study guides18 min read

Circle Theorems Practice Questions

Sir Faraz Hassan

10 Jul 2026

Table of Contents▾

Grades 8-9Fourteen exam-style questions with full worked solutions and mark schemes, covering every GCSE circle theorem from the angle in a semicircle to intersecting circles

Circle theorems reward students who can spot the right rule and chain a few steps together. These fourteen questions build from single-theorem angle chases up to full multi-theorem proofs. Each one has a complete worked solution and mark scheme, so you can see exactly how every mark is earned.

What are circle theorems?

Circle theorems are a set of rules that link the angles, chords, tangents and radii of a circle. The eight main GCSE circle theorems are: the angle in a semicircle is 90 degrees; the angle at the centre is twice the angle at the circumference; angles in the same segment are equal; opposite angles of a cyclic quadrilateral add up to 180 degrees; a tangent meets a radius at 90 degrees; the two tangents from a point are equal in length; the perpendicular from the centre bisects a chord; and the alternate segment theorem, which says the angle between a tangent and a chord equals the angle in the alternate segment. Most exam questions combine two or more of these, so the real skill is recognising which theorem applies from the diagram and chaining the steps together.

How to use this page

Try each question on paper first

Give yourself a real attempt before looking at the solution, because that is where the learning happens.

Check the calculator icon

A crossed-out calculator means non-calculator; a plain calculator means a calculator is allowed.

Reveal the worked solution

Open the solution under each question and compare it with your own method, step by step.

Read the mark scheme

See exactly where each method mark (M1) and accuracy mark (A1) is awarded, and remember that at GCSE the theorem reasons carry marks too.

Come back and redo it

Repeat any question you got wrong a few days later to lock the method in.

Download printable PDF

14 questions with full worked solutions and mark schemes - free PDF

Practice questions

Work through each question, then open the worked solution to check your method. Single-theorem questions come first, then multi-theorem problems and proofs.

$P$ , $Q$ , $R$ and $S$ are points on a circle, centre $O$ . $POR$ is a straight line, so $PR$ is a diameter. $\angle SPR = 54^{\circ}$ . Work out the size of the angle marked $x$ ( $\angle PQS$ ).

x =

^{\circ}

[Total 3 marks]

GRADE8

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Question 1 - Exam Solution

Understanding the Question

Given

P, Q, R, S

lie on a circle, centre

O

;

POR

is a straight line, so

PR

is a diameter.

\angle SPR = 54^{\circ}

Find

The size of the angle marked

x

(

\angle PQS

Plan the Solution

$PR$ is a diameter, so $\angle PQR$ is the angle in a semicircle and equals $90^{\circ}$ .
Use angles in the same segment to find $\angle SQR$ .
Subtract to find $x$ .

Worked Solution [3 marks]

Rule - Angle chasing: apply the angle in a semicircle and angles in the same segment, then subtract to reach the required angle.

Step 1: Use the diameter (angle in a semicircle).

\angle PQR = 90^{\circ}

(Reason:

PR

is a diameter, and the angle in a semicircle is

90^{\circ}

Step 2: Angles in the same segment.

\angle SQR = \angle SPR = 54^{\circ}

(Reason:

\angle SQR

and

\angle SPR

both stand on the chord

SR

, so angles in the same segment are equal.)

Step 3: Subtract to find x.

x = \angle PQR - \angle SQR = 90^{\circ} - 54^{\circ} = 36^{\circ}

(Reason:

x

is the part of

\angle PQR

that remains after

\angle SQR

is removed.)

x = 36^{\circ}

Verification

Check 1:

\angle PQS + \angle SQR = 36^{\circ} + 54^{\circ} = 90^{\circ} = \angle PQR

Check 2:

x = 36^{\circ}

is acute and smaller than the right angle at

Q

, which fits the diagram.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle SQR = 54^{\circ}$	M1	Method - angles in the same segment	✓
$\angle PQR = 90^{\circ}$	M1	Method - the angle in a semicircle	✓
$x = 36^{\circ}$	A1	Accuracy - correct final answer	✓

Full marks: 3/3

The diagram shows a circle, centre $O$ . $P$ , $Q$ and $S$ are points on the circle. $TP$ and $TS$ are tangents to the circle, and $PQR$ is a straight line. $\angle STO = 34^{\circ}$ . Work out the size of $\angle SQR$ , giving reasons.

\angle SQR =

^{\circ}

[Total 5 marks]

GRADE8Problem solving

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Question 2 - Exam Solution

Understanding the Question

Given

Circle, centre

O

;

P, Q, S

on the circle.

TP

and

TS

are tangents from

T

;

PQR

is a straight line.

\angle STO = 34^{\circ}

Find

The size of

\angle SQR

Plan the Solution

Tangent meets radius: $\angle TSO = 90^{\circ}$ , then find $\angle TOS$ in triangle $TSO$ .
The two equal tangents double the angle at the centre: $\angle POS = 2 \times \angle TOS$ .
Angle at the centre is twice the angle at the circumference: find $\angle PQS$ .
Angles on a straight line: find $\angle SQR$ .

Worked Solution [5 marks]

Rule - Work in from the tangent: use the tangent-radius right angle, the symmetry of the two equal tangents, and the centre-to-circumference link, then finish on the straight line.

Step 1: Tangent meets radius.

\angle TSO = 90^{\circ}

(Reason:

OS

is a radius and

TS

is a tangent; a tangent meets a radius at

90^{\circ}

Step 2: Angles in triangle TSO.

\angle TOS = 180^{\circ} - 90^{\circ} - 34^{\circ} = 56^{\circ}

(Reason: The angles in triangle

TSO

add up to

180^{\circ}

Step 3: Two equal tangents (angle at the centre).

\angle POS = 2 \times 56^{\circ} = 112^{\circ}

(Reason:

TP

and

TS

are equal tangents from

T

, so triangles

TPO

and

TSO

are congruent; this doubles the angle at

O

Step 4: Angle at the centre is twice the angle at the circumference.

