Revision notes for AQA GCSE Maths Circle Theorems. Open the guide for explanations and worked examples. Written against the AQA GCSE Maths (8300) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for AQA GCSE Maths Circle Theorems. Open the guide for explanations and worked examples. Written against the AQA GCSE Maths (8300) specification, so the content matches what's examinable rather than general Maths background.
Circle vocabulary

All radii in the same circle are equal. So if you join the centre to two points on the circumference, you often create an isosceles triangle, meaning a triangle with two equal sides.
Look for equal radii
If two sides of a triangle are radii of the same circle, the two base angles are equal.
Finding an angle at the centre
A and B are on a circle with centre O. In triangle AOB, angle ABO is 42°. Find angle AOB.

OA and OB are both radii, so OA=OBOA = OBOA=OB.
Triangle AOB is isosceles, so the base angles are equal: ∠OAB=42∘\angle OAB = 42^\circ∠OAB=42∘.
Angles in a triangle add to 180°, so ∠AOB=180∘−42∘−42∘=96∘\angle AOB = 180^\circ - 42^\circ - 42^\circ = 96^\circ∠AOB=180∘−42∘−42∘=96∘.
Perpendicular means meeting at a right angle, 90°. The radius drawn to the point where a tangent touches the circle is perpendicular to the tangent.
Tangent-radius theorem
A tangent to a circle is perpendicular to the radius at the point of contact.
Two tangents and a centre angle
A and C are points on a circle with centre O. Lines AB and CB are tangents, and angle ABC is 50°. Find angle OAC.

OA meets tangent AB at A, so ∠OAB=90∘\angle OAB = 90^\circ∠OAB=90∘. OC meets tangent CB at C, so ∠OCB=90∘\angle OCB = 90^\circ∠OCB=90∘.
Angles in quadrilateral AOCB add to 360°, so ∠AOC=360∘−90∘−90∘−50∘=130∘\angle AOC = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ∠AOC=360∘−90∘−90∘−50∘=130∘.
OA and OC are radii, so OA=OCOA = OCOA=OC. Triangle AOC is isosceles.
The base angles are equal, so ∠OAC=180∘−130∘2=25∘\angle OAC = \frac{180^\circ - 130^\circ}{2} = 25^\circ∠OAC=2180∘−130∘=25∘.
Wrong radius
The 90° angle is only between the tangent and the radius drawn to the exact point of contact.
If two angles stand on the same chord or arc, the angle at the centre is twice the angle at the circumference.
Using the centre-circumference rule
B and C are points on a circle with centre O. The angle BOC at the centre is 74°. A is another point on the circumference. Find angle BAC.

Both angles stand on chord BC.
The angle at the centre is twice the angle at the circumference.
Therefore ∠BAC=74∘2=37∘\angle BAC = \frac{74^\circ}{2} = 37^\circ∠BAC=274∘=37∘.
Check for reflex angles
A reflex angle is bigger than 180°. If the reflex centre angle is labelled, halve that reflex angle; if you need the smaller centre angle, subtract the reflex angle from 360° first.
Two very useful chord facts are:
Same segment and cyclic quadrilateral
A, B, C and D lie on a circle. C and D are in the same segment with chord AB. Angle ADB is 47°, and angle ADC is 78°. Find angles ACB and ABC.

Angles ADB and ACB both stand on chord AB.
Angles in the same segment are equal, so ∠ACB=47∘\angle ACB = 47^\circ∠ACB=47∘.
ABCD is a cyclic quadrilateral, so opposite angles add to 180°.
Therefore ∠ABC=180∘−78∘=102∘\angle ABC = 180^\circ - 78^\circ = 102^\circ∠ABC=180∘−78∘=102∘.
The alternate segment theorem connects tangents and chords.
It says: the angle between a tangent and a chord is equal to the angle in the opposite segment.
Tangent and chord angle
A, B and C are on a circle. A tangent touches the circle at C. Angle ABC is 58°, angle ACB is 71°, and x is the angle between the tangent and chord CB. Find x.

First find the missing angle in triangle ABC: ∠BAC=180∘−58∘−71∘=51∘\angle BAC = 180^\circ - 58^\circ - 71^\circ = 51^\circ∠BAC=180∘−58∘−71∘=51∘.
The angle x is between the tangent and chord CB.
By the alternate segment theorem, this equals the angle standing on chord CB in the opposite segment, which is angle BAC.
So x=51x = 51x=51.
Match the chord carefully
If the tangent angle uses chord CB, look for the angle on the circumference made by chord CB. If it uses chord CA, look for the angle made by chord CA.
Circle theorem questions can include ordinary geometry too. If you get a tangent and a radius, you often get a right-angled triangle, so you may need Pythagoras’ theorem.
Pythagoras’ theorem says that in a right-angled triangle, the square of the longest side equals the sum of the squares of the other two sides.
Finding a length using a tangent
AC is a tangent at A. O is the centre, and O, B and C lie on a straight line. OA is 6 cm and AC is 8 cm. Find BC.

OA is a radius to the tangent at A, so triangle OAC is right-angled at A.
Use Pythagoras’ theorem to find OC:
OC2=62+82OC2=100OC=10\begin{aligned} OC^2 &= 6^2 + 8^2 \\ OC^2 &= 100 \\ OC &= 10 \end{aligned}OC2OC2OC=62+82=100=10OB is a radius, so OB is 6 cm.
Since O, B and C are in a straight line, BC is 10 - 6 = 4 cm.
In the exam
Mark all equal radii and all tangent-radius right angles on the diagram first.
Identify the chord or arc involved before choosing a theorem.
Write a reason after each calculation, such as “radii are equal” or “opposite angles in a cyclic quadrilateral add to 180°”.
If a question has several marks, expect a chain of two or three facts, not just one theorem.
Check yourself
Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.
Test yourself on this topic, or move on to the next guide.
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