HIGHER MATHEMATICS Unit 2 - Outcome 4 The Circle

Distance Formula (a,b) (x,y) d (x,b) (x – a) (y – b) By Pythagoras, The distance from (a,b) to (x,y) is given by Proof d 2 = (x-a) 2 + (y-b) 2

Suppose P(x,y) is any point on the circumference of a circle with centre C(a,b) and radius r. X Y P(x,y) C(a,b) r CP = r So CP 2 = r 2 applying the distance formula (x-a) 2 + (y-b) 2 = r 2

The equation of a circle with centre (a,b) and radius r is Example 1 State the centres and radii of the following circles: (x-2) 2 + (y-5) 2 = 49 centre (2,5) radius = 7 (x+5) 2 + (y-1) 2 = 13 centre (-5,1) radius =  13 (x-3) 2 + y 2 = 20 centre (3,0) radius =  20 =  4  5 = 2  5 NOTE The Equation Of A Circle (x-a) 2 + (y-b) 2 = r 2

Example 2 State the equations of the following circles: Centre (2,-3) & radius = 10 (x-2) 2 + (y+3) 2 = 100 Centre (0,6) & radius = 2  3 r 2 = (2  3) 2 = (4)(3) = 12 x 2 + (y-6) 2 = 12

P Q C Centre C = ( (5-1) / 2, (2-6) / 2 ) = (2,-2) r 2 = (5-2) 2 + ( = = 25 Equation is (x-2) 2 + (y+2) 2 = 25 Example 3 Find the equation of the circle that has PQ as diameter where P is (5,2) and Q is (-1,-6). ( (5-1) / 2, (2-6) / 2 ) (5-2) 2 + (2+2) 2

Example 4 Two circles are concentric (they have the same centre). The larger circle has equation (x+3) 2 + (y-5) 2 = 12 The radius of the smaller circle is half that of the larger. Find its equation. Centres are at Larger radius =  = 43= 43 = 2  3 Smaller radius =  3 so r 2 = 3 (x+3) 2 + (y-5) 2 = 3 (-3,5)  12 33

When a circle has equation (x-a) 2 + (y-b) 2 = r 2 If (x,y) lies on the circumference then (x-a) 2 + (y-b) 2 = r 2 If (x,y) lies inside the circumference then (x-a) 2 + (y-b) 2 < r 2 If (x,y) lies outside the circumference then (x-a) 2 + (y-b) 2 > r 2 Inside,Outside or On the Circumference NOTE

Example A circle has equation (x+1) 2 + (y-4) 2 = 100 Determine where the following points lie K(-7,12), L(10,5), M(4,9) At K(-7,12) = (x+1) 2 + (y-4) 2 = (-7+1) 2 + (12-4) 2 = (-6) = So point K is on the circumference. = 100 = (x+1) 2 + (y-4) 2 = (10+1) 2 + (5-4) 2 = = Since 122 > 100 Point L is outside the circumference. At L(10,5) = 122

= (x+1) 2 + (y-4) 2 = (4+1) 2 + (9-4) 2 = = Since 50 < 100 Point M is inside the circumference. At M(4,9) = 50

The General Equation Of A Circle We know, for example, that (x - 4) 2 + (y - 3) 2 = 16 is the equation of a circle Centre (4,3), Radius 4 By multiplying out the brackets, we can alter the appearance of the equation.

(x - 4) 2 + (y - 3) 2 = 16 x 2 - 8x y 2 - 6y + 9 = 16 x 2 + y 2 - 8x- 6y+ 9 = 0 Although it looks different, this equation still represents the same circle. This works the other way, too. x 2 + y 2 + 2x - 6y + 1 = 0 x 2 + 2x + y 2 - 6y + 1 = 0 (x + 1) 2 + (y - 3) = 0 (x + 1) 2 + (y - 3) 2 = 9 Circle, Centre (-1,3), Radius 3 We want to be able to find the radius and centre quickly with our new version of the equation

In general, x 2 + y 2 + (2g)x + (2f)y + c = 0 x 2 + (2g)x + y 2 + (2f)y + c = 0 (x + g) 2 + (y + f) 2 + c - g 2 - f 2 = 0 (x + g) 2 + (y + f) 2 = r 2 (x + g) 2 + (y + f) 2 = g 2 + f 2 - c So, x 2 + y 2 + 2gx + 2fy + c = 0 Represents a circle, centre (-g,-f)  (g 2 + f 2 - c) (x + g) 2 + (y + f) 2 + c = 0- g 2 - f 2 (-g,-f)  (g 2 + f 2 - c), and radius

x 2 + y 2 + 2gx + 2fy + c = 0 where the centre is (-g,-f), and the radius is  (g 2 + f 2 - c) NOTE The General Equation Of A Circle The equation of any circle can be written in the form

