Circle Theorems and Geometric Proof Revision Notes | GCSE AQA Higher Maths

1. Why Circle Theorems Matter

Circle theorems are one of the highest-value topics on the Higher-tier paper. A single diagram can unlock four or five marks — but only if you know which theorem to reach for and how to write a watertight reason.

There are eight theorems to master. Each one is a provable fact about angles, radii, tangents, and chords. This lesson teaches you every theorem, shows you how to prove them, and trains you to chain them together the way the examiner expects.

2. The Eight Circle Theorems — A Reference Bank

Before working through proofs and examples, here is every theorem in one place. Return to this bank whenever you need a quick reminder during practice.

3. Theorems 1–4: Angles and Arcs

The first group of theorems all concern angles formed by arcs and chords inside the circle. They share a common root: every radius is equal, so triangles formed with two radii are always isosceles.

To see why this is true, draw radius $OC$. Triangle $OAC$ is isosceles (since $OA = OC$ = radius), so the base angles are equal: call them both $\alpha$. The exterior angle of triangle $OAC$ at $O$ is $2\alpha$. The same argument on triangle $OBC$ gives exterior angle $2\beta$ at $O$. Adding: $\angle AOB = 2\alpha + 2\beta = 2(\alpha + \beta) = 2\angle ACB$.

The angle in a semicircle is a direct consequence: when $AB$ is a diameter, $\angle AOB = 180^\circ$, so $\angle ACB = 90^\circ$. This is the result the examiner tests most often — and the one students most often forget to justify with a reason.

Angles in the same segment follow directly from the angle at the centre theorem. If two angles at the circumference are both subtended by the same arc, they are each half the central angle — so they must be equal to each other.

4. Theorems 5–7: Tangents, Radii, and Chords

The next group connects the geometry of tangents and chords to the centre. These theorems are frequently tested in multi-step problems alongside the angle theorems above.

The tangent–radius theorem states that the tangent at any point on a circle is perpendicular to the radius at that point. The proof uses the fact that the radius is the shortest distance from the centre to the tangent line — any other line from the centre to the tangent would be longer than the radius, so the perpendicular must be the radius itself.

Tangents from an external point are equal in length. If $P$ is outside the circle and $PA$, $PB$ are tangents to the circle at $A$ and $B$ respectively, then $PA = PB$. The proof uses congruent triangles: $OA = OB$ (radii), $OP$ is common, and $\angle OAP = \angle OBP = 90^\circ$ (tangent–radius). By RHS congruence, triangles $OAP$ and $OBP$ are congruent, so $PA = PB$.

The perpendicular from the centre to a chord bisects the chord. Draw the perpendicular from centre $O$ to chord $AB$, meeting it at $M$. Triangles $OAM$ and $OBM$ share $OM$, have $OA = OB$ (radii), and both have a right angle at $M$. By RHS congruence, $AM = MB$.

5. Theorem 8: The Alternate Segment Theorem

The alternate segment theorem is the one that trips students up most often — partly because the name sounds abstract, and partly because the diagram looks busy. Here is how to read it clearly.

Draw a tangent to a circle at point $T$, and draw a chord $TQ$ from the same point. The chord divides the circle into two segments. The angle between the tangent and the chord on one side equals the inscribed angle in the other (alternate) segment — the angle subtended by chord $TQ$ at any point on the arc on the far side.

To prove the alternate segment theorem, let the angle between the tangent and chord $TQ$ be $\theta$. Draw the diameter from $T$ through $O$ to point $S$ on the circle. The tangent is perpendicular to $TS$, so the angle between the tangent and $TS$ is $90^\circ$. The angle $\angle QTS = 90^\circ - \theta$ (angles on the tangent line). Since $TS$ is a diameter, $\angle TQS = 90^\circ$ (angle in a semicircle). In triangle $TQS$, the angles sum to $180^\circ$, giving $\angle QST = \theta$. By the angles in the same segment theorem, any inscribed angle subtending arc $TQ$ in the alternate segment also equals $\theta$.

6. Worked Examples: Applying and Chaining Theorems

Knowing each theorem individually is only half the skill. Exam questions almost always require you to chain two or three theorems together, giving a clear reason at every step. The worked examples below model exactly how to do this.

