Using Circle Theorems to Find Angles Revision Notes | GCSE AQA Higher Maths

1. Why Circle Theorems Matter

Every angle in a circle diagram is connected to every other angle by a small set of rules. Once you know those rules, a diagram that looks impossible to solve becomes a straightforward chain of reasoning. That is the power of circle theorems.

GCSE Higher expects you to know eight theorems, name the one you are using, and show the angle calculation clearly. This lesson walks through all eight, with worked examples and exam technique built in throughout.

2. Key Vocabulary and Circle Parts

The centre of a circle is labelled $O$ by convention. A radius runs from $O$ to any point on the circumference. A chord connects two circumference points without passing through $O$. A tangent just grazes the circle at one point.

Understanding what 'subtended' means is essential, because every circle theorem is described using this word. When an arc subtends an angle at the centre, the vertex of that angle is at $O$. When it subtends an angle at the circumference, the vertex sits on the circle itself.

3. The Eight Circle Theorems

Here are all eight theorems you need for GCSE Higher. Each one is a rule you can quote by name in an exam to earn the reasoning mark. Work through them in order — the first two are the foundation for several of the others.

3a. Theorem 1 — Angle at the Centre

Draw any arc on a circle. The angle it makes at the centre $O$ is exactly double the angle it makes at any point on the major arc (the larger part of the circumference). This relationship holds wherever you place the circumference point, as long as it stays on the same side of the chord.

A common way to remember this: the centre 'sees' the arc from closer in, so its angle is larger — precisely twice as large. This theorem is the parent of Theorem 2, as you will see next.

3b. Theorem 2 — Angle in a Semicircle

When the chord is a diameter, the arc at the centre spans $180^\circ$ (a straight line). Applying Theorem 1: the angle at the circumference is half of $180^\circ$, which is $90^\circ$. So any angle inscribed in a semicircle is a right angle — every time, without exception.

Worked example: Using Theorem 1 and Theorem 2 Together

In a circle with centre $O$, points $A$, $B$, and $C$ lie on the circumference. The angle $AOB = 140^\circ$. Find angle $ACB$, where $C$ is on the major arc.

1. Angle at centre $= 2 \times$ angle at circumference (Theorem 1) — Arc $AB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at the circumference. State the theorem to earn the reasoning mark [M1].
2. $\angle ACB = \dfrac{\angle AOB}{2}$ — Divide the centre angle by 2.
3. $\angle ACB = \dfrac{140^\circ}{2} = 70^\circ$ — $140 \div 2 = 70$. Answer with units (degrees) [A1].

Answer: $\angle ACB = 70^\circ$

3c. Theorem 3 — Angles in the Same Segment

Take any chord $AB$. Every angle drawn from the major arc to $A$ and $B$ is identical, no matter where on the major arc you place the vertex. This is because all those angles are each half of the same central angle (by Theorem 1), so they must all be equal to each other.

In exam diagrams, you will often see two triangles sharing the same base chord. If both opposite vertices sit on the same arc, their angles at those vertices are equal. Mark them with matching arc symbols to show the examiner you have spotted the relationship.

3d. Theorem 4 — Cyclic Quadrilateral

When all four vertices of a quadrilateral sit on a circle, opposite angles add up to $180^\circ$. The reasoning: each pair of opposite angles is subtended by the two arcs that together make the full circle ($360^\circ$), and each angle is half its arc's central angle, so the two opposite angles together account for half of $360^\circ$, which is $180^\circ$.

3e. Theorems 5 and 6 — Tangent Properties

A tangent meets the circle at exactly one point. At that point, the radius and the tangent form a right angle ($90^\circ$). This is Theorem 5, and it is the starting point for many multi-step problems involving tangents.

Theorem 6 extends this idea to two tangents drawn from the same external point $P$. Both tangents reach the circle and, because of the symmetry of the circle, they are always equal in length. This also means the line from $P$ to the centre $O$ bisects the angle between the two tangents.

Worked example: Tangent from an External Point — Finding an Angle

From external point $P$, two tangents $PA$ and $PB$ touch a circle with centre $O$. Angle $APB = 50^\circ$. Find angle $AOB$.

1. $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ — Tangent is perpendicular to radius at the point of contact (Theorem 5). State this theorem [M1].
2. $\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^\circ$ — Angles in any quadrilateral sum to $360^\circ$.
3. $90^\circ + 50^\circ + 90^\circ + \angle AOB = 360^\circ$ — Replace each known angle.
4. $230^\circ + \angle AOB = 360^\circ \Rightarrow \angle AOB = 130^\circ$ — $360 - 230 = 130$. Final answer with degree symbol [A1].

Answer: $\angle AOB = 130^\circ$

3f. Theorem 7 — Perpendicular from Centre to Chord

Draw a chord $AB$ anywhere in a circle. Drop a perpendicular from the centre $O$ to the chord — call the foot of the perpendicular $M$. Theorem 7 tells you that $M$ is the midpoint of $AB$, so $AM = MB$. This works because the two right-angled triangles $OAM$ and $OBM$ are congruent (they share $OM$, and $OA = OB$ as radii).

