Circle theorems appear on the Higher tier of every GCSE board — Edexcel, AQA, OCR — as part of geometry and measures. There are only a handful of theorems to learn, but questions combine them, sometimes with two or three theorems needed in sequence, and mark schemes are strict about the reasoning you write alongside each angle you find.
This guide states each theorem clearly, gives the exact reason wording that scores marks, and works through a multi-step angle-chasing example of the kind that appears near the end of a Higher paper.
The theorems, one by one
1. Angle in a semicircle
The angle in a semicircle is always 90°. If AB is a diameter and C is any other point on the circle, angle ACB = 90°. The reason to write is: 'angle in a semicircle is 90°'.
2. Angle at the centre
The angle at the centre of a circle is twice the angle at the circumference, when both angles are subtended by the same arc. The reason to write is: 'angle at the centre is twice the angle at the circumference'.
3. Angles in the same segment
Angles subtended by the same arc, at the circumference, on the same side of it, are equal. The reason to write is: 'angles in the same segment are equal'.
4. Cyclic quadrilaterals
Opposite angles in a cyclic quadrilateral (a quadrilateral with all four vertices on a circle) sum to 180°. The reason to write is: 'opposite angles in a cyclic quadrilateral sum to 180°'.
5. Tangent and radius
A tangent to a circle meets the radius at that point at exactly 90°. The reason to write is: 'tangent meets radius at 90°' (or 'angle between tangent and radius is 90°').
6. Two tangents from an external point
If two tangents are drawn from the same external point to a circle, they are equal in length, and the line from the external point to the centre bisects the angle between the tangents. The reason to write is: 'tangents from a point to a circle are equal in length'.
7. Alternate segment theorem
The angle between a tangent and a chord drawn from the point of contact equals the angle in the alternate segment (the angle subtended by the chord on the far side of the circle). The reason to write is: 'alternate segment theorem'.
| Theorem | Reason wording |
|---|---|
| Angle in a semicircle | Angle in a semicircle is 90° |
| Angle at the centre | Angle at the centre is twice the angle at the circumference |
| Angles in the same segment | Angles in the same segment are equal |
| Cyclic quadrilateral | Opposite angles in a cyclic quadrilateral sum to 180° |
| Tangent and radius | Tangent meets radius at 90° |
| Two tangents from a point | Tangents from a point to a circle are equal in length |
| Alternate segment theorem | Alternate segment theorem |
General angle-chasing principles that combine with the theorems
- Angles on a straight line sum to 180°; angles around a point sum to 360°.
- Angles in a triangle sum to 180°; base angles of an isosceles triangle are equal — and a radius to any point on the circle is always the same length, so any triangle formed by two radii and a chord is isosceles.
- Vertically opposite angles are equal.
- Alternate and corresponding angles are equal where lines are parallel — occasionally relevant when a tangent is given as parallel to a chord.
Worked example: a multi-step angle chase
A, B, C and D lie on a circle with centre O. AC is a diameter. Angle BAC = 34°. Find angle BDC, giving reasons.
- Since AC is a diameter, angle ABC = 90° (angle in a semicircle).
- In triangle ABC, angles sum to 180°: angle ACB = 180° − 90° − 34° = 56°.
- Angle BDC and angle BAC are both subtended by arc BC from the circumference, on the same side, so angle BDC = angle BAC = 34° (angles in the same segment).
Notice the structure: one circle theorem opens the question (semicircle), ordinary angle facts do the middle step (angle sum of a triangle), and a second circle theorem finishes it (same segment). Most higher-mark circle theorem questions are built this way — a chain of two or three facts, not one big theorem in isolation.
Worked example: the alternate segment theorem
PT is a tangent to a circle at point T. TQ is a chord, and angle PTQ = 62°. R is a point on the circle in the alternate segment. Find angle TRQ.
- The angle between the tangent PT and the chord TQ is 62°.
- By the alternate segment theorem, this equals the angle subtended by TQ in the alternate segment.
- So angle TRQ = 62° (alternate segment theorem).
Common mistakes
- Assuming any triangle with two sides drawn from the centre is equilateral rather than isosceles — it is isosceles because both sides are radii, not necessarily equilateral.
- Applying 'angle at centre = 2 × angle at circumference' to angles that are not subtended by the same arc.
- Forgetting that a cyclic quadrilateral's rule applies to opposite angles, not adjacent ones.
- Missing that a diagram implies a diameter (and therefore a right angle) without it being labelled explicitly — check whether two points and the centre are stated to be collinear.
- Giving only a numerical answer where the question explicitly asks for reasons, losing marks that were entirely available for a sentence.
How to revise this topic
- Write all seven theorems and their exact reason wording from memory, without notes, until you can do it without hesitation.
- Practise identifying which theorem applies from a diagram alone, before calculating anything.
- Work through multi-step past paper questions, writing every reason in full sentence form exactly as the mark scheme would expect.
- Revisit any question where you got the right angle but the wrong (or missing) reason — that gap is just as important as a wrong number.