Example 1 State the centres and radii of the following circles: x 2 + y x + 8y - 8 = 0 (a) g = 5, f = 4, c = -8 Centre (-5,-4) =  ( ) = 7 =  49 r =  [5²+ 4²- (-8)]

x 2 + y 2 + 6x - 2y - 6 = 0 (b) g = 3, f = -1, c = -6 Centre = (-3,1) =  ( ) x 2 + y 2 - 2x - 14y + 10 = 0 (c) g = -1, f = -7, c = 10 Centre = (1,7) =  ( ) =  16 = 4 = 2  10 =  40 r =  [3²+ (-1)²- (-6)] r =  [(-1)²+ (-7)²- 10]

x 2 + y 2 + 2x - 4y + 11 = 0 (d) g = 1, f = -2, c = 11 Centre = ( -1, 2 ) r =  ( ) To have a valid equation of a circle, g 2 + f 2 - c must be greater than zero. The equation is not a circle, since g 2 + f 2 - c < 0 =  (-6) NOT POSSIBLE r =  [1²+ (-2)²- 11]

EXAMPLE 2 For what values of k does x²+ y²- 2kx - 4y + k² + k – 4 = 0 represent a circle? To represent a circle g² + f² - c > 0 ( g² + f² - c > 0 ) (-k)²+ (-2)² – (k²+ k – 4) > 0 ((k²+ 4 – k²- k + 4 > 0 -k + 8 > 0 ) -k > -8 k < 8 g = -k, f = -2, c = k²+ k - 4

Example 3 Frosty the Snowman’s lower body section can be represented by the equation x 2 + y 2 – 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! x 2 + y 2 – 6x + 2y – 26 = 0 centre (3,-1) radius =  [(-3) – (-26)] =  ( ) =  36 = 6 g = -3, f = 1, c = -26

radius of head = Equation of head is Ha Ha Ha ! (3,-1) 2 (x-3) 2 + (y-19) 2 = 4 (3,19)

Circle 1x 2 + y 2 + 4x - 2y - 5 = 0 Circle 2x 2 + y x + 6y + 19 = 0 Circle 1 g = 2, f = -1, c = -5 centre (-2,1) radius =  [2 2 + (-1) 2 – (-5)] =  ( ) =  10 By considering centres and radii prove that the following two circles touch each other externally. Example 4 Circle 2 g = -10, f = 3, c = 19 centre (10,-3) radius =  [(-10) – 19)] =  ( ) =  90 = 3  10 ie.

Let d be the distance between the centres, then: =  160 =  16  10 = 4  10 =   10 = 4  10 ie. d = [(x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 ]  = [(10 + 2) 2 + (-3 - 1) 2 ]  = ( )  r + r 1 2 Since d = r + r, the circles touch externally 1 2 centre (-2,1) 1 centre (10,-3) 2 r 1 r 2

Find the equation of the tangent to the circle x 2 + y 2 - 4x - 6y + 8 = 0 at the point A(3,5). Centre of circle =C(2,3) A(3,5) Equations of Tangents EXAMPLE 1 C (2,3) g = -2, f = -3, c = 8

m AC = = 2 So, m tang. = -½ y - b = m(x - a) Equation Of Tangent A(3,5) y - 5 = -½(x - 3) 2y - 10 = -x + 3 2y + x = 13 A(3,5) C (2,3) 2y - 10 = -(x – 3)

Prove that the point A(-4,4) lies on the circle x 2 + y 2 – 12y + 16 = 0 And find the equation of the tangent at A. At A(-4,4) = – = 0 So A(-4,4) must lie on the circle. x 2 + y 2 – 12y + 16 = 0 g = 0, f = -6, c = 16 Centre of circle = C(0,6) EXAMPLE 2 = x 2 + y 2 – 12y + 16 = (-4) 2 + (4) 2 – 12(4) + 16

m AC = = ½ So, m tang. =-2 y - b = m(x - a) Equation Of Tangent A(-4,4) y - 4 = -2x - 8 y + 2x = -4 y – 4 = -2(x + 4) A(-4,4) C(0,6)

Calculate the length of the tangent from the point P(8,-3) to the circle with equation x 2 + y 2 – 4x – 6y + 4 = 0 EXAMPLE 3 Centre of circle = C(2,3) x 2 + y 2 – 4x – 6y + 4 = 0 g = -2, f = -3, c = 4 r =  ( ) =  9 = 3 A C (2,3) P (8,-3) 3 ?

PC =  [(8 – 2) 2 + (-3 – 3) 2 ] PC =  ( ) PC =  72 PC = 6  2 6262 AP 2 = (  72) AP 2 = AP 2 = 63 AP =  63 AP = 3  7 3 (2,3) (8,-3) C P A

The circle below, with centre C, has the equation x 2 + y 2 – 12x + 4y + 20 = 0 EXAMPLE 4 A P (8,2) C (a) Find the equation of the tangent to the circle at the point P. (b) Find the coordinates of A, the point where the tangent cuts the x-axis. (c) Find the equation of the circle which passes through the points P, A and C.

