Equation of a Circle Centre at the Origin

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

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

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

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

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

Inside / Outside or On Circumference If (x, y) lies on the circumference then (x 2 + (y – b)2 = r 2 – a) When a circle has equation (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 Circumference Consider the circle (x + 1)2 + (y – 2 = 100 4) Determine where the following points lie; 222= 22 circumference 2+ Centre , radius, r = 10 CK CL CM = 50 100 122 K L M K(– 7, 12) =outside = on inside 11 65 C(– 1, 4) ++85122, 2 L(10, 5) , M(4, 9) 2

Another version of Circle Equation is a circle x 2 (+x: y– 2 centre a) + 22 g + x(y+(a– 2 f, b)yb)2+=radius, cr=2 0 r

Equation x 2 + y 2 + 2 gx + 2 fy + c = 0 (a + b)2 Write the equation (x – 5)2 + (y + 3)22 = 49 without = a + 2 ab + b 2 brackets. (x – 5)2 + (y + 3)2 = 49 x 2 – 10 x + 25 + y 2 + 6 y + 9 – 49 = 0 x 2 + y 2 – 10 x + 6 y – 15 = 0 This takes the form given above where 2 g = – 10 , 2 f = 6 and

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

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

We now have 2 ways on finding the centre and 2 + y 2 + 2 gx + 2 fy + c = 0 x radius of a circle 2 depending on the form we have. 2 x + y – 10 x + 6 y – 15 = 0 2 g = – 10 g = – 5 centre = (-g, -f) = (5, -3) 2 f = 6 f=3 c=– 15 radius = (g 2 + f 2 – c) = (25 + 9 – (15)) = 497 =

x 2 + y 2 + 2 gx + 2 fy + c = 0 x 2 + y 2 – 6 x + 2 y – 71 = 0 2 g = -6 g = -3 centre = (-g, -f) = (3, 1) 2 f = 2 f=1 c = -71 radius = (g 2 + f 2 – c) = (9 + 1 – (71)) = 81 = 9

x 2 + y 2 + 2 gx + 2 fy + c = 0 Find the centre & radius of x 2 + y 2 – 10 x + 4 y – 5 =0 x 2 + y 2 – 10 x + 4 y – 5 = 0 2 g = -10 g = -5 centre = (-g, -f) = (5, -2) 2 f = 4 f=2 c = -5 radius = (g 2 + f 2 – c) = (25 + 4 – (5)) 34 =

The circle x 2 + y 2 – 10 x – 8 y + 7 = 0 cuts the y- axis at A & B. Find the length of AB. At A & B x = 0 so the equation becomes – 8 y + 7 = 0 (y – 1)(y – 7) = 0 y = 1 or y = 7 A is (0, 7) & B is (0, 1) So AB = 6 units y 2

Frosty the Snowman’s lower body section can be represented by the equation x 2 + y 2 – 6 x + 2 y – 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 – 6 x + 2 y – 26 = 0 2 g = -6 g = -3 2 f = 2 c = -26 f=1 centre = (-g, -f) = (3, -1) radius = (g 2 + f 2 – c) = (9 + 1 + = 36 26) = 6

(3, 19 )2 radius of head = 1/3 of 6 = 2 6 Using (x – a)2 + (y – b)2 = r 2 (3, 1 6 1) Equation is (x – 3)2 + (y – 19)2 = 4 6 (3, 1)

By considering centres and radii prove that the following two circles touch each other. Circle 1 x 2 + y 2 + 4 x – 2 y – 5 = 0 Circle 2 x 2 + y 2 – 20 x + 6 y + 19 = 0 Circle 1 2 g = = -2 42 fso g =so 2 f = -1 c = -5 centre = (-g, = (-f) radius = (g 2 +2, 1) f 2 – c)= (4 + 1 = + 5) Circle 2 f= 2 g = -20 2 fso= 6 g so = -10 3 c= centre = (-g, = (10, -19 -f) radius = (g 2 +3) f 2 – c)= (100 + 9 – = 90 19) =

(-2, 1) √ 10 (10, -3) 3√ 10 If d is the distance between the centres then by Pythagoras 2 2 2 d = 12 + 4 = 144 + 160 = 16 16 × 10 d = 4 10 radius 1 + radius 2= 10 + 3 10 = 4 10 It now follows that the circles must touch ! = distance between

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

Equations of Tangents At the point of contact a tangent and radius /diameter are perpendicular. Tangen radius t This means we make use of m 1 m 2 = -1.

