Equation of a Circle

The number on the right-hand side of x^2+y^2=r^2 is the radius squared.

1. A circle with centre (0, 0) and radius r has equation x^2+y^2=r^2
2. r^2=9^2=81
3. Answer: x^2+y^2=81

Compare with (x-a)^2+(y-b)^2=r^2: x^2 is the same as (x-0)^2.

1. Compare with (x-a)^2+(y-b)^2=r^2
2. x^2=(x-0)^2, so a=0
3. (y+5)^2=(y-(-5))^2, so b=-5
4. Answer: (0, -5)

The right-hand side is the radius squared, so square root it and simplify the surd.

1. The right-hand side is r^2, so r^2=72
2. r=\sqrt{72}
3. 72=36\times2, so \sqrt{72}=\sqrt{36}\times\sqrt2
4. Answer: 6\sqrt2

The radius is the distance from the centre to the point on the circle.

1. r^2=(2-(-1))^2+(5-3)^2
2. r^2=9+4=13
3. (x-a)^2+(y-b)^2=r^2 with a=2, b=5
4. Answer: (x-2)^2+(y-5)^2=13

Work out (x-2)^2+(y+1)^2 for each point and compare it with 20.

1. Substitute each point into (x-2)^2+(y+1)^2
2. (-3, 2): 25+9=34, more than 20, so outside
3. (6, 1): 16+4=20, so on the circle
4. (4, -4): 4+9=13, less than 20, so inside
5. (5, 3): 9+16=25, more than 20, so outside
6. Answer: (4, -4)

Use the circumference to find the radius, then remember that k is the radius squared.

1. Circumference =2\pi r=12\pi
2. So r=6
3. k=r^2=6^2
4. Answer: k=36

If the circle just touches the y-axis, the radius is the distance from the centre to the y-axis.

1. The distance from (-4, 3) to the y-axis is 4 (the size of the x-coordinate).
2. The circle touches the y-axis, so the radius is 4 (not 3, which would be the distance to the x-axis).
3. (x-a)^2+(y-b)^2=r^2 with a=-4, b=3, r=4
4. (x+4)^2+(y-3)^2=4^2
5. Answer: (x+4)^2+(y-3)^2=16

Substitute x=a and y=5 into the equation, then remember that a square has a positive and a negative square root.

1. Substitute: (a+4)^2+(5-2)^2=34
2. (a+4)^2+9=34, so (a+4)^2=25
3. a+4=5 or a+4=-5
4. Answer: a=1 or a=-9

OP is a radius of length 6: use \cos30^\circ for the x-coordinate and \sin30^\circ for the y-coordinate.

1. OP is a radius, so OP=\sqrt{36}=6
2. x=6\cos30^\circ=6\times\dfrac{\sqrt3}{2}=3\sqrt3
3. y=6\sin30^\circ=6\times\dfrac12=3
4. Answer: P=(3\sqrt3, 3)

The tangent at P is perpendicular to the radius from the centre to P.

1. The centre is (-1, 2)
2. Gradient of the radius from (-1, 2) to P(1, 8): \dfrac{8-2}{1-(-1)}=\dfrac62=3
3. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of 3.
4. Answer: -\dfrac13

The centre is the midpoint of AB, and the radius is half the length of AB.

1. Centre = midpoint of AB: \left(\dfrac{-2+6}{2}, \dfrac{3+(-1)}{2}\right)=(2, 1)
2. AB^2=8^2+4^2=80, so AB=\sqrt{80}
3. The radius is half the diameter: r^2=\left(\dfrac{\sqrt{80}}{2}\right)^2=\dfrac{80}{4}=20
4. Check: from (2, 1) to A(-2, 3), r^2=4^2+2^2=20
5. Answer: (x-2)^2+(y-1)^2=20

The tangent at P is perpendicular to the radius OP.

1. Gradient of OP=\dfrac{6-0}{-3-0}=-2
2. The tangent is perpendicular to OP, so its gradient is \dfrac12
3. y-6=\dfrac12(x+3)
4. Multiply by 2: 2y-12=x+3
5. Answer: x-2y=-15

AB is a chord, and the centre lies on the perpendicular bisector of a chord.

1. AB is horizontal, so its perpendicular bisector is the vertical line through its midpoint
2. Midpoint of AB: x=\dfrac{-3+5}{2}=1, so the centre lies on x=1
3. The centre also lies on y=x, so the centre is (1, 1)
4. r^2=(1-(-3))^2+(1-4)^2=16+9=25
5. Answer: (x-1)^2+(y-1)^2=25

The centre is (5, -4): find the gradient of the radius to P, then the tangent is perpendicular to it.

1. The centre is (5, -4)
2. Gradient of the radius: \dfrac{-1-(-4)}{9-5}=\dfrac34
3. Gradient of the tangent: -\dfrac43
4. Tangent: y+1=-\dfrac43(x-9), which is y=-\dfrac43x+11
5. At the y-axis x=0, so y=11
6. Answer: Q=(0, 11)

Join the centre C to P and to T: the angle between a tangent and a radius is 90^\circ, so use Pythagoras in triangle CPT.

