Higher Circle Notes

1. Higher Circle Notes

2. Equation of a Circle Centre at the Origin P(x,y) y x By Pythagoras Theorem y r x O

3. General Equation of a Circle P(x,y) y b a x y Circle centre C(a ,b) Radius r r C(a,b) y-b x-a x O By Pythagoras Theorem When you are asked to determine the equation of a circle, this is the equation you should use!

4. The Circle examples centre (2,5) radius = 7 (x – 2)2 + (y – 5)2 = 49 radius = 13 centre (-5,1) (x + 5)2 + (y – 1)2 = 13 centre (3,0) radius = 20 (x – 3)2 + y2 = 20 = 25 Centre (2,– 3) & radius = 10 Equation is (x – 2 )2 + (y + 3)2 = 100 Centre (0,6) & radius = 23 Equation is x2 + (y – 6 )2 = 12

5. Find the equation of the circle that has PQ as diameter P(5 , 2) 4 3 C (2, – 2) C, the centre is the mid point of PQ Q (– 1, – 6) C (2,– 2) CP2, Radius, r2 = 32 + 42 = 25 Using (x – a)2 + (y – b)2 = r2 Equation is (x – 2)2 + (y + 2)2 = 25

6. Two circles are concentric, same centre The larger has equation (x + 3)2 + (y – 5)2 = 12 The radius of the smaller is half that of the larger. Find its equation. From (x – a)2 + (y – b)2 = r2 Centres is (– 3, 5) = 2 3 Larger radius = 12 so r2 = 3 Smaller radius = 3 Required equation is (x + 3)2 + (y – 5 )2 = 3

7. Inside / Outside or On Circumference When a circle has equation (x – a)2 + (y – b)2 = r2 If (x,y) lies on the circumference then (x – a)2 + (y – b)2 = r2 If (x,y) lies inside the circumference then (x – a)2 + (y – b)2 < r2 If (x,y) lies outside the circumference then (x – a)2 + (y – b)2 > r2

8. Inside / Outside or On Circumference Consider the circle (x + 1)2 + (y – 4)2 = 100 Determine where the following points lie; K(–7,12) , L(10,5) , M(4,9) Centre C(–1,4) , radius, r = 10  r2 = 100 CK2 = 62 + 82 = 100  K on circumference  L outside = 122 CL2 = 112 + 12  M inside = 50 CM2 = 52 + 52

9. Another version of Circle Equation is a circle : centre (– g , – f) radius, r = √g2 + f2 - c (x – a)2 + (y – b)2 = r2 is a circle : centre (a , b) radius, r x2 + y2 + 2gx + 2fy + c = 0

10. Equation x2 + y2 + 2gx + 2fy + c = 0 Write the equation (x – 5)2 + (y + 3)2 = 49 without brackets. (a + b)2 (x – 5)2 + (y + 3)2 = 49 = a2 + 2ab + b2 x2 – 10x + 25 + y2 + 6y + 9 – 49 = 0 x2 + y2 – 10x + 6y – 15 = 0 This takes the form given above where 2g = – 10 , 2f = 6 and c = – 15

11. Show that the equation x2 + y2 - 6x + 2y - 71 = 0 represents a circle and find the centre and radius. Circle exists if radius exists x2 + y2 – 6x + 2y – 71 = 0 x2 + y2 + 2gx + 2fy + c = 0 So g = – 3, f = 1, c = – 71 So equation represents a circle radius 9 Centre is (– g, – f)

12. Show that the equation x2 + y2 - 6x + 2y - 71 = 0 represents a circle and find the centre and radius. Circle exists if radius exists and the coefficients of x2 and y2 are equal x2 + y2 – 6x + 2y – 71 = 0 x2 + y2 + 2gx + 2fy + c = 0 So g = – 3, f = 1, c = – 71 So equation represents a circle radius 9  Centre is (3, – 1) Centre is (– g, – f)

13. We now have 2 ways on finding the centre and radius of a circle depending on the form we have. x2 + y2 – 10x + 6y – 15 = 0 x2 + y2 + 2gx + 2fy + c = 0 2g = – 10 2f = 6 c = – 15 g = – 5 f = 3 radius = (g2 + f2 – c) centre = (-g,-f) = (25 + 9 – (-15)) = (5,-3) = 49 = 7

