Teach Me Maths

Equation of a Circle

How to recognise circle equations, find centres and radii, and solve problems involving diameters, chords and tangents.

The equation of a circle

A circle is the set of points a fixed distance from its centre. Pythagoras gives the equation of a circle with centre (a,b) and radius r:

(x-a)^2+(y-b)^2=r^2

For centre (0,0) this becomes:

x^2+y^2=r^2

The right-hand side is the square of the radius. Take its positive square root to get the radius.

Watch the signs: (x+2)^2 means the centre has x-coordinate -2. To check whether a point is on a circle, substitute both coordinates into its equation.

Worked example 1

Find the centre and radius of x^2+y^2=45. Give the radius in exact form.

  1. There are no shifts inside the squares, so the centre is (0,0)
  2. r^2=45, so r=\sqrt{45}

  3. Your turn. Simplify \sqrt{45}

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    \sqrt{45}=\sqrt{9\times5}=3\sqrt5

  4. Answer: centre (0,0), radius 3\sqrt5. The radius is not 45

Explore the centre and radius

Try it: change a from -2 to 2 and watch the sign in the equation. Then change r: doubling the radius multiplies the right-hand side by 4

Worked example 2

A circle has centre C(-2,3) and passes through P(4,-5). Find its equation.

  1. The radius is CP. Its squared length is (4-(-2))^2+(-5-3)^2

  2. Your turn. Work out 6^2+(-8)^2

    r^2=36+64=100

  3. Use the centre (-2,3): (x+2)^2+(y-3)^2=r^2
  4. Answer: (x+2)^2+(y-3)^2=100. Substituting (4,-5) gives 36+64=100, as required.

When the equation is expanded

Complete the square separately for the x terms and the y terms. Move the remaining constant to the other side, then read off the centre and radius.

If the coefficients of x^2 and y^2 are equal but not 1, divide the whole equation by that coefficient first. A positive value of r^2 is needed for a circle of non-zero radius.

Worked example 3

Find the centre and radius of x^2+y^2-6x+8y-11=0.

  1. Group the terms: (x^2-6x)+(y^2+8y)=11
  2. Complete both squares: (x-3)^2-9+(y+4)^2-16=11

  3. Your turn. What is 11+9+16?

    (x-3)^2+(y+4)^2=36

  4. Answer: centre (3,-4) and radius 6

Diameters and chords

The centre is the midpoint of a diameter, and the radius is half its length. If you work with squared lengths, r^2 is one quarter of the diameter squared.

The perpendicular from the centre to a chord bisects the chord. This creates two right-angled triangles, so Pythagoras can find half the chord length. An angle subtended by a diameter at the circumference is 90^\circ

Worked example 4

The endpoints of a diameter are A(-5,1) and B(3,7). Find the equation of the circle.

  1. Your turn. Find the midpoint of AB

    Write your answer as (x, y)

    The centre is \left(\frac{-5+3}{2},\frac{1+7}{2}\right)=(-1,4)

  2. The diameter squared is 8^2+6^2=100. Hence r^2=\frac{100}{4}=25
  3. Answer: (x+1)^2+(y-4)^2=25

Tangents to circles

The tangent at a point on a circle is perpendicular to the radius there. Find the radius gradient from the centre to the point, take the negative reciprocal, then use the point in a line equation.

A horizontal radius gives a vertical tangent; a vertical radius gives a horizontal tangent. There is no need to divide by zero.

Tangents from the same external point are equal in length. The radius to a point of contact forms a right angle, so if the external point is distance d from the centre, the tangent length is \sqrt{d^2-r^2}

Worked example 5

Find the tangent to (x-1)^2+(y+2)^2=25 at P(4,2). Give the equation in the form ax+by=c.

  1. The centre is C=(1,-2). Check P is on the circle: (4-1)^2+(2+2)^2=25
  2. The radius gradient is \frac{2-(-2)}{4-1}=\frac43

  3. Your turn. What is the tangent gradient?

    The tangent gradient is -\frac34

  4. Through P, y-2=-\frac34(x-4). Multiplying by 4 gives 4y-8=-3x+12
  5. Answer: 3x+4y=20

Worked example 6

The line y=3 cuts the circle x^2+y^2=25 at A and B. Find both points and the length of the chord AB.

  1. Substitute y=3: x^2+9=25, so x^2=16

  2. Your turn. Solve x^2=16. Enter both values.

    Give every value, separated by commas

    x=-4 or x=4

  3. The endpoints are (-4,3) and (4,3). Their horizontal distance is 4-(-4)=8
  4. Answer: the points are (-4,3) and (4,3), and AB=8. The chord midpoint (0,3) is directly above the centre, as expected.

Common mistakes

  • Reading the centre with the wrong signs. In (x-a)^2+(y-b)^2=r^2, the centre is (a,b).
  • Giving the squared radius. Take the positive square root to find the radius.
  • Halving the squared diameter. Use r^2=d^2/4, because r=d/2.
  • Finding a radius gradient from the origin. Use the actual centre and the point of contact.
  • Using the radius gradient for the tangent. The tangent is perpendicular to the radius; use the negative reciprocal when defined.
  • Keeping only one intersection. Keep both square-root signs and find each matching coordinate.

Now try it: Equation of a Circle practice questions

More on this topic: Equation of a Circle worksheet with full solutions