This theorem is particularly useful in problems where you need to find the length of a chord or the distance from the centre to a chord. You can combine it with Pythagoras' theorem to calculate missing lengths.

Worked example: Chord Length Using Theorem 7 and Pythagoras

A circle has centre $O$ and radius $10$ cm. A chord $AB$ is $6$ cm from the centre (measured perpendicularly). Find the length of chord $AB$.

1. Let $M$ be the foot of the perpendicular from $O$ to $AB$. Then $OM = 6$ cm and $OA = 10$ cm. — The perpendicular from the centre bisects the chord (Theorem 7), so triangle $OAM$ is right-angled at $M$ [M1].
2. $OA^2 = OM^2 + AM^2$ — Right angle is at $M$, so the hypotenuse is $OA$.
3. $10^2 = 6^2 + AM^2 \Rightarrow 100 = 36 + AM^2$ — $10 \times 10 = 100$ and $6 \times 6 = 36$.
4. $AM^2 = 100 - 36 = 64 \Rightarrow AM = 8$ cm — $\sqrt{64} = 8$.
5. $AB = 2 \times AM = 2 \times 8 = 16$ cm — Theorem 7 tells us $M$ bisects $AB$, so the full chord is double $AM$ [A1].

Answer: $AB = 16$ cm

3g. Theorem 8 — Alternate Segment Theorem

This is the theorem students find most surprising. Draw a tangent to a circle at point $T$, and draw a chord $TA$ from the point of tangency. The chord divides the circle into two segments. The angle between the tangent and the chord (on one side) equals the angle in the alternate (opposite) segment — that is, any angle subtended by chord $TA$ from the other arc.

A useful way to picture it: the tangent 'remembers' the angle of the arc on the other side. In exam questions, look for a tangent line with a chord drawn from the point of contact — that is the signal to apply the alternate segment theorem.

Worked example: Alternate Segment Theorem — Exam Question

A tangent touches a circle at $T$. Chord $TA$ is drawn. The angle between the tangent and chord $TA$ is $65^\circ$. Point $B$ lies on the arc on the opposite side of chord $TA$. Find angle $TBA$.

1. Alternate segment theorem: angle between tangent and chord $=$ angle in the alternate segment. — State the theorem by name — this earns the reasoning mark in an exam question [M1].
2. $\angle TBA = 65^\circ$ — The angle in the alternate segment equals the tangent-chord angle [A1].

Answer: $\angle TBA = 65^\circ$

4. Multi-Step Problems — Combining Theorems

Exam questions rarely test just one theorem. The real skill is chaining two or three theorems together, labelling each step with the theorem name. Here is a typical multi-step question that combines Theorem 1, Theorem 5, and angle facts.

Worked example: Multi-Step Circle Theorem Problem

In a circle with centre $O$, $A$ and $B$ are points on the circumference. A tangent at $B$ meets the line $OA$ extended at point $P$. Angle $OBP = 90^\circ$ and angle $AOB = 100^\circ$. Find angle $ABP$.

1. $\angle OBP = 90^\circ$ (tangent perpendicular to radius at $B$, Theorem 5) — This is given and confirmed by Theorem 5. State the theorem [M1].
2. $OA = OB$ (both radii), so triangle $OAB$ is isosceles. — Equal radii mean the base angles of triangle $OAB$ are equal.
3. $\angle OAB = \angle OBA = \dfrac{180^\circ - 100^\circ}{2} = \dfrac{80^\circ}{2} = 40^\circ$ — Angles in a triangle sum to $180^\circ$. $180 - 100 = 80$, then $80 \div 2 = 40$ [M1].
4. $\angle ABP = \angle OBP - \angle OBA = 90^\circ - 40^\circ = 50^\circ$ — $\angle OBP$ is the full right angle at $B$; subtracting $\angle OBA$ gives the remaining angle [A1].

Answer: $\angle ABP = 50^\circ$

5. Deliberate Practice

These questions cover each theorem. For each one, identify the theorem first, then calculate the angle. Check your answers below each question.

6. Exam Technique and Common Mistakes

Circle theorem questions on GCSE Higher are typically worth 3–5 marks. The marks split between the correct angle value and the correct theorem name. Losing the reasoning mark is the most common way students drop points on questions they actually understand.

When a question says 'give a reason for each step', write the theorem name in full — for example, 'angles in the same segment are equal' rather than 'same segment rule' or 'circle theorem'. Abbreviated or vague reasons do not earn the mark.

All eight theorems tested in this lesson — angle at the centre, angle in a semicircle, angles in the same segment, cyclic quadrilateral, tangent-radius, tangents from an external point, perpendicular from centre to chord, and the alternate segment theorem — are directly assessed in GCSE Higher and can appear on both questions without a calculator and calculator-allowed exam questions.