14. x2 + y2 – 6x + 2y – 71 = 0 x2 + y2 + 2gx + 2fy + c = 0 2g = -6 2f = 2 c = -71 g = -3 f = 1 radius = (g2 + f2 – c) centre = (-g,-f) = (9 + 1 – (-71)) = (3,-1) = 81 = 9

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

16. The circle x2 + y2 – 10x – 8y + 7 = 0 cuts the y- axis at A & B. Find the length of AB. y A B x At A & B x = 0 so the equation becomes y2 – 8y + 7 = 0 (y – 1)(y – 7) = 0 y = 1 or y = 7 A is (0,7) & B is (0,1) So AB = 6 units

17. Frosty the Snowman’s lower body section can be represented by the equation x2 + y2 – 6x + 2y – 26 = 0 His middle section is the same size as the lower but his head is only 1/3 the size of the other two sections. Find the equation of his head ! x2 + y2 – 6x + 2y – 26 = 0 2g = -6 2f = 2 c = -26 g = -3 f = 1 radius = (g2 + f2 – c) centre = (-g,-f) = (9 + 1 + 26) = (3,-1) = 36 = 6

18. (3,19) radius of head = 1/3 of 6 = 2 2 6 Using (x – a)2 + (y – b)2 = r2 (3,11) 6 Equation is (x – 3)2 + (y – 19)2 = 4 6 (3,-1)

19. By considering centres and radii prove that the following two circles touch each other. Circle 1 x2 + y2 + 4x – 2y – 5 = 0 Circle 2 x2 + y2 – 20x + 6y + 19 = 0 Circle 22g = -20 so g = -10 Circle 12g = 4 so g = 2 c = 19 2f = 6 so f = 3 2f = -2 so f = -1 c = -5 centre = (-g, -f) = (10,-3) = (-2,1) centre = (-g, -f) radius = (g2 + f2 – c) radius = (g2 + f2 – c) = (100 + 9 – 19) = (4 + 1 + 5) = 90 = 10 = 310

20. (-2,1) (10,-3) √10 If d is the distance between the centres then by Pythagoras 3√10 d2 = 122 + 42 = 144 + 16 = 160 radius 1 + radius 2 = 16 × 10 = 10 + 310 d = 410 = 410 = distance between centres It now follows that the circles must touch !

21. Intersection of Lines & Circles There are 3 possible scenarios 2 points of contact 1 point of contact 0 points of contact (b2- 4ac > 0) (b2- 4ac = 0) (b2- 4ac < 0) line is tangent To determine where the line and circle meet we solve simultaneous equations and the discriminant tells us how many solutions we have.

22. Equations of Tangents At the point of contact a tangent and radius /diameter are perpendicular. radius Tangent This means we make use of m1m2 = -1.

23. Prove that the point (-4,4) lies on the circle x2 + y2 – 12y + 16 = 0 (0,6) (-4,4) Find the equation of the tangent here. = 0 = 16 + 16 – 48 + 16 At (-4,4) x2 + y2 – 12y + 16 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)

24. Prove that the point (-4,4) lies on the circle x2 + y2 – 12y + 16 = 0 Find the equation of the tangent here. mradius = 2/4 = 1/2 (0,6) m1× m2= – 1 mtangent = – 2 (-4,4) y – b = m(x – a) -4 -2 4 y – 4 = -2(x + 4) y – 4 = -2x – 8 2x +y + 4 = 0

25. 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 x2 – 8x + 16 + 4x2 + 8x + 4 = 20 5x2 = 0 x = 0 one solution  tangent Using y = 2x + 1, if x = 0 then y = 1 Point of contact is (0,1)

26. 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 + (2x + 6)2 + 10x – 2(2x+ 6) + 1 = 0 x2 + 4x2 + 24x + 36 + 10x– 4x – 12 + 1 = 0 ( 5 ) 5x2 + 30x + 25 = 0 x2 + 6x + 5 = 0 (x + 5)(x + 1) = 0  x = -5 or x = -1 Using y = 2x + 6 Points of contact (-5,-4) and (-1,4) x = -5 y = -4 x = -1 y = 4