\angle PQS = \tfrac{1}{2} \times 112^{\circ} = 56^{\circ}

(Reason:

\angle POS

is at the centre and

\angle PQS

is at the circumference, both standing on arc

PS

Step 5: Angles on a straight line.

\angle SQR = 180^{\circ} - 56^{\circ} = 124^{\circ}

(Reason:

PQR

is a straight line, so the angles at

Q

add up to

180^{\circ}

\angle SQR = 124^{\circ}

Verification

Check 1: In triangle

TSO

90^{\circ} + 34^{\circ} + 56^{\circ} = 180^{\circ}

Check 2: Quadrilateral

OPTS

90^{\circ} + 68^{\circ} + 90^{\circ} + 112^{\circ} = 360^{\circ}

(with

\angle PTS = 2 \times 34^{\circ} = 68^{\circ}

Check 3: At

Q

\angle PQS + \angle SQR = 56^{\circ} + 124^{\circ} = 180^{\circ}

, matching the straight line.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle TSO = 90^{\circ}$	M1	Method - tangent meets a radius at 90 degrees	✓
$\angle TOS = 56^{\circ}$	M1	Method - angles in a triangle	✓
$\angle POS = 112^{\circ}$	M1	Method - equal tangents give the angle at the centre	✓
$\angle PQS = 56^{\circ}$	M1	Method - angle at the centre is twice the circumference	✓
$\angle SQR = 124^{\circ}$	A1	Accuracy - correct final answer	✓

Full marks: 5/5

$P$ , $Q$ , $R$ and $S$ lie on a circle. $NPQ$ is a straight line and $HK$ is the tangent to the circle at $Q$ . $\angle HQP = 108^{\circ}$ and $\angle NPR = 152^{\circ}$ . Find the size of $\angle RSP$ , giving a reason for each step.

\angle RSP =

^{\circ}

[Total 4 marks]

GRADE9

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Question 3 - Exam Solution

Understanding the Question

Given

P, Q, R, S

lie on a circle.

NPQ

is a straight line and

HK

is the tangent at

Q

\angle HQP = 108^{\circ}

and

\angle NPR = 152^{\circ}

Find

The size of

\angle RSP

, giving a reason for each step.

Plan the Solution

Alternate segment theorem: the tangent-chord angle $\angle HQP$ equals $\angle QRP$ .
Angles on a straight line at $P$ : find $\angle RPQ$ .
Angles in triangle $PQR$ : find $\angle PQR$ .
Angles in the same segment: $\angle RSP = \angle PQR$ .

Worked Solution [4 marks]

Rule - Transfer angles round the circle: use the alternate segment theorem, a straight line and a triangle, then move the angle to the required point with the same-segment rule.

Step 1: Alternate segment theorem.

\angle QRP = 108^{\circ}

(Reason:

\angle HQP

is the angle between the tangent

HK

and the chord

QP

; by the alternate segment theorem it equals

\angle QRP

in the alternate segment.)

Step 2: Angles on a straight line at P.

\angle RPQ = 180^{\circ} - 152^{\circ} = 28^{\circ}

(Reason:

NPQ

is a straight line, so

\angle NPR

and

\angle RPQ

add up to

180^{\circ}

Step 3: Angles in triangle PQR.

\angle PQR = 180^{\circ} - 108^{\circ} - 28^{\circ} = 44^{\circ}

(Reason: The angles in triangle

PQR

add up to

180^{\circ}

Step 4: Angles in the same segment.

\angle RSP = \angle PQR = 44^{\circ}

(Reason:

\angle RSP

and

\angle PQR

both stand on the chord

RP

, so angles in the same segment are equal.)

\angle RSP = 44^{\circ}

Verification

Check 1: Triangle

PQR

108^{\circ} + 28^{\circ} + 44^{\circ} = 180^{\circ}

Check 2: Straight line

NPQ

P

\angle NPR + \angle RPQ = 152^{\circ} + 28^{\circ} = 180^{\circ}

Check 3: The same-segment angles agree:

\angle RSP = \angle PQR = 44^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle QRP = 108^{\circ}$	M1	Method - alternate segment theorem	✓
$\angle RPQ = 28^{\circ}$	M1	Method - angles on a straight line	✓
$\angle PQR = 44^{\circ}$	M1	Method - angles in a triangle	✓
$\angle RSP = 44^{\circ}$	A1	Accuracy - correct final answer	✓

Full marks: 4/4

The diagram shows a circle, centre $O$ . $P$ , $Q$ , $R$ and $S$ are points on the circle. $PT$ and $RT$ are tangents from $T$ . $\angle PTR = 72^{\circ}$ . Find the size of $\angle PQR$ , giving reasons.

\angle PQR =

^{\circ}

[Total 4 marks]

GRADE8Problem solving

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Question 4 - Exam Solution

Understanding the Question

Given

Circle, centre

O

;

P, Q, R, S

on the circle.

PT

and

RT

are tangents from

T

\angle PTR = 72^{\circ}

Find

The size of

\angle PQR

, giving reasons.

Plan the Solution

Tangent meets radius: both $\angle OPT$ and $\angle ORT$ are $90^{\circ}$ .
Angles in quadrilateral $OPTR$ add to $360^{\circ}$ : find the central angle $\angle POR$ .
Angle at the centre is twice the angle at the circumference: find $\angle PSR$ .
Opposite angles of a cyclic quadrilateral: find $\angle PQR$ .

Worked Solution [4 marks]

Rule - From the tangents to the far angle: use the two tangent-radius right angles inside the quadrilateral, then the centre-to-circumference link, and finish with the cyclic-quadrilateral rule.

Step 1: Tangent meets radius (two right angles).

\angle OPT = \angle ORT = 90^{\circ}

(Reason:

OP

and

OR

are radii and

PT

RT

are tangents; a tangent meets a radius at

90^{\circ}

Step 2: Angles in quadrilateral OPTR.

