The equation of a circle describes every point equidistant from a fixed centre. At GCSE you use two forms: x² + y² = r² when the centre is the origin, and (x − a)² + (y − b)² = r² for any centre (a, b). Both appear regularly in Paper 3 exams.
Why does the equation of a circle take this form?
The equation comes directly from Pythagoras' theorem. Any point P = (x, y) on a circle of radius r centred at the origin satisfies:
$$x^2 + y^2 = r^2$$
This is simply Pythagoras applied to the right-angled triangle formed by the x-distance, the y-distance, and the line from the origin to P. Every point on the circle is exactly r units from the centre, so the equation captures that condition precisely.
What is the general equation when the centre is not at the origin?
When the centre is at the point (a, b), the horizontal distance from the centre to any point (x, y) is (x − a) and the vertical distance is (y − b). Applying Pythagoras:
$$(x - a)^2 + (y - b)^2 = r^2$$
Key reading rule: the centre coordinates come from reversing the signs inside the brackets. If the equation is (x − 3)² + (y + 5)² = 49, the centre is (3, −5) and the radius is √49 = 7.
| Equation | Centre | Radius |
|---|---|---|
| x² + y² = 25 | (0, 0) | 5 |
| (x − 4)² + (y − 1)² = 16 | (4, 1) | 4 |
| (x + 2)² + (y − 3)² = 36 | (−2, 3) | 6 |
| (x − 7)² + y² = 50 | (7, 0) | √50 = 5√2 |
Notice the last row: when there is no bracket around y, the y-part of the centre is 0.
How do you write the equation given the centre and radius?
Worked example 1: Write the equation of a circle with centre (5, −2) and radius 3.
Substitute directly into (x − a)² + (y − b)² = r²:
$$(x - 5)^2 + (y - (-2))^2 = 3^2$$
$$(x - 5)^2 + (y + 2)^2 = 9$$
Worked example 2: A circle has centre (0, 4) and passes through (3, 0). Find the equation.
First find the radius using the distance formula (Pythagoras): $$r^2 = (3 - 0)^2 + (0 - 4)^2 = 9 + 16 = 25$$
So the equation is: (x − 0)² + (y − 4)² = 25, which simplifies to x² + (y − 4)² = 25.
How do you find where a circle and a straight line intersect?
This is a common GCSE question that combines circle equations with simultaneous equations. The method is substitution:
Worked example: Find the intersections of x² + y² = 25 and y = x + 1.
- Substitute y = x + 1 into the circle equation:
$$x^2 + (x + 1)^2 = 25$$
- Expand:
$$x^2 + x^2 + 2x + 1 = 25$$ $$2x^2 + 2x - 24 = 0$$ $$x^2 + x - 12 = 0$$
-
Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3.
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Find y-values: when x = −4, y = −3; when x = 3, y = 4.
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Intersection points: (−4, −3) and (3, 4).
You can check: 16 + 9 = 25 ✓ and 9 + 16 = 25 ✓.
How do you find the equation of a tangent to a circle?
A tangent at point P on a circle is perpendicular to the radius at P. The method uses the circle theorem "tangent is perpendicular to radius."
Worked example: Find the equation of the tangent to x² + y² = 50 at the point (5, 5).
- Find the gradient of the radius from (0, 0) to (5, 5): gradient = 5/5 = 1.
- The tangent is perpendicular, so its gradient is −1.
- Use y − y₁ = m(x − x₁) with point (5, 5) and gradient −1:
$$y - 5 = -1(x - 5)$$ $$y = -x + 10$$
The tangent is y = −x + 10.
What does the expanded form of the circle equation look like?
Sometimes GCSE questions give the equation in expanded form. You need to recognise it and complete the square to return to the standard form.
Worked example: Show that x² + y² − 6x + 4y − 12 = 0 is a circle, and state its centre and radius.
-
Group x-terms and y-terms: $$(x^2 - 6x) + (y^2 + 4y) = 12$$
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Complete the square: $$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$$ $$(x - 3)^2 + (y + 2)^2 = 25$$
Centre: (3, −2), radius: 5.
Frequently asked questions
Why do I reverse the signs when reading the centre from the equation?
Because the equation is (x − a)² + (y − b)² = r². The centre coordinates are a and b, but they appear with a minus sign in the equation. So (x − 3)² gives centre x-coordinate +3, not −3. Think of it as "what value of x makes the bracket equal zero?" — that is the centre x-value.
Can the equation of a circle have a coefficient in front of x² or y²?
In the standard GCSE form, no — both x² and y² have coefficient 1. If you see 4x² + 4y² = 100, divide everything by 4 to get x² + y² = 25 before reading off the radius. The examiners sometimes present it in scaled form to test this step.
How many marks does this topic typically carry?
Circle equation questions at GCSE Higher are worth 3–5 marks each. Writing the equation from centre and radius is usually 2 marks. Finding an intersection with a line is typically 4–5 marks because it involves a quadratic solve. Finding a tangent is usually 3–4 marks. Completing the square to find centre and radius is 3 marks.
Does this appear on Foundation tier?
No. The equation of a circle is a Higher tier topic only in all major specifications (AQA, OCR, Edexcel). It is one of the topics that can separate A from A* at GCSE.
Let Professor Pi guide you through every step of circle equations in a live conversation — add the AI Tutors connector at aitutors.me.