27. Tangency Prove that the line 2x + y = 19 is a tangent to the circle x2 + y2 - 6x + 4y - 32 = 0 , and also find the point of contact. 2x + y = 19 so y = 19 – 2x Replace y by (19 – 2x) in the circle equation. x2 + (19 – 2x)2 – 6x + 4(19– 2x) - 32 = 0 x2 + 361 – 76x + 4x2- 6x + 76 – 8x – 32 = 0 ( 5) 5x2 – 90x + 405 = 0 Using y = 19 – 2x x2 – 18x + 81 = 0 x = 9  y = 1 (x – 9)(x – 9) = 0 x = 9 twice  tangent Point of contact is (9,1)

28. 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. For x2 – 18x + 81 = 0 , a =1, b = -18 and c = 9 So b2 – 4ac = (-18)2 – 4 × 1 × 81 = 364 - 364 = 0 Since discriminant = 0 the equation has only one root so there is only one point of contact so line is a tangent. The next example uses discriminants in a slightly different way.

29. Find the equations of the tangents to the circle x2 + y2 – 4y – 6 = 0 from the point (0,-8). Each tangent takes the form y = mx – 8 Replace y by (mx – 8) in the circle equation x2 + (mx – 8)2 – 4(mx – 8) – 6 = 0 x2 + m2x2 – 16mx + 64– 4mx + 32 – 6 = 0 (m2+ 1)x2 – 20mx + 90 = 0 Tangency  b2 – 4ac =0 see next slide

30. Find the equations of the tangents to the circle x2 + y2 – 4y – 6 = 0 from the point (0,-8). a = (m2 + 1) b = -20m c = 90 (m2+ 1)x2 – 20mx + 90 = 0 Tangency  b2 – 4ac =0 (-20m)2 – 4 × (m2 + 1) × 90 = 0 400m2 – 360m2 – 360 = 0 40m2 – 360 = 0 Equations of tangents y = -3x – 8 y = 3x - 8 m2 = 9 m= ± 3

31. Special case

32. Past Paper Questions

33. Find the equation of the circle with centre(–3 , 4) and passing through the origin. Find radius (distance formula): You know the centre: Write down equation:

34. Explain why this equation does not represent a circle. Evaluate Consider the 2 conditions 1. Coefficients of x2 and y2 must be the same. 2. Radius must be > 0 Calculate g and f: Deduction: Equation does not represent a circle

35. Find the equation of the circle which has P(–2, –1) and Q(4, 5) as the end points of a diameter. Q(4, 5) C P(-2, -1) Make a sketch (1 , 2) Calculate mid-point for centre: Calculate radius CQ: r2 = 32 + 32 = 18 Using (x – a)2 + (y – b)2 = r2 Write down equation:

36. Find the equation of the tangent at the point (3, 4) on the circle P(3, 4) O(-1, 2) x2 + y2 + 2gx + 2fy + c = 0 (-g , -f) Find centre of circle: Make a sketch mOP (radius to tangent) Gradient of tangent: y – b = m(x – a) y = -2x + 10 Equation of tangent: y – 4 = -2(x – 3)

37. O, A and B are the centres of the three circles shown in the diagram. The two outer circles are congruent, each touches the smallest circle. Circle centre A has equation (x – 12)2 + (y + 5)2 = 25 The three centres lie on a parabola whose axis of symmetry is shown the by broken line through A. a) i) State coordinates of A and find length of line OA. ii) Hence find the equation of the circle with centre B. b) The equation of the parabola can be written in the form y = px(x + q) find p and q.