\angle POR = 360^{\circ} - 90^{\circ} - 90^{\circ} - 72^{\circ} = 108^{\circ}

(Reason:

OPTR

is a quadrilateral, and its angles add up to

360^{\circ}

Step 3: Angle at the centre is twice the angle at the circumference.

\angle PSR = \tfrac{1}{2} \times 108^{\circ} = 54^{\circ}

(Reason:

\angle POR

is at the centre and

\angle PSR

is at the circumference, both standing on arc

PR

Step 4: Opposite angles of a cyclic quadrilateral.

\angle PQR = 180^{\circ} - 54^{\circ} = 126^{\circ}

(Reason:

PQRS

is a cyclic quadrilateral, so the opposite angles

\angle PQR

and

\angle PSR

add up to

180^{\circ}

\angle PQR = 126^{\circ}

Verification

Check 1: Quadrilateral

OPTR

90^{\circ} + 90^{\circ} + 72^{\circ} + 108^{\circ} = 360^{\circ}

Check 2: Opposite angles:

\angle PQR + \angle PSR = 126^{\circ} + 54^{\circ} = 180^{\circ}

Check 3: The angle at the centre (

108^{\circ}

) is exactly twice the angle at the circumference (

54^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle OPT = \angle ORT = 90^{\circ}$	M1	Method - tangent meets a radius at 90 degrees	✓
$\angle POR = 108^{\circ}$	M1	Method - angles in a quadrilateral add to 360 degrees	✓
$\angle PSR = 54^{\circ}$	M1	Method - angle at the centre is twice the circumference	✓
$\angle PQR = 126^{\circ}$	A1	Accuracy - correct final answer	✓

Full marks: 4/4

$P$ , $Q$ , $R$ , $S$ and $T$ lie on a circle. $MN$ is the tangent to the circle at $T$ . $\angle PQT = 34^{\circ}$ and $\angle SRT = 56^{\circ}$ . Prove that the chord $PS$ passes through the centre of the circle.

[Total 3 marks]

GRADE9Problem solving

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Question 5 - Exam Solution

Understanding the Question

Given

P, Q, R, S, T

lie on a circle.

MN

is the tangent at

T

\angle PQT = 34^{\circ}

and

\angle SRT = 56^{\circ}

Prove

That the chord

PS

passes through the centre of the circle.

Plan the Solution

Alternate segment theorem: find the two tangent-chord angles $\angle PTM$ and $\angle STN$ .
Angles on the straight line $MTN$ : find $\angle PTS$ .
If $\angle PTS = 90^{\circ}$ , then $PS$ is a diameter (angle in a semicircle), so it passes through the centre.

Worked Solution [3 marks]

Rule - A right angle means a diameter: show that the angle at

T

90^{\circ}

; the converse of the angle in a semicircle then forces

PS

to be a diameter through the centre.

Step 1: Alternate segment theorem.

\angle PTM = \angle PQT = 34^{\circ} \quad \text{and} \quad \angle STN = \angle SRT = 56^{\circ}

(Reason:

\angle PTM

is the tangent-chord angle for chord

TP

, so it equals

\angle PQT

in the alternate segment; similarly

\angle STN = \angle SRT

Step 2: Angles on a straight line.

\angle PTS = 180^{\circ} - 34^{\circ} - 56^{\circ} = 90^{\circ}

(Reason:

MTN

is a straight line, so

\angle PTM + \angle PTS + \angle STN = 180^{\circ}

Step 3: Angle in a semicircle (converse).

\angle PTS = 90^{\circ} \Rightarrow PS \text{ is a diameter}

(Reason: The angle in a semicircle is

90^{\circ}

; conversely, if an inscribed angle is

90^{\circ}

then the chord opposite it is a diameter, which passes through the centre.)

Therefore

PS

is a diameter, so it passes through the centre of the circle.

Verification

Check 1: Angles on the straight line at

T

34^{\circ} + 90^{\circ} + 56^{\circ} = 180^{\circ}

Check 2: The two given angles satisfy

34^{\circ} + 56^{\circ} = 90^{\circ}

, which is why

\angle PTS = 180^{\circ} - 90^{\circ} = 90^{\circ}

Check 3:

\angle PTS = 90^{\circ}

is the angle in a semicircle, confirming

PS

is a diameter.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle PTM = 34^{\circ} \text{ and } \angle STN = 56^{\circ}$	M1	Method - alternate segment theorem	✓
$\angle PTS = 90^{\circ}$	M1	Method - angles on a straight line	✓
$PS$ is a diameter through the centre	A1	Accuracy - angle in a semicircle, so PS is a diameter	✓

Full marks: 3/3

$P$ , $Q$ , $R$ and $S$ lie on a circle, centre $O$ . $MN$ is the tangent at $S$ . $\angle QSN = 58^{\circ}$ .

(a) Work out the size of $\angle SOQ$ . [2 marks]

(b) Explain why $\angle ROQ$ is half the size of $\angle SOQ$ . [2 marks]

(a)

\angle SOQ =

^{\circ}

[Total 4 marks]

GRADE8

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Question 6 - Exam Solution

Understanding the Question

Given

P, Q, R, S

lie on a circle, centre

O

;

MN

is the tangent at

S

\angle QSN = 58^{\circ}

Find

(a) The size of

\angle SOQ

(b) Why

\angle ROQ

is half the size of

\angle SOQ

Plan the Solution

Part (a): alternate segment theorem, then the centre-to-circumference link.
Part (b): the radius perpendicular to a chord bisects both the chord and the angle at the centre.

Part (a): Work out

\angle SOQ

[2 marks]

Rule - Alternate segment, then double at the centre: transfer the tangent-chord angle to the circumference, then use the centre-to-circumference link.

Step 1: Alternate segment theorem.

\angle QPS = \angle QSN = 58^{\circ}

(Reason:

\angle QSN

is the tangent-chord angle for chord

SQ

, equal to

\angle QPS

in the alternate segment.)

