Equation of a Circle –

The equation of a circle is a Higher tier GCSE Maths topic that connects algebra and geometry. You need to recognise the standard form, find the radius, and solve problems involving tangents to a circle at a given point.

What Is the Equation of a Circle?

A circle with centre at the origin (0, 0) and radius r has the equation x² + y² = r². Every point (x, y) on the circle satisfies this equation. For example, the equation x² + y² = 25 describes a circle centred at the origin with radius 5, because √25 = 5.

You should be able to identify whether an equation represents a circle, state the radius given the equation, and write the equation given the radius. Common exam tasks include checking whether a point lies on a circle and finding where a line intersects a circle.

A key related skill is finding the tangent to a circle at a given point. The tangent is perpendicular to the radius at that point. So you find the gradient of the radius (from the origin to the point), take the negative reciprocal, and use it to write the equation of the tangent line.

Key Formulas

Step-by-Step Method

1. To identify a circle equation, check that it has x² + y² on one side and a positive constant on the other, with no xy term.
2. To find the radius, take the square root of the constant: r = √(right-hand side).
3. To check if a point lies on the circle, substitute its x and y values — if x² + y² equals r², the point is on the circle.
4. To find the tangent at a point, first calculate the gradient of the radius from (0, 0) to the point.
5. The tangent gradient is the negative reciprocal. Use y - y₁ = m(x - x₁) to write the tangent equation.

Worked Example 1 — Foundation Level

Question: This is a Higher only topic. Here is a basic entry. A circle has the equation x² + y² = 36. State the centre and radius.

Working:

Step 1 — The equation is in the form x² + y² = r², so the centre is at the origin (0, 0).

Step 2 — r² = 36, so r = √36 = 6.

Answer: Centre (0, 0), radius 6.

Worked Example 2 — Higher Level

Question: Does the point (3, 4) lie on the circle x² + y² = 25?

Working:

Step 1 — Substitute x = 3 and y = 4: 3² + 4² = 9 + 16 = 25.

Step 2 — Since 25 = 25, the point satisfies the equation.

Answer: Yes, (3, 4) lies on the circle.

Worked Example 3 — Exam Style

Question: Find the equation of the tangent to the circle x² + y² = 50 at the point (5, -5). (4 marks)

Working:

Step 1 — Check the point is on the circle: 5² + (-5)² = 25 + 25 = 50. Yes.

Step 2 — Find the gradient of the radius from (0, 0) to (5, -5): gradient = (-5 - 0) / (5 - 0) = -1.

Step 3 — The tangent is perpendicular, so its gradient is the negative reciprocal: m = -1/(-1) = 1.

Step 4 — Use y - y₁ = m(x - x₁): y - (-5) = 1(x - 5), so y + 5 = x - 5, which gives y = x - 10.

Answer: y = x - 10

Common Mistakes

• Confusing r and r². If x² + y² = 49, the radius is 7 (not 49). Always take the square root.
• Forgetting the tangent is perpendicular to the radius. Students sometimes use the radius gradient as the tangent gradient instead of the negative reciprocal.
• Assuming the centre is not at the origin. At GCSE, the equation x² + y² = r² always has its centre at (0, 0). If the equation includes (x - a)² + (y - b)² = r², the centre is (a, b) — but this is rare at GCSE.

Exam Tips

• If you see x² + y² = a number, immediately think "circle, centre origin, radius = square root."
• To find where a line intersects a circle, substitute the equation of the line into the circle equation and solve the resulting quadratic.
• For tangent questions, always start by confirming the point lies on the circle.

Practice Questions

Q1 (Higher): A circle has equation x² + y² = 100. State the radius.

Q2 (Higher): Does the point (1, 7) lie on the circle x² + y² = 50?

Q3 (Higher): Find the equation of the tangent to x² + y² = 20 at (4, 2).

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Related Topics

• Finding the Equation of a Line
• Pythagoras' Theorem
• Linear Graphs and Equation of a Line

Summary

• The equation x² + y² = r² represents a circle centred at the origin with radius r.
• Find the radius by square-rooting the right-hand side.
• To check if a point is on the circle, substitute and see if x² + y² equals r².
• The tangent at any point is perpendicular to the radius at that point.
• Use the negative reciprocal of the radius gradient to find the tangent gradient.