Equation of a circle: worked examples and method

Pythagoras on a grid

A point (x, y) is a distance r from the origin when x^2 + y^2 = r^2. The distance is the longest side of a right-angled triangle with sides x and y.

Centre at the origin

The circle with centre (0, 0) and radius r is x^2 + y^2 = r^2. The number on the right is r squared, not r.

Move the centre

For the centre (a, b) the equation is (x - a)^2 + (y - b)^2 = r^2. Each bracket subtracts the coordinate of the centre.

Read the signs

Set each bracket to zero to find the centre. The bracket (x + 4) gives x = -4, so the sign you read is the opposite of the sign you see.

Radius from a point

Find the horizontal and vertical steps from the centre to a point on the circle. Square both and add them to get r squared. You do not need to square-root.

On, inside or outside

Substitute a point into the left-hand side. Equal to r squared means on the circle, smaller means inside, larger means outside.

Circles centred on the origin

Take any point (x, y) and draw a straight line from the origin to it. Drop a vertical line to the x-axis and you have a right-angled triangle with shorter sides of length x and y (use the positive lengths; squaring takes care of the signs). Pythagoras says the length of the line from the origin, squared, is x^2 + y^2. If that distance is the radius r, the point is on the circle. So every point of the circle satisfies x^2 + y^2 = r^2, and every point that satisfies it is on the circle.

Reading an equation is the quickest skill to gain. For x^2 + y^2 = 36, the centre is the origin and the radius squared is 36, so the radius is √(36) = 6. For x^2 + y^2 = 20 the radius is √(20), which is 2√(5) in simplest surd form. Leave the radius as a surd unless the question asks for a decimal, because a surd is exact. Writing the equation is the same step in reverse: a circle of radius 9 has r^2 = 81, so its equation is x^2 + y^2 = 81.

Method: 1. Check the equation has the form x^2 + y^2 = number with no brackets: the centre is (0, 0). 2. Square-root the number on the right to get the radius, as a surd if it is not a perfect square. 3. To write an equation, square the radius first, then put it on the right of x^2 + y^2 =.

Tip: If you are told a radius, square it. If you are told an equation, square-root it.

Moving the centre to (a, b)

Sliding a circle sideways or up and down does not change its radius, only the position of its centre. Moving the centre to (a, b) replaces x with x - a and y with y - b, because the triangle that measures the distance now starts at the centre instead of the origin. The equation is (x - a)^2 + (y - b)^2 = r^2. For the centre (4, 1) and radius 3 you write (x - 4)^2 + (y - 1)^2 = 9.

The sign of each bracket is the opposite of the coordinate. Centre (-5, 2) gives (x + 5)^2 + (y - 2)^2. To read an equation, set each bracket to zero: x + 5 = 0 gives x = -5, and y - 2 = 0 gives y = 2. That tells you the centre without any memorising. A bracket that is just x^2 means the x-coordinate of the centre is 0, so x^2 + (y - 3)^2 = 16 has centre (0, 3) and radius 4.

Method: 1. To read the centre, set each bracket equal to zero and solve: (x + 5) gives x = -5. 2. To write the equation, put the opposite sign of each centre coordinate inside its bracket. 3. Whatever the centre is, finish with the radius squared on the right.

Tip: Zero the brackets to find the centre. It works every time and it cannot give you the sign backwards.

Writing the equation from other information

Often you are not given the radius. If you know the centre and one point on the circle, the radius is the distance between them, and you only need its square. Find the horizontal difference and the vertical difference, square both and add. For centre (1, 2) and a point (4, 6) the differences are 3 and 4, so r^2 = 9 + 16 = 25. You never take the square root, because the equation wants r^2 anyway.

If you are given the two ends of a diameter, the centre is the midpoint, found by averaging the x-coordinates and the y-coordinates. Then use the centre and either end to find r^2 as before. Check your answer by substituting the other end of the diameter. If both ends satisfy the equation you wrote, the equation is right. This check takes one line and catches a sign slip.

Method: 1. Find the centre: it is given, or it is the midpoint of a diameter. 2. Work out r^2 as (horizontal difference)^2 + (vertical difference)^2 from the centre to a point on the circle. 3. Write (x - a)^2 + (y - b)^2 = r^2 with the signs reversed inside the brackets. 4. Substitute a second point on the circle to check the equation.

Points on, inside and outside

To test a point, substitute its coordinates into the left-hand side and compare the result with r^2. If the two are equal the point is on the circle. If the left-hand side is smaller than r^2 the point is closer to the centre than the radius, so it is inside. If it is larger, the point is farther away, so it is outside. For x^2 + y^2 = 25, the point (3, 4) gives 25 and is on the circle, (1, 2) gives 5 and is inside, and (5, 5) gives 50 and is outside.

The same idea finds where a circle crosses an axis. On the x-axis y = 0, so substituting gives an equation in x alone. For (x - 2)^2 + y^2 = 25 you set y = 0 and get (x - 2)^2 = 25, so x - 2 = 5 or x - 2 = -5, giving x = 7 or x = -3. Remember both signs when you square-root. A circle can cross an axis twice, touch it once, or miss it completely, depending on whether the number left to square-root is positive, zero or negative.

Method: 1. Substitute the point into the left-hand side of the equation. 2. Compare the value with r^2: equal is on, smaller is inside, larger is outside. 3. For an axis crossing, set y = 0 or x = 0 and solve, keeping both square roots.

Tip: Say the conclusion in words: the point is on, inside or outside the circle. A bare number does not answer the question.