A P (8,2) C x 2 + y 2 – 12x + 4y + 20 = 0 g = -6, f = 2, c = -20 Centre of circle = C(6,-2) (a) m C P = = 2 So, m tang. =-½-½ y - b = m(x - a) Equation Of Tangent 2y - 4 = -(x – 8) 2y – 4 = -x + 8 y – 2 = -½(x - 8) 2y + x = 12

A P (8,2) C 2y + x = 12 At y = 0 x = 12 (b) A (12,0)

A P (8,2) C (c) A P C Triangle CPA is right angled at P, so the line CA must be adiameter. = (9,-1) r 2 = (5-2) 2 + ( = = 10 Equation is (x-9) 2 + (y+1) 2 = 10 (12-9) 2 + (0+1) 2 Centre = ( (5-1) / 2, (2-6) / 2 ) ( (6+12) / 2, (-2+0) / 2 ) (12,0) (6,-2)

There are 3 possible scenarios 2 points of contact1 point of contact0 points of contact ie. line is a tangent Intersection of a line and a circle NOTE

EXAMPLE 1 Find the coordinates of the points of intersection between the line 5y – x + 7 = 0 and the circle x 2 + y 2 + 2x – 2y - 11 = 0. 5y – x + 7 = 0  x = 5y + 7 x 2 + y 2 + 2x – 2y - 11 = 0 (5y + 7) 2 + y 2 + 2(5y + 7) – 2y - 11 = 0 25y y y y y - 11 = 0 26y y+ 52 = 0 y 2 + 3y + 2 = 0  26 (y + 1)(y + 2)= 0 y = -1 or y = -2 x = 5y + 7 At y = -1 x = 5(-1) + 7 At y = -2 Points of contact: (2,-1), (-3,-2) x = 2 x = 5(-2) + 7 x = -3

If the quadratic equation formed after substituting gives: 2 distinct, real roots b 2 – 4ac > 0 Line meets circle at two points No real roots b 2 – 4ac < 0 Line does not meet the circle Equal roots (one repeated real root) b 2 – 4ac = 0 Line meets circle at one point Line is a tangent to the circle NOTE Circles, lines and the discriminant (b 2 – 4ac)

EXAMPLE 2 Prove that the line y = x + 6 is a tangent to the circle x 2 + y 2 + 2x – 2y – 6 = 0. x 2 + y 2 + 2x – 2y - 6 = 0 x 2 + (x + 6) 2 + 2x – 2(x + 6) - 6 = 0 + x x + 36x2x2 + 2x - 2x – = 0 2x x+ 18 = 0 x 2 + 6x + 9 = 0 (x + 3)(x + 3) = 0 x = -3 = 6 2 – 4(1)(9) b 2 – 4ac or Only 1 point of contact Since b 2 – 4ac = 0 Line is a tangent to the circle. = 0 So line is a tangent to the circle. Use this method if you need the point(s) of intersection!

EXAMPLE 3 Find the values of k, given that the line y = x + k is a tangent to the circle x 2 + y 2 = 8. x 2 + y 2 = 8 x 2 + (x + k) 2 = 8 + x 2 + 2kx + k 2 = 8x2x2 2x 2 + 2kx + k 2 – 8 = 0 a = 2b = 2kc = k For tangency, b 2 – 4ac = 0 b 2 – 4ac = 0 (2k) 2 – 4(2)(k 2 – 8) = 0 4k 2 – 8k = 0 -4k = 0 -4k 2 = -64 k 2 = 16 k =  4

Bonus Question Find the equations of the tangents to the circle x 2 + y 2 – 4y – 6 = 0 from the point (0,-8). x 2 + y 2 – 4y – 6 = 0 g = 0, f = -2, c = -6 Centre (0,2) Y Each tangent takes the form y = mx - 8 x 2 + y 2 – 4y – 6 = 0 x 2 + (mx – 8) 2 – 4(mx – 8) – 6 = 0 x 2 + m 2 x 2 – 16mx + 64 – 4mx + 32 – 6 = 0 (m 2 + 1)x 2 – 20mx + 90 = a = (m 2 + 1)b = -20mc = 90 For tangency, b 2 – 4ac = 0

b 2 – 4ac = 0 (-20m) 2 – 4(m 2 + 1)(90) = 0 400m 2 – 360m 2 – 360 = 0 40m 2 – 360 = 0 40m 2 = 360 m 2 = 9 m = 3 or -3 So the two tangents are y = -3x – 8 and y = 3x - 8