A circle with centre (a, b) and radius r has equation (x − a)² + (y − b)² = r², which is the distance formula squared. When the equation is given expanded, as x² + y² + 2fx + 2gy + c = 0, complete the square in x and in y to recover the centre and radius. The angle in a semicircle is a right angle, which gives a quick way to spot or prove a diameter.
Circle equations
- Centre (a, b), radius r: (x − a)² + (y − b)² = r²
- Expanded form: x² + y² + 2fx + 2gy + c = 0
- Point inside the circle: (x − a)² + (y − b)² < r²; outside: > r²
The centre has the opposite sign to the numbers in the brackets, and the right-hand side is r², not r.
What is the equation of a circle?
Every point on a circle is the same distance, r, from the centre (a, b). Applying the distance formula and squaring both sides gives (x − a)² + (y − b)² = r².
Reading it back needs care with two traps. The centre has the opposite sign to the number in the bracket, so (x + 4)² means a = −4. And the right-hand side is the radius squared, so 25 means a radius of 5.
To test where a point lies, substitute it into the left-hand side: less than r² means inside, equal means on the circle, greater means outside.
Completing the square to find the centre and radius
Expanding the brackets gives the form x² + y² + 2fx + 2gy + c = 0, which hides the centre and radius. You can recognise it because the x² and y² coefficients are equal and there is no xy term.
Complete the square in x and in y separately: halve the coefficient of x, square it and subtract it, and do the same for y. Collect the constants on the right, then read off the centre and radius.
Finding a circle from a diameter or three points
Given the ends of a diameter, the centre is the midpoint and the radius is half the length. Given three points on the circle, the centre is where the perpendicular bisectors of two chords cross, because it must be equidistant from all three.
Some questions want the answer in expanded form with integer coefficients, so be ready to multiply out.
The angle in a semicircle is a right angle
If BC is a diameter and A is any other point on the circle, then angle BAC = 90°. In coordinates, that means the chords AB and AC are perpendicular, so their gradients multiply to −1.
It also works in reverse: if you can show the angle at A is 90°, then BC must be a diameter, which gives you the centre and radius straight away. Watch for the word “diameter” in a question.
Where does a circle cross the axes?
Put x = 0 into the equation to find where it crosses the y-axis, or y = 0 for the x-axis, and solve the resulting quadratic, keeping the ±. A circle in expanded form passes through the origin only if c = 0. A quick sketch with the centre marked tells you whether your answers are plausible.
Worked examples
Example 1. Write down an equation of the circle with centre (3, −2) and radius 5.
- Use (x − a)² + (y − b)² = r² with a = 3, b = −2, r = 5.
- (x − 3)² + (y − (−2))² = 5².
- The centre's −2 becomes +2 in the bracket, and the radius is squared.
Answer: (x − 3)² + (y + 2)² = 25.
Marks: B1 for the brackets, B1 for r².
Example 2. The circle C has equation x² + y² − 6x + 4y − 12 = 0. Find its centre and radius.
- Group the terms: (x² − 6x) + (y² + 4y) − 12 = 0.
- x² − 6x = (x − 3)² − 9.
- y² + 4y = (y + 2)² − 4.
- So (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0, giving (x − 3)² + (y + 2)² = 25.
Answer: Centre (3, −2), radius 5.
Marks: M1 for completing both squares, A1 for the centre, A1 for the radius.
Example 3. A(−3, 1) and B(3, −7) are the ends of a diameter of a circle. Find its equation.
- Centre = midpoint of AB = (0, −3).
- AB = √(6² + 8²) = 10, so the radius is 5.
- Equation: x² + (y + 3)² = 25. Check with A: 9 + 16 = 25.
Answer: x² + (y + 3)² = 25.
Marks: M1 for the centre, M1 for the radius, A1 for the equation.
Example 4. B(0, 6) and C(4, −1) are the ends of a diameter, and A(−2, 3) is on the circle. Show that angle BAC = 90°.
- Gradient of AB: (6 − 3) / (0 − (−2)) = 3/2.
- Gradient of AC: (−1 − 3) / (4 − (−2)) = −4/6 = −2/3.
- Product: 3/2 × (−2/3) = −1, so AB is perpendicular to AC and angle BAC = 90°.
Marks: M1 for both gradients, A1 for the product −1, A1 for the conclusion.
Example 5. Find where the circle (x − 3)² + (y − 2)² = 25 crosses the y-axis.
- On the y-axis x = 0: 9 + (y − 2)² = 25.
- (y − 2)² = 16, so y − 2 = ±4.
- y = 6 or y = −2.
Answer: (0, 6) and (0, −2).
Marks: M1 for substituting, M1 for the ±, A1 for both points.
Common mistakes
- Getting the centre's signs wrong. (x − 3)² + (y + 2)² = 25 has centre (3, −2), not (−3, 2).
- Giving r² as the radius. If the right-hand side is 25, the radius is 5.
- Forgetting to subtract the squared term. When completing the square, x² − 6x = (x − 3)² − 9. Leaving out the −9 gives the wrong radius.
- Dropping the ± when solving. Crossing points come in pairs: (y − 2)² = 16 gives y = 6 and y = −2.
- Mixing up a chord and a diameter. The semicircle rule only applies when the chord is a diameter.
How the equation of a circle is examined
Circles are Pure 2 content in Edexcel International A-Level (paper WMA12), in the coordinate geometry chapter, and part of first-year Pure in Edexcel A-Level (9MA0). Typical questions:
- write the equation from a centre and radius
- complete the square to find the centre and radius
- find a circle from a diameter, or from three points
- use the semicircle property to prove a right angle or identify a diameter
Try the Pure 2 coordinate geometry exam questions, which have full mark schemes, and the Pure 2 coordinate geometry worksheet. The coordinate geometry revision engine generates unlimited circle questions.
Next: tangents and chords. If completing the square feels rusty, revisit the completing the square course.
Frequently asked questions
How do you find the centre and radius of a circle?
If the equation is (x − a)² + (y − b)² = r², the centre is (a, b) and the radius is r. If it is expanded, complete the square in x and y first.
How can you tell an equation is a circle?
The x² and y² terms have equal coefficients, and there is no xy term. After completing the square, the right-hand side must be positive.
Can r² be negative?
No. If completing the square gives a negative right-hand side, the equation does not represent a circle.