + 1 + 1)2 = 20 (x – 4)2 + (2x + 2)2 = 20 x 2 – 8x + 16 + 4x 2 + 8x + 4 = 20 5x 2 = 0 x 2 = 0 x = 0 one solution tangent point Using y = 2x + 1, if x = 0 then y = 1 Point of contact is (0,1) Thursday, 18 September 2014

8. (-1,4) (-5,-4) Points of contact are (-5,-4) and (-1,4). Circle Unit 2 Outcome 4 Intersection of a Line and a circle Find where the line y = 2x + 6 meets the circle x2 + y2 + 10x – 2y + 1 = 0 Replace y by 2x + 6 in the circle equation x2 + y2 + 10x – 2y + 1 = 0 becomes x2 + (2x + 6)2+ 10x – 2(2x + 6) + 1 = 0 x 2 + 4x2 + 24x + 36 + 10x – 4x - 12 + 1 = 0 5x2 + 30x + 25 = 0 ( 5 ) x 2 + 6x + 5 = 0 (x + 5)(x + 1) = 0 x = -5 or x = -1 Using y = 2x + 6 if x = -5 then y = -4 if x = -1 then y = 4 Thursday, 18 September 2014

9. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Using Discriminants At the line x2 – 18x + 81 = 0 we can also show there is only one solution by showing that the discriminant is zero. ax2 – bx + c = 0 a =1, b = -18 c = 9 For x2 – 18x + 81 = 0 , So b2 – 4ac = (-18)2 – 4 X 1 X 81 = 0 = 364 - 364 Since disc = 0 then equation has only one root so there is only one point of contact so line is a tangent. The next question uses discriminants in a slightly different way. Thursday, 18 September 2014

10. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Using Discriminants Example Find the equations of the tangents to the circle x2 + y2 – 4y – 6 = 0 from the point (0,-8). x2 + y2 – 4y – 6 = 0 2g = 0 so g = 0 Each tangent takes the form y = mx -8 2f = -4 so f = -2 Replace y by (mx – 8) in the circle equation Centre is (0,2) to find where they meet. This gives us … Y x2 + y2 – 4y – 6 = 0 (0,2) x2 + (mx – 8)2 – 4(mx – 8) – 6 = 0 x2 + m2x2 – 16mx + 64 –4mx + 32 – 6 = 0 (m2+ 1)x2 – 20mx + 90 = 0 -8 c =90 In this quadratic a = (m2+ 1) b = -20m

11. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Tangency For tangency we need discriminate = 0 b2 – 4ac = 0 (-20m)2 – 4 X (m2+ 1) X 90 = 0 400m2 – 360m2 – 360 = 0 40m2 – 360 = 0 40m2 = 360 m = -3 or 3 m2 = 9 So the two tangents are y = -3x – 8 and y = 3x - 8 and the gradients are reflected in the symmetry of the diagram. Thursday, 18 September 2014

12. Equations of Tangents Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle NB: At the point of contact a tangent and radius/diameter are perpendicular. Tangent radius This means we make use of m1m2 = -1. Thursday, 18 September 2014

13. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Equations of Tangents Example Prove that the point (-4,4) lies on the circle x2 + y2 – 12y + 16 = 0 NAB Find the equation of the tangent here. At (-4,4) x2 + y2 – 12y + 16 = 16 + 16 – 48 + 16 = 0 So (-4,4) must lie on the circle. x2 + y2 – 12y + 16 = 0 2g = 0 so g = 0 2f = -12 so f = -6 Centre is (-g,-f) = (0,6) Thursday, 18 September 2014

14. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Equations of Tangents y2 – y1 x2 – x1 Gradient of radius = = (6 – 4)/(0 + 4) (0,6) = 2/4 (-4,4) = 1/2 So gradient of tangent = -2 ( m1m2 = -1) Using y – b = m(x – a) We get y – 4 = -2(x + 4) y – 4 = -2x - 8 y = -2x - 4 Thursday, 18 September 2014

15. Higher Circle Unit 2 Outcome 4 Tangents to circles Tangents to a circle Thursday, 18 September 2014

16. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Tangents to a circle Page 174 Exercise 4 Q1 Intersections of lines and circles Page 176 Q2 to Build skills Thursday, 18 September 2014

17. Higher Circle Unit 2 Outcome 4 Intersection of a Line and a circle Arithmetic sequences Thursday, 18 September 2014

18. Equation x2 + y2 + 2gx + 2fy + c = 0 Example Find the centre & radius of x2 + y2 - 10x + 4y - 5 = 0 x2 + y2 - 10x + 4y - 5 = 0 NAB c = -5 2g = -10 2f = 4 g = -5 f = 2 radius = (g2 + f2 – c) centre = (-g,-f) = (5,-2) = (25 + 4 – (-5)) = 34