Step 2: Angle at the centre is twice the angle at the circumference.

\angle SOQ = 2 \times 58^{\circ} = 116^{\circ}

(Reason:

\angle SOQ

is at the centre and

\angle QPS

is at the circumference, both standing on arc

SQ

\angle SOQ = 116^{\circ}

Mark scheme

Step	Mark	Description	Got it?
$\angle QPS = 58^{\circ}$	M1	Method - alternate segment theorem	✓
$\angle SOQ = 116^{\circ}$	A1	Accuracy - angle at centre is twice the circumference	✓

Full marks: 2/2

Part (b): Explain why

\angle ROQ = \tfrac{1}{2}\angle SOQ

[2 marks]

Rule - Isosceles symmetry: the radius that is perpendicular to a chord bisects both the chord and the angle at the centre.

Step 1: Isosceles triangle and perpendicular bisector.

OQ = OS

(radii), so triangle

OSQ

is isosceles.

OR

meets chord

SQ

at right angles, so it bisects

SQ

(Reason: All radii are equal, and the perpendicular from the centre of a circle to a chord bisects the chord.)

Step 2: Bisecting the base bisects the angle at O.

Bisecting the base of the isosceles triangle also bisects the apex angle, so

\angle ROQ = \tfrac{1}{2} \times \angle SOQ

(Reason: In an isosceles triangle, the line from the apex perpendicular to the base bisects the apex angle.)

Therefore

\angle ROQ = \tfrac{1}{2} \times \angle SOQ

Mark scheme

Step	Mark	Description	Got it?
Triangle $OSQ$ isosceles and $OR$ bisects $SQ$	M1	Reason - perpendicular from the centre bisects a chord	✓
$\angle ROQ = \tfrac{1}{2}\angle SOQ$	A1	Reason - the bisector halves the apex angle	✓

Full marks: 2/2

Verification

Check 1 (a):

\angle SOQ = 2 \times \angle QPS = 2 \times 58^{\circ} = 116^{\circ}

Check 2 (b):

OR

splits triangle

OSQ

symmetrically, so

\angle ROQ = \angle SOR = \tfrac{1}{2} \times 116^{\circ} = 58^{\circ}

, which is half of

\angle SOQ

$P$ , $Q$ , $R$ and $S$ lie on a circle. $Y$ is a point inside the circle, joined to $S$ and $R$ . $\angle PQR = 118^{\circ}$ , $\angle YSP = 29^{\circ}$ and $\angle YRS = 31^{\circ}$ . Show that $Y$ is not the centre of the circle.

[Total 3 marks]

GRADE9Problem solving

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Question 7 - Exam Solution

Understanding the Question

Given

P, Q, R, S

lie on a circle.

Y

is inside, joined to

S

and

R

\angle PQR = 118^{\circ}

\angle YSP = 29^{\circ}

\angle YRS = 31^{\circ}

Show

That

Y

is not the centre of the circle.

Plan the Solution

Cyclic quadrilateral $PQRS$ : find $\angle PSR$ .
Subtract to find $\angle RSY$ .
If $Y$ were the centre, $YS$ and $YR$ would be radii, giving equal base angles. Check whether $\angle RSY = \angle SRY$ .

Worked Solution [3 marks]

Rule - Assume and contradict: suppose

Y

is the centre; two radii would force equal base angles, so if the angles differ,

Y

cannot be the centre.

Step 1: Opposite angles of a cyclic quadrilateral.

\angle PSR = 180^{\circ} - 118^{\circ} = 62^{\circ}

(Reason:

PQRS

is a cyclic quadrilateral, so the opposite angles

\angle PQR

and

\angle PSR

add up to

180^{\circ}

Step 2: Subtract to find angle RSY.

\angle RSY = \angle PSR - \angle PSY = 62^{\circ} - 29^{\circ} = 33^{\circ}

(Reason:

\angle PSY

is part of

\angle PSR

, so subtracting it leaves

\angle RSY

Step 3: Test the "centre" assumption.

Y

were the centre,

YS

and

YR

would be radii, so triangle

RSY

would be isosceles with

\angle RSY = \angle SRY

. But

\angle RSY = 33^{\circ}

and

\angle SRY = 31^{\circ}

(Reason: A triangle made from two radii is isosceles, so its base angles would be equal - but here they are not.)

Since

\angle RSY = 33^{\circ} \ne 31^{\circ} = \angle SRY

Y

is not the centre of the circle.

Verification

Check 1: Cyclic quadrilateral:

\angle PQR + \angle PSR = 118^{\circ} + 62^{\circ} = 180^{\circ}

Check 2: The parts add up:

\angle PSY + \angle RSY = 29^{\circ} + 33^{\circ} = 62^{\circ} = \angle PSR

Check 3: Two radii would force

\angle RSY = \angle SRY

, but

33^{\circ} \ne 31^{\circ}

, confirming

Y

is not the centre.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle PSR = 62^{\circ}$	M1	Reason - opposite angles in a cyclic quadrilateral	✓
$\angle RSY = 33^{\circ}$	M1	Method - angle subtraction	✓
$33^{\circ} \ne 31^{\circ}$ , so $Y$ is not the centre	A1	Reason - two radii would give equal base angles	✓

Full marks: 3/3

$P$ , $Q$ and $R$ lie on a circle with $PQ = PR$ . $ST$ is the tangent at $R$ , and $ST$ is parallel to $PQ$ . $\angle SRP = x$ . Prove that triangle $PQR$ is an equilateral triangle.

[Total 4 marks]

GRADE8Problem solving

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Question 8 - Exam Solution

Understanding the Question

Given

Triangle

PQR

with

P, Q, R

on a circle;

PQ = PR

ST

is a tangent at

R

, parallel to

PQ

;

\angle SRP = x

Prove

That triangle

PQR

is an equilateral triangle.

