Circle Theorems Revision Notes | GCSE AQA Higher Further Maths

1. Why Circle Theorems Matter

Every angle in a circle diagram is connected to every other angle by a small set of rules. Master those rules and multi-step angle problems that look impossible become straightforward — because you always have a reason for every line of working.

Examiners test circle theorems heavily at GCSE because they require both calculation and justification. You will not get full marks by writing the correct number alone — you must name the theorem you used. This lesson builds both skills together.

2. The Five Core Circle Theorems

Think of these five theorems as a toolkit. Each one describes a specific relationship between angles in a circle. Once you can recognise which tool fits the diagram, the calculation follows directly.

A useful special case of this theorem: when the arc is a semicircle (the chord is a diameter), the angle at the centre is $180^\circ$, so the angle at the circumference must be $90^\circ$. This is sometimes called the angle in a semicircle theorem.

This theorem is the one students most often overlook in diagrams. When you see two angles that both sit on the same side of a chord and both touch the circumference, they are equal — no calculation needed.

3. Cyclic Quadrilaterals and the Alternate Segment Theorem

The next two theorems involve more complex configurations, but they follow the same logic: angles in a circle are always connected by the arc they share.

A good way to anchor this: each pair of opposite angles together subtends the full circle ($360^\circ$) at the centre, and the angle at the circumference is half the angle at the centre, so each pair sums to $180^\circ$. You do not need to reproduce that reasoning in an exam — naming the theorem is enough.

This theorem catches students out because the two equal angles look completely unrelated at first glance — one is between a tangent and a chord, the other is inside the circle. The key is to identify the chord, find the tangent at one of its endpoints, and then look for an inscribed angle on the far side of that chord.

4. Two Further Theorems: Radius and Tangent

Two more results appear regularly in GCSE circle theorem questions, often combined with the theorems above. Both involve the radius and are worth knowing precisely.

• Theorem 5a — Radius to tangent: A tangent to a circle is perpendicular to the radius at the point of tangency. This means the angle between them is always $90^\circ$.
• Theorem 5b — Perpendicular from centre to chord: The perpendicular from the centre of a circle to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to the chord.

These two results are frequently used as stepping stones in longer problems. For example, if a question gives you a tangent and asks for an angle, your first move is almost always to draw the radius to the point of tangency and mark the $90^\circ$ angle.

5. Worked Examples

Theory is only useful when you can apply it under exam conditions. The following walkthroughs show how to identify the correct theorem, set up the equation, and write the justification that earns the marks.

Worked example: Using the Angle at the Centre Theorem

O is the centre of a circle. Points A, B, and C lie on the circumference. Angle AOB $= 130^\circ$. Find angle ACB, where C is on the major arc.

1. O is the centre; A, B, C are on the circumference; both angles subtend arc AB. — This matches the angle at the centre theorem exactly. [M1 — correct theorem identified]
2. $\angle AOB = 2 \times \angle ACB$ — The centre angle is twice the circumference angle for the same arc.
3. $130^\circ = 2 \times \angle ACB$ — Replace $\angle AOB$ with the given value $130^\circ$. [M1 — correct substitution]
4. $\angle ACB = \dfrac{130^\circ}{2} = 65^\circ$ — Divide both sides by 2. [A1 — correct answer with justification]

Answer: $\angle ACB = 65^\circ$ (angle at the centre is twice the angle at the circumference)

Worked example: Finding a Missing Angle in a Cyclic Quadrilateral

ABCD is a cyclic quadrilateral. Angle DAB $= 112^\circ$ and angle ABC $= 74^\circ$. Find angles BCD and CDA.

1. ABCD has all four vertices on the circle, so it is a cyclic quadrilateral. — Opposite angles in a cyclic quadrilateral sum to $180^\circ$. [M1]
2. $\angle DAB + \angle BCD = 180^\circ$ $112^\circ + \angle BCD = 180^\circ$ $\angle BCD = 180^\circ - 112^\circ = 68^\circ$ — DAB and BCD are opposite angles. [A1]
3. $\angle ABC + \angle CDA = 180^\circ$ $74^\circ + \angle CDA = 180^\circ$ $\angle CDA = 180^\circ - 74^\circ = 106^\circ$ — ABC and CDA are the other pair of opposite angles. [A1]
4. $112^\circ + 74^\circ + 68^\circ + 106^\circ = 360^\circ$ ✓ — All four angles in any quadrilateral sum to $360^\circ$. This confirms both answers are correct.

Answer: $\angle BCD = 68^\circ$ and $\angle CDA = 106^\circ$ (opposite angles in a cyclic quadrilateral sum to $180^\circ$)

Worked example: Applying the Alternate Segment Theorem

A tangent touches a circle at point T. Chord TA is drawn. The angle between the tangent and chord TA (measured on one side) is $63^\circ$. Point B lies on the arc on the opposite side of TA. Find angle TBA.

1. Tangent at T, chord TA, angle between tangent and chord $= 63^\circ$. Point B is in the alternate segment. — This is the alternate segment theorem setup. [M1]
2. Angle in alternate segment $=$ angle between tangent and chord — The alternate segment theorem states these two angles are equal.
3. $\angle TBA = 63^\circ$ — B is in the alternate segment, so angle TBA equals the tangent-chord angle. [A1]

Answer: $\angle TBA = 63^\circ$ (alternate segment theorem)

6. Multi-Step Problems

Exam questions rarely test one theorem in isolation. More often, you need to chain two or three theorems together, using the result from one step as the input for the next. The strategy is always the same: identify what you know, choose the theorem that connects it to something new, and work step by step.

Worked example: Multi-Step: Centre Theorem then Isosceles Triangle

O is the centre of a circle. A, B, and C are points on the circumference. OA and OB are radii. Angle ACB $= 34^\circ$. Find angle OAB.

1. $\angle AOB = 2 \times \angle ACB = 2 \times 34^\circ = 68^\circ$ — Angle at the centre is twice the angle at the circumference. [M1, A1]
2. OA $=$ OB (both radii), so triangle OAB is isosceles. — Radii of the same circle are always equal in length. This is a key linking fact.
3. $\angle OAB + \angle OBA + \angle AOB = 180^\circ$ $\angle OAB + \angle OBA = 180^\circ - 68^\circ = 112^\circ$ Since $\angle OAB = \angle OBA$: $\angle OAB = \dfrac{112^\circ}{2} = 56^\circ$ — Base angles of an isosceles triangle are equal. [M1, A1]

Answer: $\angle OAB = 56^\circ$

7. Practice Problems

Now apply the theorems yourself. For each question, write down the theorem name before calculating — this mirrors exactly what the mark scheme rewards.

8. Exam Technique and Common Mistakes

Circle theorem questions are among the most mark-efficient on the paper — if you know the rules, you can score full marks quickly. The mistakes below are the ones that cost students marks most often.