1MA1 · G9–G10Geometry & MeasuresHigherGrades 6–9

Circle Theorems

Draw a circle, mark some points on it, join them up — and hidden angle relationships appear, true for every circle ever drawn. These $\textbf{circle theorems}$ turn diagrams full of chords, tangents and radii into angle puzzles, and Edexcel Higher papers expect you to both use them and name them precisely.

55 min Angles & Geometric Reasoning

The big picture

Almost every theorem follows from one fact about radii: they are all equal, so any triangle with two radii as sides is isosceles. The key result — the angle at the centre is twice the angle at the circumference — is proved with two such isosceles triangles, and several other theorems follow from it.

Exam questions usually need two or three theorems combined with basic angle facts, and "give reasons" means naming each theorem in full. Spotting the shapes — the arrowhead, the bow tie, the cyclic quadrilateral — is the practical skill.

What you'll learn

State and use the circle theorems involving angles at the centre and circumference
Use cyclic quadrilateral, tangent and alternate segment theorems
Combine circle theorems with other angle facts, giving reasons
Prove circle theorems using angle facts and isosceles triangles

Angles from the same arc

$\textbf{The angle at the centre is twice the angle at the circumference}$ subtended by the same arc. Look for an arrowhead or "V" shape.

$\textbf{The angle in a semicircle is a right angle}$ : if one side of a triangle is a diameter, the angle opposite it is $90^{\circ}$ . (A special case of the centre theorem, since the angle at the centre is $180^{\circ}$ .)

$\textbf{Angles in the same segment are equal}$ : angles subtended by the same chord, on the same side, are equal. Look for a "bow tie".

Example 1Points

A

B

and

C

lie on a circle with centre

O

. Angle

AOB = 124^{\circ}

. Find angle

ACB

, where

C

is on the major arc.

1The angle at the centre is twice the angle at the circumference.

2Angle

ACB = 124^{\circ} \div 2

Answer

62^{\circ}

Your turnSame method, new numbers — try it before moving on.

Angle $AOB$ at the centre is $150^\circ$ . Find angle $ACB$ at the circumference, with $C$ on the major arc.

Your answer

Show worked solution

Solution

1The angle at the centre is twice the angle at the circumference.
2 $150 \div 2 = 75^\circ$ .

Answer: $75^\circ$

Tip — Check both angles stand on the same arc before halving. The centre angle and the circumference angle must "look at" the same two points.

Cyclic quadrilaterals

A $\textbf{cyclic quadrilateral}$ has all four vertices on a circle.

$\textbf{Opposite angles in a cyclic quadrilateral sum to }180^{\circ}$ .

Not every quadrilateral is cyclic — check all four corners touch the circle.

Example 1

PQRS

is a cyclic quadrilateral with angle

PQR = 108^{\circ}

. Find angle

PSR

PQR

and

PSR

are opposite angles.

2Opposite angles in a cyclic quadrilateral sum to

180^{\circ}

AnswerAngle

PSR = 72^{\circ}

Your turnSame method, new numbers — try it before moving on.

In a cyclic quadrilateral, one angle is $115^\circ$ . Find the opposite angle.

Your answer

Show worked solution

Solution

1Opposite angles of a cyclic quadrilateral sum to $180^\circ$ .
2 $180 - 115 = 65^\circ$ .

Answer: $65^\circ$

Tangents

$\textbf{A tangent meets a radius at }90^{\circ}$ at the point of contact.

$\textbf{Tangents from an external point are equal in length}$ , forming a kite with the two radii.

$\textbf{The perpendicular from the centre to a chord bisects the chord}$ .

$\textbf{Alternate segment theorem}$ : the angle between a tangent and a chord equals the angle in the alternate segment — the angle at the circumference on the opposite side of the chord.

Example 1Two tangents from point

T

touch a circle with centre

O

A

and

B

. Angle

ATB = 46^{\circ}

. Find angle

AOB

1Angles

OAT

and

OBT

are

90^{\circ}

: a tangent meets a radius at

90^{\circ}

2Angles in the quadrilateral

OATB

sum to

360^{\circ}

3Angle

AOB = 360 - 90 - 90 - 46

Answer

134^{\circ}

Your turnSame method, new numbers — try it before moving on.

Two tangents from $T$ touch a circle with centre $O$ at $A$ and $B$ . Angle $ATB = 50^\circ$ . Find angle $AOB$ .

Your answer

Show hint

Angles $OAT$ and $OBT$ are both $90^\circ$ .

Show worked solution

Solution

1 $OATB$ is a quadrilateral with two right angles.
2 $360 - 90 - 90 - 50 = 130^\circ$ .

Answer: $130^\circ$

Why this works

The alternate segment theorem is the hardest to spot. Find the tangent and the chord from the point of contact, then look across the chord to the triangle in the other segment.