Plan the Solution

Alternate segment theorem: $\angle PQR = x$ .
Isosceles triangle ( $PQ = PR$ ): $\angle PRQ = x$ .
Alternate angles ( $ST$ parallel to $PQ$ ): $\angle QPR = x$ .
All three angles equal, so the triangle is equilateral.

Worked Solution [4 marks]

Rule - Show all three angles are equal: a triangle is equilateral when its three angles are equal; find each angle in turn and show they are all

x

Step 1: Alternate segment theorem.

\angle PQR = x

(Reason:

\angle SRP

is the angle between the tangent

ST

and the chord

RP

, so by the alternate segment theorem it equals

\angle PQR

in the alternate segment.)

Step 2: Isosceles triangle.

\angle PRQ = x

(Reason:

PQ = PR

, so triangle

PQR

is isosceles; its base angles are equal, so

\angle PRQ = \angle PQR = x

Step 3: Alternate angles.

\angle QPR = x

(Reason:

ST

is parallel to

PQ

, so

\angle SRP

and

\angle QPR

are alternate angles and are equal.)

Step 4: All three angles are equal.

\angle PQR = \angle PRQ = \angle QPR = x

(Reason: A triangle with three equal angles is equilateral, so each angle is

60^{\circ}

All three angles equal

x

, so triangle

PQR

is equilateral (each angle is

60^{\circ}

Verification

Check 1: The three equal angles sum to

180^{\circ}

x + x + x = 3x = 180^{\circ}

, so

x = 60^{\circ}

Check 2: Equal angles give equal sides, so

PQ = QR = RP

, which is exactly what equilateral means.

Check 3: With

x = 60^{\circ}

, the tangent-chord angle is

60^{\circ}

, consistent with an equilateral triangle inscribed in the circle.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle PQR = x$	M1	Reason - alternate segment theorem	✓
$\angle PRQ = x$	M1	Reason - isosceles triangle (PQ = PR)	✓
$\angle QPR = x$	M1	Reason - alternate angles (ST parallel to PQ)	✓
All three angles equal, so equilateral	A1	Reason - three equal angles mean equilateral	✓

Full marks: 4/4

The diagram shows a circle, centre $O$ . $P$ , $Q$ , $R$ and $S$ lie on the circle. $\angle POS = 118^{\circ}$ , $\angle PQO = 56^{\circ}$ and $\angle ORQ = 42^{\circ}$ . Find the size of $\angle RSO$ .

\angle RSO =

^{\circ}

[Total 4 marks]

GRADE8

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Question 9 - Exam Solution

Understanding the Question

Given

Circle, centre

O

;

P, Q, R, S

on the circle.

\angle POS = 118^{\circ}

\angle PQO = 56^{\circ}

\angle ORQ = 42^{\circ}

Find

The size of

\angle RSO

Plan the Solution

Triangle $POS$ is isosceles (two radii): find $\angle OSP$ .
Triangle $QOR$ is isosceles (two radii): find $\angle OQR$ .
Combine at $Q$ to get $\angle PQR$ , then use the cyclic quadrilateral for $\angle RSP$ .
Subtract to find $\angle RSO$ .

Worked Solution [4 marks]

Rule - Radii make isosceles triangles: every pair of radii forms an isosceles triangle with equal base angles; use these, then the cyclic-quadrilateral rule, and subtract.

Step 1: Isosceles triangle POS.

\angle OSP = \tfrac{1}{2}(180^{\circ} - 118^{\circ}) = \tfrac{1}{2} \times 62^{\circ} = 31^{\circ}

(Reason:

OP = OS

are radii, so triangle

POS

is isosceles; the base angles are equal, each being half of what is left after the apex angle.)

Step 2: Isosceles triangle QOR.

\angle OQR = \angle ORQ = 42^{\circ}

(Reason:

OQ = OR

are radii, so triangle

QOR

is isosceles; its base angles are equal.)

Step 3: Cyclic quadrilateral PQRS.

\angle PQR = \angle PQO + \angle OQR = 56^{\circ} + 42^{\circ} = 98^{\circ}

\angle RSP = 180^{\circ} - 98^{\circ} = 82^{\circ}

(Reason: Adding the two parts at

Q

gives

\angle PQR

; then

PQRS

is a cyclic quadrilateral, so the opposite angles

\angle PQR

and

\angle RSP

add up to

180^{\circ}

Step 4: Subtract to find angle RSO.

\angle RSO = \angle RSP - \angle OSP = 82^{\circ} - 31^{\circ} = 51^{\circ}

(Reason:

\angle OSP

is part of

\angle RSP

, so subtracting it leaves

\angle RSO

\angle RSO = 51^{\circ}

Verification

Check 1: Triangle

POS

118^{\circ} + 31^{\circ} + 31^{\circ} = 180^{\circ}

Check 2: Cyclic quadrilateral:

\angle PQR + \angle RSP = 98^{\circ} + 82^{\circ} = 180^{\circ}

Check 3: The parts add up:

\angle RSO + \angle OSP = 51^{\circ} + 31^{\circ} = 82^{\circ} = \angle RSP

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle OSP = 31^{\circ}$	M1	Method - isosceles triangle (two radii)	✓
$\angle OQR = 42^{\circ}$	M1	Method - isosceles triangle (two radii)	✓
$\angle RSP = 82^{\circ}$	M1	Method - cyclic quadrilateral opposite angles	✓
$\angle RSO = 51^{\circ}$	A1	Accuracy - correct final answer	✓

Full marks: 4/4

The diagram shows a circle with centre $O$ . $PQ$ is a diameter and $R$ is a point on the circumference. Prove that $\angle PRQ = 90^{\circ}$ .

[Total 3 marks]

GRADE8Problem solving

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Question 10 - Exam Solution

Understanding the Question

Given

A circle with centre

O

PQ

is a diameter and

R

is a point on the circumference.