1. The centre is C(2, -3) and the radius is 5
2. CT^2=(8-2)^2+(5-(-3))^2=36+64=100
3. Angle CPT=90^\circ (tangent and radius), so TP^2=CT^2-CP^2
4. TP^2=100-25=75
5. TP=\sqrt{75}=\sqrt{25}\times\sqrt3
6. Answer: 5\sqrt3

A radius drawn to the point where the circle touches the x-axis is vertical, so the centre is directly above (3, 0). Answer part (a) before attempting part (b).

Part (a)

1. The radius to (3, 0) is perpendicular to the x-axis, so the centre is (3, r), where r is the radius
2. The distance from (3, r) to (-1, 8) is r: 4^2+(8-r)^2=r^2
3. 16+64-16r+r^2=r^2
4. 80=16r, so r=5 and the centre is (3, 5)
5. Answer: (x-3)^2+(y-5)^2=25

Part (b)

1. On the y-axis x=0: 9+(y-5)^2=25
2. (y-5)^2=16, so y=1 or y=9
3. Distance =9-1
4. Answer: 8

The tangent is perpendicular to the radius OP, so you know the gradient of OP. Answer part (a) before attempting part (b).

Part (a)

1. The radius OP is perpendicular to the tangent, so the gradient of OP is 1
2. So P lies on the line y=x
3. Substitute into the circle: x^2+x^2=50, so x^2=25
4. P is in the first quadrant, so x=5 and y=5
5. Answer: P=(5, 5)

Part (b)

1. Tangent through (5, 5) with gradient -1: y=-x+10
2. So A=(10, 0) and B=(0, 10)
3. Area of triangle OAB=\dfrac12\times10\times10=50
4. The radius is \sqrt{50}, so the quarter circle has area \dfrac14\times\pi\times50=12.5\pi
5. Shaded area =50-12.5\pi=10.730\ldots
6. Answer: 10.7

Join the centre O to M: the radius through the midpoint of a chord meets the chord at right angles.

Part (a)

1. The centre of the circle is O(0, 0)
2. The line from the centre to the midpoint of a chord is perpendicular to the chord.
3. Gradient of OM=\dfrac{-4-0}{4-0}=-1
4. So the gradient of AB is the negative reciprocal: 1
5. AB passes through M(4, -4): y-(-4)=1(x-4)
6. Answer: y=x-8

Part (b)

1. Method 1 (Pythagoras): OM^2=4^2+4^2=32 and OA^2=r^2=130
2. Triangle OMA has a right angle at M, so MA^2=130-32=98 and MA=\sqrt{98}=7\sqrt2
3. M is the midpoint, so AB=2\times7\sqrt2=14\sqrt2
4. Method 2 (intersection): substitute y=x-8 into x^2+y^2=130: 2x^2-16x+64=130
5. x^2-8x-33=0, so (x-11)(x+3)=0, giving A(11, 3) and B(-3, -11)
6. AB=\sqrt{14^2+14^2}=\sqrt{392}=14\sqrt2
7. Answer: 14\sqrt2

The tangent at a point is perpendicular to the radius to that point. Answer part (a) before attempting part (b).

Part (a)

1. The centre is C(-1, 2)
2. Gradient of CP=\dfrac{3-2}{2-(-1)}=\dfrac13
3. The tangent is perpendicular to CP, so its gradient is -3
4. y-3=-3(x-2)
5. Answer: y=-3x+9

Part (b)

1. P and Q are reflections of each other in the line x=-1, which passes through C
2. So T also lies on x=-1: y=-3(-1)+9=12, so T=(-1, 12)
3. Check: gradient of CQ=-\dfrac13, so the tangent at Q is y=3x+15
4. At x=-1 this also gives y=12
5. CPTQ is a kite with diagonals CT=12-2=10 and PQ=2-(-4)=6
6. Area =\dfrac12\times10\times6
7. Answer: 30

Rearrange the line to make y the subject and substitute into the circle; then check whether the line passes through the centre. Answer part (a) before attempting part (b).

Part (a)

1. Rearrange the line: y=\dfrac{-2-3x}{4}, so y-1=\dfrac{-6-3x}{4}=-\dfrac34(x+2)
2. Substitute: (x+2)^2+\dfrac{9}{16}(x+2)^2=25
3. \dfrac{25}{16}(x+2)^2=25, so (x+2)^2=16
4. x+2=4 or x+2=-4, so x=2 or x=-6
5. x=-6: y=\dfrac{-2+18}{4}=4
6. Answer: A=(-6, 4)

Part (b)

1. x=2: y=\dfrac{-2-6}{4}=-2
2. Answer: B=(2, -2)

Part (c)

1. The line passes through the centre (-2, 1), since 3(-2)+4(1)=-2, so AB is a diameter
2. The angle in a semicircle is 90^\circ, so angle ADB=90^\circ
3. AD=\sqrt{9^2+3^2}=\sqrt{90} and BD=\sqrt{1^2+3^2}=\sqrt{10}
4. Area =\dfrac12\times\sqrt{90}\times\sqrt{10}=\dfrac12\times\sqrt{900}
5. Answer: 15