Proving a theorem

Proofs use radii to create isosceles triangles, then basic angle facts.

To prove the angle at the centre is twice the angle at the circumference: join the circumference point to the centre and extend the line. This creates two isosceles triangles, each with an exterior angle at the centre equal to twice a base angle.

Example 1Prove that the angle at the centre is twice the angle at the circumference (for the case where the centre lies inside the angle).

1Let the circle have centre

O

, with

A

B

on the arc and

C

on the circumference. Join

CO

and extend it to

D

OA = OC

(radii), so triangle

OAC

is isosceles with base angles

x

. The exterior angle

AOD = 2x

3Similarly,

OB = OC

gives base angles

y

and exterior angle

BOD = 2y

4So angle

AOB = 2x + 2y = 2(x + y) = 2 \times

angle

ACB

AnswerAngle

AOB = 2 \times

angle

ACB

Think like an examiner

Name each theorem in full as a reason.
Mark radii as equal and use isosceles triangles.
Check the correct arc or segment before applying a theorem.

Circle theorems

\angle \text{centre} = 2 \times \angle \text{circumference}

Same arc.

\angle \text{in a semicircle} = 90^{\circ}

Diameter.

\text{opposite angles} = 180^{\circ}

Cyclic quadrilateral.

\text{tangent} \perp \text{radius}

Tangent meets radius at 90°.

Watch out for these

Halving the centre angle when it stands on a different arc.

Both angles must be subtended by the same arc.

Using the cyclic quadrilateral rule when one vertex is the centre.

All four vertices must lie on the circle.

Writing "tangent rule" as a reason.

Write "a tangent meets a radius at

90^{\circ}

Forgetting that triangles with two radii are isosceles.

Equal radii give equal base angles.

Stretch yourself

$A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ . $AC$ is a diameter. Angle $BAC = 34^{\circ}$ and angle $CAD = 22^{\circ}$ . Find angle $BCD$ and angle $BOD$ , giving reasons.

Hint — Use the angle in a semicircle for triangles $ABC$ and $ADC$ , or use the cyclic quadrilateral $ABCD$ .

Reveal solution

SolutionAngle

BAD = 34^{\circ} + 22^{\circ} = 56^{\circ}

ABCD

is a cyclic quadrilateral, and opposite angles in a cyclic quadrilateral sum to

180^{\circ}

, so angle

BCD = 180^{\circ} - 56^{\circ} = 124^{\circ}

. Angle

BOD

at the centre stands on arc

BD

— the same arc as angle

BAD

at the circumference. The angle at the centre is twice the angle at the circumference, so angle

BOD = 2 \times 56^{\circ} = 112^{\circ}

\blacksquare

Questions students ask

How many circle theorems do I need?

About eight: centre and circumference, semicircle, same segment, cyclic quadrilateral, tangent–radius, equal tangents, chord bisector, and alternate segment.

How do I spot which theorem to use?

Look for shapes: an arrowhead (centre), a bow tie (same segment), four points on the circle (cyclic quadrilateral), a tangent (right angle or alternate segment).

Do I need to prove the theorems?

Edexcel can ask for proofs of the main theorems, so learn the isosceles-triangle argument.

Key takeaways

Angle at the centre is twice the angle at the circumference on the same arc.
Angle in a semicircle is $90^{\circ}$ ; angles in the same segment are equal.
Opposite angles of a cyclic quadrilateral sum to $180^{\circ}$ .
Tangent ⊥ radius; tangents from a point are equal; alternate segment theorem.
Name every theorem fully as a reason.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 63 marks

$AB$ is a diameter of a circle and $C$ is a point on the circle. Angle $CAB = 34^\circ$ .

Work out angle $ABC$ . Give reasons for your answer.[3]
Show mark scheme
- B1Angle $ACB = 90^\circ$ (angle in a semicircle)
- M1 $180 - 90 - 34$
- A1 $56^\circ$ with angles in a triangle sum to $180^\circ$
Answer: $56^\circ$

Question 2Grade 72 marks

A tangent touches a circle at $A$ . $AB$ is a chord, and the angle between the tangent and $AB$ is $63^\circ$ . $C$ is a point on the major arc.

Work out angle $ACB$ . Give a reason.[2]
Show mark scheme
- B1 $63^\circ$
- B1Alternate segment theorem
Answer: $63^\circ$

Question 3Grade 83 marks

$A$ , $B$ and $C$ lie on a circle with centre $O$ . Angle $OAB = 25^\circ$ , and $C$ is on the major arc.

Work out angle $ACB$ . Give reasons for each step.[3]
Show mark scheme
- B1Triangle $OAB$ is isosceles ( $OA = OB$ are radii), so angle $OBA = 25^\circ$
- M1Angle $AOB = 130^\circ$
- A1Angle $ACB = 65^\circ$ , the angle at the centre is twice the angle at the circumference
Answer: $65^\circ$

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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