Prove that the point (-4, 4) lies on the 2 + y 2 – 12 y circle x (-4, 4) must lietangent on the here. circle. Find the So equation of the + 16 = 0 At (-4, 4) x 2 + y 2 – 12 y + = 16 + 16 – 48 + = 0 16 16 (0, 6) x 2 + y 2 – 12 y + 16 = 0 2 g = 0 so g(-4, 4) =0 2 f = -12 so f = -6 Centre is (-g, -f) = (0, 6)

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

Find where the line y = 2 x + 1 meets the circle (x – 4)2 + (y + 1)2 = 20 and comment on the answer Replace y by 2 x + 1 in the circle (x – 4)2 + (y + 1)2 equation = 20 2 becomes (x – 4) + (2 x + 1 + 2 + (2 x + 2)2 2 ( x – 4) 1) = 20 x 2 – 8 x =+ 20 16 + 4 x 2 + 8 x + x 5 x=2 0=one solution tangent 4 = 20 0 Point of contact is (0, 1) Using y = 2 x + 1, if x = 0 then y

Find where the line y = 2 x + 6 meets the circle x 2 + y 2 + 10 x – 2 y + 1 = 0 Replace y by 2 x + 6 in the circle equation x 2 + (2 x + 6)2 + 10 x – 2(2 x + 6) + 1 = 0 x 2 + 4 x 2 + 24 x + 36 + 10 x – 4 x – 12 + 1 = 5 x 02 + 30 x + 25 ( 5 = 0 x 2 + 6 x + 5 ) x = -5 or x = -1 (x= +0 5)(x + 1) =0 Points of contact Using y = 2 x x = -5 y = -4 (-5, -4) and (-1, 4) +6 x = -1 y = 4

Tangency Prove that the line 2 x + y = 19 is a tangent to the circle x 2 + y 2 - 6 x + 4 y - 32 = 0 , and also find the 2 x + y = 19 so ypoint = 19 of – 2 contact. x Replace y by (19 – 2 x) in the circle equation. x 2 + (19 – 2 x)2 – 6 x + 4(19 – 2 x) - 32 = x 02 + 361 – 76 x + 4 x 2 - 6 x + 76 – 8 x – 32 = 05 x 2 – 90 x + 405 = (0 5) Using y = 19 – 2 xx= 9 y = 1 x 2 – 18 x + 81 = x = 9 twice Point of contact 0(x – 9) = 0 tangent is (9, 1)

Using Discriminants At the line x 2 – 18 x + 81 = 0 we can also show there is only one solution by showing that the discriminant is zero. For x 2 – 18 x + 81 = 0 , a =1, b = -18 and c = 9 2 2 Since. So discriminant = 0 the one b – 4 ac (-18) – 4 equation × 1 × 81=has 364 only - 364 = 0 root so there is only = one point of contact so line is a tangent. The next example uses discriminants in a slightly different way.

Find the equations of the tangents to the circle x 2 + y 2 – 4 y – 6 = 0 from the point (0, -8). Each tangent takes the form y = Replace mx – 8 y by (mx – 8) in the circle equation Tangency b 2 x 2 + (mx – 8)2 – 4(mx – 8) –x 26+=m 02 x 2 – 16 mx + 64 – 4 mx + 32 (m 2–+61)=x 02 – 20 mx + 90 = 0

a = (m 2 + 1) Find the equations of the tangents to the circle b = -20 m x 2 + y 2 – 4 y – 6 = 0 from the point (0, -8). c = 90 Equations of tangents y = -3 x – 8 Tangency b 2 (m 2+ 1)x 2 – 20 mxy+= 3 x - 8 90 = 0 (-20 m) 400 m m m=2 =± 2292– 3–– 40 m

Special case

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:

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

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) Make a sketch C (1 , 2) P(-2, -1) Calculate mid-point for centre: Calculate radius CQ: Write down equation: Using (x – a)2 + (y – b)2 = r 2 = 32 + 32 = 18

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

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.

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

Circle P has equation x 2 + y 2 – 8 x – 10 y + 9 = 0. centre (– 2, – 1) and radius 2 2. Circle Q has 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 x 2 + y 2 + 2 gx + 2 fy + c = 0 x 2 + y 2 – 8 x – 10 y + 9 = 0

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

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

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

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) 10 C(24 , 12) r=5 C(24 , 12) r = 10 5 A(-12 , -15)

AC 2 = 362 + 272 10 C(24 , 12) 45 AC 2 = 2025 AC = 45 Diameter of B is 45 – 15 = 30 15 5 15 radius = 15 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 x – coordinate: -12 to 24 = 36 y – coordinate: -15 to 12 = 27 B( 4 , -3) 4/ of 36 = 16 9 4/ of 27 = 12 9 x. B = -12 + 16 = 4 y. B = -15 + 12 = -3 Equation (x – 4)2 + (y + 3)2 = 152

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) Tangent and radius are perpendicular. m 1 m 2 = – 1 mtangent = 1/2 – 3 1/ 2 2 y + 3 = 1/2(x – 2)

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