Prove

That

\angle PRQ = 90^{\circ}

Plan the Solution

Draw the radius $OR$ , splitting triangle $PQR$ into two isosceles triangles.
Label the base angles $x$ and $y$ .
Use the angle sum of triangle $PQR$ to show $2x + 2y = 180^{\circ}$ , hence $x + y = 90^{\circ}$ .

Worked Solution [3 marks]

Rule - Two radii, two isosceles triangles: the radius to

R

creates two isosceles triangles; adding their base angles and using the triangle angle sum forces the angle at

R

to be

90^{\circ}

Step 1: Split into two isosceles triangles.

Draw

OR

. Since

OP = OR

and

OQ = OR

(radii):

\angle OPR = \angle ORP = x \quad \text{and} \quad \angle OQR = \angle ORQ = y

(Reason:

OP

OQ

and

OR

are all radii, so triangles

OPR

and

OQR

are isosceles, with equal base angles.)

Step 2: Angle sum of triangle PQR.

The angle at

R

\angle PRQ = x + y

, so:

x + x + y + y = 180^{\circ}

(Reason: The angles in triangle

PQR

- namely

x

y

and

x + y

- add up to

180^{\circ}

Step 3: Solve for the angle at R.

2x + 2y = 180^{\circ}, \text{ so } x + y = 90^{\circ}

\angle PRQ = x + y = 90^{\circ}

(Reason: Dividing by 2 gives

x + y = 90^{\circ}

, and

\angle PRQ

equals

x + y

Therefore

\angle PRQ = 90^{\circ}

(the angle in a semicircle is a right angle).

Verification

Check 1:

x + x + y + y = 2(x + y) = 2 \times 90^{\circ} = 180^{\circ}

, matching the triangle angle sum.

Check 2: The proof uses no fixed value for

x

y

, so

\angle PRQ = 90^{\circ}

for any point

R

on the semicircle.

Check 3: A right angle at

R

means the diameter

PQ

subtends

90^{\circ}

, which is exactly the angle in a semicircle.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two isosceles triangles, base angles $x$ and $y$	M1	Method - split using the radius OR	✓
$x + x + y + y = 180^{\circ}$	M1	Method - angles in a triangle	✓
$x + y = 90^{\circ}$ , so $\angle PRQ = 90^{\circ}$	A1	Reason - the angle in a semicircle is 90 degrees	✓

Full marks: 3/3

The diagram shows two intersecting circles with centres $O$ and $P$ . The circles intersect at $G$ and $H$ . $W$ , $X$ and $Y$ are points on the circle with centre $O$ , and $Z$ is a point on the circle with centre $P$ . $XZ$ and $ZY$ are straight lines. $\angle GWY = 74^{\circ}$ and $\angle GPH = 84^{\circ}$ . Find the size of $\angle XYH$ , giving reasons.

\angle XYH =

^{\circ}

[Total 3 marks]

GRADE9

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Question 11 - Exam Solution

Understanding the Question

Given

Two intersecting circles, centres

O

and

P

, meeting at

G

and

H

W, X, Y

on circle

O

;

Z

on circle

P

XZ

and

ZY

are straight lines.

\angle GWY = 74^{\circ}

and

\angle GPH = 84^{\circ}

Find

The size of

\angle XYH

, giving reasons.

Plan the Solution

Circle $P$ : $\angle GPH$ is at the centre, so the inscribed angle $\angle GZH$ is half of it.
Circle $O$ : $\angle YXG$ equals $\angle GWY$ (same segment).
These are two angles of triangle $XZY$ (as $G$ lies on $XZ$ and $H$ lies on $ZY$ ). Use the triangle angle sum for $\angle XYH$ .

Worked Solution [3 marks]

Rule - Work through each circle: use the centre-to-circumference rule in circle

P

and the same-segment rule in circle

O

to fill in two angles of triangle

XZY

, then use the angle sum.

Step 1: Circle P - angle at the centre.

\angle GZH = \tfrac{1}{2} \times \angle GPH = \tfrac{1}{2} \times 84^{\circ} = 42^{\circ}

(Reason: In circle

P

\angle GPH

is at the centre and

\angle GZH

is at the circumference, both standing on arc

GH

Step 2: Circle O - angles in the same segment.

\angle YXG = \angle GWY = 74^{\circ}

(Reason: In circle

O

\angle YXG

and

\angle GWY

both stand on the chord

GY

, so angles in the same segment are equal.)

Step 3: Angles in triangle XZY.

G

lies on

XZ

and

H

lies on

ZY

, so

\angle XZY = \angle GZH = 42^{\circ}

and

\angle XYH = \angle XYZ

\angle XYH = 180^{\circ} - 74^{\circ} - 42^{\circ} = 64^{\circ}

(Reason: The angles in triangle

XZY

add up to

180^{\circ}

;

\angle XYH

is the angle at

Y

because

H

lies on

ZY

\angle XYH = 64^{\circ}

Verification

Check 1: Triangle

XZY

74^{\circ} + 42^{\circ} + 64^{\circ} = 180^{\circ}

Check 2:

\angle GZH = 42^{\circ}

is exactly half of

\angle GPH = 84^{\circ}

(centre is twice the circumference).

Check 3:

\angle YXG = \angle GWY = 74^{\circ}

, confirming the same-segment step.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle GZH = 42^{\circ}$	M1	Method - angle at the centre is twice the circumference (circle P)	✓
$\angle YXG = 74^{\circ}$	M1	Method - angles in the same segment (circle O)	✓
$\angle XYH = 64^{\circ}$	A1	Accuracy - angles in a triangle	✓

Full marks: 3/3

$P$ , $Q$ , $R$ and $S$ lie on a circle. $MN$ is a tangent at $S$ , and $PR$ and $QS$ are straight lines meeting at $T$ . $\angle QTR = 70^{\circ}$ and $\angle NSR = 58^{\circ}$ . Show that $T$ is not the centre of the circle.

[Total 4 marks]

GRADE9Problem solving

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Question 12 - Exam Solution

Understanding the Question

Given

P, Q, R, S

lie on a circle.

