How to find the centre of a circle

Method 1: by construction, with two chords

A chord is a straight line joining two points on a circle. The key fact: the perpendicular bisector of any chord passes through the centre, because every point on it is the same distance from both ends of the chord, and so is the centre.

1. Draw any two chords that aren't parallel.
2. Construct the perpendicular bisector of the first chord: with compasses set to more than half the chord's length, draw arcs from each end on both sides, then join the two crossing points.
3. Do the same for the second chord.
4. The point where the two bisectors cross is the centre.

Two chords are enough, but a third bisector through the same point is a good check on accuracy. If you only need a quick answer for a round object, such as a plate, a shortcut is to draw a chord, mark where a right angle at one end of it meets the circle again, and join that point to the other end of the chord: that line is a diameter, and its midpoint is the centre.

Method 2: from the equation of a circle

A circle with centre (a, b) and radius r has the equation:

So (x − 4)² + (y + 1)² = 25 has centre (4, −1) and radius 5. Watch the signs: (y + 1) means b = −1. A circle centred at the origin, x² + y² = r², is the special case a = b = 0, which is the form GCSE Maths uses.

When the equation is expanded

A-level questions often give the circle expanded, like x² + y² − 8x + 2y − 8 = 0. Complete the square for the x terms and the y terms separately. Our guide to completing the square explains the method.

The centre is (4, −1) and the radius is √25 = 5. A quick check: the centre's coordinates are always minus half the coefficients of x and y. Here −(−8) ÷ 2 = 4 and −(2) ÷ 2 = −1.

Example: another expanded circle

Find the centre and radius of x² + y² + 10x − 4y + 20 = 0. Completing the square: (x + 5)² − 25 + (y − 2)² − 4 + 20 = 0, so (x + 5)² + (y − 2)² = 9. The centre is (−5, 2) and the radius is 3. If the number on the right had come out negative, the equation wouldn't be a real circle at all.

Method 3: from three points on the circle

Three points on a circle are enough to fix it. This is method 1 done with coordinates: find the perpendicular bisectors of two chords and solve for where they meet. Take A(1, 7), B(7, 7) and C(8, 0), the points in the diagram above.

1. Chord AB is horizontal with midpoint (4, 7), so the line through that midpoint at right angles to AB is x = 4.
2. Chord BC has midpoint (7.5, 3.5) and gradient (0 − 7) ÷ (8 − 7) = −7, so its perpendicular bisector has gradient 1/7: y − 3.5 = (x − 7.5) ÷ 7.
3. Where they meet, x = 4, so y = 3.5 + (4 − 7.5) ÷ 7 = 3.5 − 0.5 = 3.
4. The centre is (4, 3). Check: the distance from (4, 3) to A is √(3² + 4²) = 5, and to C is √(4² + 3²) = 5.

The Midpoint and Distance Calculator finds midpoints and distances like these, and our Graphing Functions course covers the coordinate skills behind them.

From the ends of a diameter

If you know the two ends of a diameter, the centre is simply their midpoint. For a diameter from (2, 1) to (8, 9), the centre is ((2 + 8) ÷ 2, (1 + 9) ÷ 2) = (5, 5), and the radius is half the diameter's length: √(6² + 8²) ÷ 2 = 10 ÷ 2 = 5.

Common mistakes

• Getting the signs wrong. (x − 4)² + (y + 1)² = 25 has centre (4, −1), not (−4, 1).
• Forgetting to square root for the radius. If the right-hand side is 25, the radius is 5, not 25.
• Losing a constant when completing the square. Subtract both squared halves, and move the original constant across with the right sign.
• Using parallel chords in the construction: their perpendicular bisectors are the same line, so they don't give a single point.
• Using an ordinary bisector instead of a perpendicular one. It must be at right angles to the chord.

Read next

Maths · TrigonometryUnit circle: chart, radians, sin, cos, tan and how to use itThe unit circle explained: a full chart in degrees and radians, how it gives sin, cos and tan, quadrant signs, reference angles and worked examples.