Worked example: Finding an Angle Using the Angle at the Centre Theorem

O is the centre of a circle. Points A, B, and C lie on the circumference. Angle ACB = 38°. Find angle AOB. [2 marks]

1. Both $\angle ACB$ and $\angle AOB$ are subtended by arc $AB$. $C$ is on the circumference; $O$ is the centre. — This matches the angle at the centre theorem exactly. [M1 for identifying the correct theorem]
2. $\angle AOB = 2 \times \angle ACB = 2 \times 38^\circ = 76^\circ$ — The central angle is twice the inscribed angle subtending the same arc. [A1 for correct answer with reason]

Answer: $\angle AOB = 76^\circ$ (angle at the centre is twice the angle at the circumference)

Worked example: Cyclic Quadrilateral — Finding Two Angles

ABCD is a cyclic quadrilateral. Angle DAB = 112° and angle BCD = x. Find x and angle ABC if angle ADC = 74°. [3 marks]

1. $\angle DAB + \angle BCD = 180^\circ$ $112^\circ + x = 180^\circ$ $x = 68^\circ$ — Opposite angles in a cyclic quadrilateral sum to 180°. [M1 for correct theorem, A1 for x = 68°]
2. $\angle ABC + \angle ADC = 180^\circ$ $\angle ABC + 74^\circ = 180^\circ$ $\angle ABC = 106^\circ$ — ABC and ADC are the other pair of opposite angles. [A1 for 106°]

Answer: $x = 68^\circ$, $\angle ABC = 106^\circ$

Worked example: Alternate Segment Theorem — Multi-Step Problem

A tangent to a circle at point T makes an angle of 55° with chord TQ. Point P lies on the major arc TQ. Find angle TPQ. [2 marks]

1. The angle between the tangent and chord $TQ$ is $55^\circ$. Point $P$ is in the major arc, which is the alternate segment to the angle of $55^\circ$. — The alternate segment is on the opposite side of chord TQ from the tangent angle. [M1 for correct identification]
2. $\angle TPQ = 55^\circ$ — By the alternate segment theorem, the inscribed angle in the alternate segment equals the tangent–chord angle. [A1]

Answer: $\angle TPQ = 55^\circ$ (alternate segment theorem)

Worked example: Prove: Tangents from an External Point are Equal

O is the centre of a circle. PA and PB are tangents from external point P to the circle, touching at A and B respectively. Prove that PA = PB. [4 marks]

1. $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ — The tangent at any point is perpendicular to the radius at that point. [M1 — must state this reason]
2. $OA = OB$ (both radii of the same circle) — [M1 — must state 'radii of the same circle']
3. $OP$ is common to both triangles $OAP$ and $OBP$. — This gives us three pieces of information about the two triangles.
4. Triangles $OAP$ and $OBP$ are congruent by RHS (right angle, hypotenuse $OP$, side $OA = OB$). Therefore $PA = PB$. $\square$ — [A1 for RHS congruence stated, A1 for conclusion PA = PB with QED or equivalent]

Answer: $PA = PB$ — proved by RHS congruence of triangles $OAP$ and $OBP$.

7. Practice: Try These Yourself

These problems cover every theorem from the lesson. Work through each one before checking the solution — the habit of writing a reason alongside every angle is what separates full-mark answers from partial ones.

8. Exam Technique: Proofs, Reasons, and Common Mistakes

Circle theorem questions on Higher-tier papers fall into two types: find-the-angle (where you apply theorems) and prove (where you construct a logical argument). Both types demand written reasons — a bare number with no justification will not earn full marks.

For find-the-angle questions, write the theorem name or a brief description next to every angle you calculate. For example: "$\angle ACB = 38^\circ$ (angles in the same segment are equal)" earns the mark; "$\angle ACB = 38^\circ$" alone may not.

For proof questions, work forwards from what you are given. Never assume the result you are trying to prove. State every theorem you use by name, and end with a clear conclusion: "Therefore [result], as required" or the QED symbol $\square$.

9. Exam-Style Proof Question

The final challenge brings together multiple theorems in a single proof — exactly the style of question that appears on Higher-tier calculator-allowed exam questions. Read the question carefully, plan which theorems you need, then work step by step.

Worked example: Prove: Angles in the Same Segment are Equal

O is the centre of a circle. A and B are points on the circumference. C and D are two further points on the same arc (the major arc AB). Prove that angle ACB = angle ADB. [4 marks]

1. By the angle at the centre theorem: $\angle AOB = 2 \times \angle ACB$ (both subtended by arc $AB$, $C$ on major arc) $\angle AOB = 2 \times \angle ADB$ (both subtended by arc $AB$, $D$ on major arc) — [M1 for applying the angle at the centre theorem to both C and D, with reason stated]
2. $2 \times \angle ACB = \angle AOB = 2 \times \angle ADB$ — Both equal the same central angle AOB. [M1 for this logical link]
3. $\angle ACB = \angle ADB$ — [A1 for the correct conclusion]
4. Therefore, angles subtended by the same arc at the circumference are equal. $\square$ — [A1 for a complete, clearly stated conclusion — do not omit this line in an exam proof]

Answer: $\angle ACB = \angle ADB$ — proved using the angle at the centre theorem applied to both inscribed angles.