MN

is a tangent at

S

;

PR

and

QS

are straight lines meeting at

T

\angle QTR = 70^{\circ}

and

\angle NSR = 58^{\circ}

Show

That

T

is not the centre of the circle.

Plan the Solution

Alternate segment theorem: find $\angle SQR$ .
Angles on a straight line at $T$ : find $\angle STR$ .
If $T$ were the centre, $\angle STR$ would equal $2 \times \angle SQR$ . Check whether this holds.

Worked Solution [4 marks]

Rule - Assume and contradict: suppose

T

is the centre; the angle at the centre would be twice the inscribed angle, so if that fails,

T

cannot be the centre.

Step 1: Alternate segment theorem.

\angle SQR = \angle NSR = 58^{\circ}

(Reason:

\angle NSR

is the tangent-chord angle for chord

SR

, equal to

\angle SQR

in the alternate segment.)

Step 2: Angles on a straight line at T.

\angle STR = 180^{\circ} - 70^{\circ} = 110^{\circ}

(Reason:

QS

is a straight line, so

\angle QTR

and

\angle STR

add up to

180^{\circ}

Step 3: Apply the "centre" assumption.

T

were the centre,

TS

and

TR

would be radii, and:

\angle STR = 2 \times \angle SQR = 2 \times 58^{\circ} = 116^{\circ}

(Reason: The angle at the centre is twice the angle at the circumference standing on the same arc

SR

Step 4: Compare.

\text{But } \angle STR = 110^{\circ} \ne 116^{\circ}.

(Reason: The actual angle at

T

does not match the value it would have if

T

were the centre.)

Since

\angle STR = 110^{\circ} \ne 116^{\circ} = 2 \times \angle SQR

T

is not the centre of the circle.

Verification

Check 1: Straight line at

T

\angle QTR + \angle STR = 70^{\circ} + 110^{\circ} = 180^{\circ}

Check 2: If

T

were the centre,

\angle STR = 2 \times 58^{\circ} = 116^{\circ}

, but the straight line gives

110^{\circ}

- the two disagree.

Check 3:

110^{\circ} \ne 116^{\circ}

confirms

T

cannot be the centre.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle SQR = 58^{\circ}$	M1	Reason - alternate segment theorem	✓
$\angle STR = 110^{\circ}$	M1	Method - angles on a straight line	✓
If centre, $\angle STR = 116^{\circ}$	M1	Reason - angle at centre is twice the circumference	✓
$116^{\circ} \ne 110^{\circ}$ , so $T$ is not the centre	A1	Reason - the values disagree	✓

Full marks: 4/4

The diagram shows a circle, centre $O$ . $P$ , $Q$ , $R$ and $S$ lie on the circle. $JP$ and $JS$ are tangents from $J$ . $PK$ and $SK$ are straight lines. $\angle PJS = 40^{\circ}$ , $\angle OPQ = 52^{\circ}$ and $\angle PQR = 116^{\circ}$ . Find the size of $\angle QKR$ .

\angle QKR =

^{\circ}

[Total 5 marks]

GRADE9

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Question 13 - Exam Solution

Understanding the Question

Given

Circle, centre

O

;

P, Q, R, S

on the circle.

JP

and

JS

are tangents from

J

;

PK

and

SK

are straight lines.

\angle PJS = 40^{\circ}

\angle OPQ = 52^{\circ}

\angle PQR = 116^{\circ}

Find

The size of

\angle QKR

Plan the Solution

Tangents from $J$ are equal, so triangle $JPS$ is isosceles: find the base angles.
Tangent meets radius, then the isosceles triangle $OPS$ : find $\angle SPO$ .
Add at $P$ to get $\angle SPQ$ , then use the cyclic quadrilateral for $\angle SRQ$ .
Angles on the straight lines $PK$ and $SK$ : find $\angle KQR$ and $\angle KRQ$ .
Angles in triangle $QKR$ : find $\angle QKR$ .

Worked Solution [5 marks]

Rule - Work from the tangents to the triangle: build up the angles from the two tangents, use the centre and the cyclic quadrilateral, then finish with the triangle at

K

Step 1: Tangents from J are equal (isosceles triangle).

\angle JPS = \angle JSP = \tfrac{1}{2}(180^{\circ} - 40^{\circ}) = \tfrac{1}{2} \times 140^{\circ} = 70^{\circ}

(Reason:

JP

and

JS

are equal tangents from

J

, so triangle

JPS

is isosceles with equal base angles.)

Step 2: Tangent-radius, then isosceles triangle OPS.

\angle JSO = 90^{\circ}, \text{ so } \angle PSO = 90^{\circ} - 70^{\circ} = 20^{\circ}, \text{ and } \angle SPO = \angle PSO = 20^{\circ}

(Reason:

OS

is a radius and

JS

a tangent, so

\angle JSO = 90^{\circ}

; and

OP = OS

are radii, so triangle

OPS

is isosceles with equal base angles.)

Step 3: Cyclic quadrilateral PQRS.

\angle SPQ = \angle SPO + \angle OPQ = 20^{\circ} + 52^{\circ} = 72^{\circ}

\angle SRQ = 180^{\circ} - 72^{\circ} = 108^{\circ}

(Reason: Adding the parts at

P

gives

\angle SPQ

; then

PQRS

is a cyclic quadrilateral, so opposite angles

\angle SPQ

and

\angle SRQ

add up to

180^{\circ}

Step 4: Angles on the straight lines.

\angle KQR = 180^{\circ} - 116^{\circ} = 64^{\circ} \quad \text{and} \quad \angle KRQ = 180^{\circ} - 108^{\circ} = 72^{\circ}

(Reason:

PQK

and

SRK

are straight lines, so the angles at

Q

and at

R

each add up to

180^{\circ}

Step 5: Angles in triangle QKR.