38. (x – 12)2 + (y + 5)2 = 25 A(12 , -5) OA2 = 122 + (-5)2 OA2 = 169 OA = 13 OB = 13 Radius small circle = 5 Radius circle B = 8 Equation of parabola is y = px(x – 24), q = -24 Centre circle B = (24 , 0), by symmetry Parabola passes through A(12 , -5) Equation circle B is (x – 24)2 + (y – 0)2 = 82 -5 = p × 12(12 – 24) -5 = p × -144 Equation circle B is (x – 24)2 + y2 = 64 p = 5/144

39. Circle P has equation x2 + y2 – 8x – 10y + 9 = 0. Circle Q has centre (–2, –1) and radius 22. a) i) Show that the radius of circle P is 42 ii) Hence show that circles P and Q touch. b) Find the equation of the tangent to circle Q at the point (–4, 1) c) The tangent in (b) intersects circle P in two points. Find the x-coordinates of the points of intersection, expressing your answers in the form a ± b√3 x2 + y2 + 2gx + 2fy + c = 0 x2 + y2 – 8x – 10y + 9 = 0

40. If circles touch the sum of the radii will equal the distance between the centres P Q (4 , 5) (-2 , -1) (-4 , 1) (-2 , -1) to (4 , 5)  d2 = 62 + 62 4√2 2√2 d2 = 36 × 2 d2 = 72 So circles touch d = 6√2 Tangent is perpendicular to radius m radius = 2/-2 = -1  m tangent = 1 y – b = m(x – a) y – 1 = 1(x + 4) -4 1 1 y = x + 5

41. Solve equations to find points of intersection x2 + y2 – 8x – 10y + 9 = 0 P Q (4 , 5) (-2 , -1) Replace y in circle equation by (x + 5) y = x + 5 (-4 , 1) 4√2 2√2 x2 + (x + 5)2 – 8x – 10(x + 5) + 9 = 0 x2 + x2 + 10x + 25 – 8x – 10x – 50 + 9 = 0 2x2 – 8x – 16 = 0 (÷ by 2)  a = 1, b = -4 and c = -8 x2 – 4x – 8 = 0 x = 2 ± 2√3

42. For what range of values of c does the equation x2 + y2 – 6x + 4y + c = 0 represent a circle ? From x2 + y2 + 2gx + 2fy + c = 0 g = 3, f = -2 and c = c Circle exists if radius exists and the coefficients of x2 and y2 are equal r2 = g2 + f2 - c r2 = 32 + (-2)2 - c r2 = 13 – c For circle to exist 13 – c > 0 c < 13

43. When newspapers were printed by lithograph, the newsprint had to run over three rollers, illustrated in the diagram by 3 circles. The centres A, B and C of the three circles are collinear. The equations of the circumferences of the outer circles are (x + 12)2 + (y + 15)2 = 25 and (x – 24)2 + (y – 12)2 = 100 Find the equation of the central circle. A(-12 , -15) r = 5 10 C(24 , 12) C(24 , 12) r = 10 5 A(-12 , -15)

44. AC2 = 362 + 272 AC2 = 2025 10 C(24 , 12) AC = 45 45 Diameter of B is 45 – 15 = 30 15 15 radius = 15 5 A(-12 , -15) B divides AC in the ratio 20 : 25 = 4 : 5 So B is 4/9 along the line from A to C xB = -12 + 16 = 4 4/9 of 36 = 16 x – coordinate: -12 to 24 = 36 4/9 of 27 = 12 yB = -15 + 12 = -3 y – coordinate: -15 to 12 = 27 B( 4 , -3) Equation (x – 4)2 + (y + 3)2 = 152

45. The point P(2, –3) lies on the circle with centre C as shown. The gradient of CP is –2. What is the equation of the tangent at P? y – b = m(x – a) – 3 1/2 2 Tangent and radius are perpendicular. m1m2 = – 1 y + 3 = 1/2(x –2) mtangent = 1/2

46. Circle C1 has equation (x + 1)2 + (y – 1)2 = 121. A circle C2 with equation x2 + y2 – 4x + 6y + p = 0 is drawn inside C1. The circles have no points of contact. What is the range of values of p? C1: centre (–1 , 1), r = 11 C2: centre (2 , –3), r = √(4 + 9 – p) = √(13 – p) (-1 , 12) (-11 , 1) (10 , 1) (-1 , 1) • (2 , -3) (-1 , -10)