\angle QKR = 180^{\circ} - 64^{\circ} - 72^{\circ} = 44^{\circ}

(Reason: The angles in triangle

QKR

add up to

180^{\circ}

\angle QKR = 44^{\circ}

Verification

Check 1: Triangle

JPS

40^{\circ} + 70^{\circ} + 70^{\circ} = 180^{\circ}

Check 2: Cyclic quadrilateral:

\angle SPQ + \angle SRQ = 72^{\circ} + 108^{\circ} = 180^{\circ}

Check 3: Triangle

QKR

64^{\circ} + 72^{\circ} + 44^{\circ} = 180^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle JPS = \angle JSP = 70^{\circ}$	M1	Method - tangents from a point are equal (isosceles)	✓
$\angle SPO = 20^{\circ}$	M1	Method - tangent-radius and isosceles triangle	✓
$\angle SRQ = 108^{\circ}$	M1	Method - cyclic quadrilateral opposite angles	✓
$\angle KQR = 64^{\circ}, \angle KRQ = 72^{\circ}$	M1	Method - angles on straight lines	✓
$\angle QKR = 44^{\circ}$	A1	Accuracy - angles in a triangle	✓

Full marks: 5/5

The diagram shows a circle, centre $O$ . $JK$ is a tangent at $P$ and $ST$ is a tangent at $R$ . Here $x$ is the tangent-chord angle at $P$ , $y$ is the tangent-chord angle at $R$ , and $b = \angle ORQ$ . Prove that $b = x + y - 90^{\circ}$ . State any circle theorems that you use.

[Total 4 marks]

GRADE9Problem solving

Show solution & mark schemeHide solution & mark scheme

Question 14 - Exam Solution

Understanding the Question

Given

Circle, centre

O

JK

is a tangent at

P

and

ST

is a tangent at

R

x

is the tangent-chord angle at

P

y

the tangent-chord angle at

R

, and

b = \angle ORQ

Prove

That

b = x + y - 90^{\circ}

Plan the Solution

Alternate segment theorem: $\angle PRQ = x$ .
Tangent meets radius at $R$ : $\angle ORS = 90^{\circ}$ .
Subtract to find $\angle ORP = 90^{\circ} - y$ .
$b$ is $\angle PRQ$ minus $\angle ORP$ ; simplify to get the result.

Worked Solution [4 marks]

Rule - Build the target from known angles: write

b

as a difference of angles found with the alternate segment theorem and the tangent-radius right angle, then simplify.

Step 1: Alternate segment theorem.

\angle PRQ = x

(Reason:

x

is the angle between the tangent at

P

and the chord

PQ

, so by the alternate segment theorem it equals

\angle PRQ

in the alternate segment.)

Step 2: Tangent meets radius.

\angle ORS = 90^{\circ}

(Reason:

OR

is a radius and

RS

is a tangent; a tangent meets a radius at

90^{\circ}

Step 3: Subtract to find angle ORP.

\angle ORP = \angle ORS - \angle SRP = 90^{\circ} - y

(Reason:

y = \angle SRP

is part of

\angle ORS

, so subtracting it gives

\angle ORP

Step 4: Form b and simplify.

b = \angle PRQ - \angle ORP = x - (90^{\circ} - y) = x + y - 90^{\circ}

(Reason:

b

is the part of

\angle PRQ

that remains after

\angle ORP

is removed; expanding the bracket gives the result.)

Therefore

b = x + y - 90^{\circ}

, as required.

Verification

Check 1: For example, if

x = 55^{\circ}

and

y = 60^{\circ}

, the formula gives

b = 55^{\circ} + 60^{\circ} - 90^{\circ} = 25^{\circ}

, which agrees with the actual

\angle ORQ

Check 2: Rearranging gives

x + y = b + 90^{\circ}

, so the two tangent-chord angles together are always

90^{\circ}

more than

b

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle PRQ = x$	M1	Reason - alternate segment theorem	✓
$\angle ORS = 90^{\circ}$	M1	Reason - tangent meets a radius at 90 degrees	✓
$\angle ORP = 90^{\circ} - y$	M1	Method - angle subtraction	✓
$b = x + y - 90^{\circ}$	A1	Accuracy - correct algebraic result	✓

Full marks: 4/4

Download printable PDF

Print it, work offline, mark yourself against the scheme.

Frequently Asked Questions

There are eight main circle theorems at GCSE: the angle in a semicircle is 90 degrees; the angle at the centre is twice the angle at the circumference; angles in the same segment are equal; opposite angles of a cyclic quadrilateral add up to 180 degrees; a tangent meets a radius at 90 degrees; the two tangents from a point are equal in length; the perpendicular from the centre bisects a chord; and the alternate segment theorem.

GCSE Maths uses eight main circle theorems. Most exam questions combine two or three of them, so it helps to know all eight and to recognise which one applies from the diagram.

The alternate segment theorem says that the angle between a tangent and a chord equals the angle in the alternate segment, which is the inscribed angle on the other side of the chord. It is often the hardest circle theorem to spot, and it appears frequently in Grade 8 and Grade 9 questions.

Circle theorems are Higher tier content in GCSE Maths for AQA, Edexcel and OCR. They are usually worth 3 to 5 marks and are graded from Grade 7 up to Grade 9, with the alternate segment theorem and multi-step proofs at the top end.

Look at the diagram, work out which theorem each given angle points to, write down every angle you find with its reason, and chain the steps until you reach the required angle. In the exam you must state the theorem at each step, because the reasons carry marks.

Yes. Circle theorems appear on almost every Higher tier GCSE Maths paper, usually as a multi-step find the angle giving reasons question or a short proof. They are a reliable source of marks once you know the eight theorems.

Keep revising

Once you are confident with circle theorems, browse the full topic lists for GCSE Maths topics and IGCSE Maths topics, and check the GCSE grade boundaries and IGCSE grade boundaries to set your target. For more exam-style practice, try the Simultaneous Equations Practice Questions.

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Circle Theorems Practice Questions

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