Circle Theorems – GCSE Maths

1 The Eight Theorems

There are eight circle theorems to learn. Every one has an exact form of words that must be quoted as your reason — a rough description will not earn the mark.

Learn the wording, not just the picture. The reason is worth a mark on its own in almost every circle-theorem question.

Theorem 1 — Angle in a semicircle

If $AB$ is a diameter and $C$ is any other point on the circle, then $\angle ACB = 90^\circ$.

Reason to write: "the angle in a semicircle is $90^\circ$"

Theorem 2 — Angle at the centre

The angle subtended at the centre is twice the angle subtended at the circumference, when both stand on the same arc.

Reason to write: "the angle at the centre is twice the angle at the circumference"

Theorem 3 — Angles in the same segment

Angles in the same segment, standing on the same arc, are equal.

Reason to write: "angles in the same segment are equal"

Theorem 4 — Cyclic quadrilateral

Opposite angles of a cyclic quadrilateral (all four vertices on the circle) add to $180^\circ$.

Reason to write: "opposite angles of a cyclic quadrilateral add up to $180^\circ$"

Theorem 5 — Tangent and radius

A tangent meets the radius at the point of contact at $90^\circ$.

Reason to write: "the angle between a tangent and a radius is $90^\circ$"

Theorem 6 — Two tangents

The two tangents from an external point are equal in length, and the line from that point to the centre bisects the angle between them.

Reason to write: "tangents from an external point are equal"

Theorem 7 — Perpendicular from the centre to a chord

The perpendicular from the centre to a chord bisects the chord.

Reason to write: "the perpendicular from the centre bisects the chord"

Theorem 8 — Alternate segment theorem

The angle between a tangent and a chord equals the angle in the alternate segment — the angle subtended by that chord on the other side.

Reason to write: "alternate segment theorem"

2 Proving the Theorems

Higher papers sometimes ask you to prove a circle theorem. Two proofs come up most often.

Worked Example 1 — Proving the angle at the centre theorem

Prove that the angle at the centre is twice the angle at the circumference.

①Let $A$, $B$ and $C$ be on the circle with centre $O$. Draw the line $CO$ and extend it to a point $D$ on the far side.

②$OA = OC$ (radii), so triangle $OAC$ is isosceles. Let $\angle OCA = \angle OAC = a$.

③$\angle AOD$ is the exterior angle of triangle $OAC$, so $\angle AOD = a + a = 2a$.

④Similarly $OB = OC$, so let $\angle OCB = \angle OBC = b$, giving $\angle BOD = 2b$.

⑤$\angle ACB = a + b$ and $\angle AOB = 2a + 2b = 2(a+b)$.

⑥Therefore $\angle AOB = 2 \times \angle ACB$ ∎

Worked Example 2 — Proving the cyclic quadrilateral theorem

Prove that opposite angles of a cyclic quadrilateral add to $180^\circ$.

①Let $ABCD$ be a cyclic quadrilateral with centre $O$. Join $OB$ and $OD$.

②The angle at the centre standing on arc $BCD$ is twice $\angle BAD$ (angle at the centre is twice the angle at the circumference).

③The reflex angle at the centre standing on arc $BAD$ is twice $\angle BCD$, for the same reason.

④The two angles at $O$ together make a full turn: $2\angle BAD + 2\angle BCD = 360^\circ$.

⑤Dividing by $2$: $\angle BAD + \angle BCD = 180^\circ$ ∎

Theorem 1 is a special case of Theorem 2. If $AB$ is a diameter, the angle at the centre is the straight angle $180^\circ$, so the angle at the circumference is $180 \div 2 = 90^\circ$.

3 How to Spot Which Theorem to Use

What you see in the diagram	Theorem to try
A line through the centre (a diameter) with a triangle on it	Angle in a semicircle $= 90^\circ$
An angle at the centre and an angle at the edge on the same arc	Centre angle is twice the other
Two angles "pointing at" the same chord from the same side	Angles in the same segment are equal
A four-sided shape with all corners on the circle	Opposite angles add to $180^\circ$
A straight line touching the circle, with a radius drawn	Tangent meets radius at $90^\circ$
Two lines from one outside point touching the circle	Equal tangents; a kite is formed
A tangent with a chord drawn from the point of contact	Alternate segment theorem
Two radii and a chord	Isosceles triangle

Always mark the radii. If the centre is shown, draw in every radius you can. Each one creates an isosceles triangle, and those often supply the missing step.

4 Worked Examples

Worked Example 3 — Angle at the centre

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

①Both angles stand on the same arc $AB$.

②The angle at the centre is twice the angle at the circumference.

③$\angle ACB = 130 \div 2 = 65^\circ$

Worked Example 4 — The reflex trap

$P$, $Q$ and $R$ are on a circle with centre $O$. Angle $PRQ = 110^\circ$, where $R$ is on the minor arc. Find the obtuse angle $POQ$.

①The angle at the centre is twice the angle at the circumference: $2 \times 110 = 220^\circ$.

②But $220^\circ$ is more than $180^\circ$ — this is the reflex angle at $O$.

③The obtuse angle $POQ = 360 - 220 = 140^\circ$

Watch for this. When the angle at the circumference is obtuse, doubling gives the reflex angle at the centre. Subtract from $360^\circ$ to get the "ordinary" one.

Worked Example 5 — Cyclic quadrilateral with algebra

In cyclic quadrilateral $ABCD$, $\angle A = 3x + 10$ and $\angle C = 2x + 20$. Find $x$ and both angles.

①$A$ and $C$ are opposite angles, so they add to $180^\circ$.

②$(3x + 10) + (2x + 20) = 180$

③$5x + 30 = 180 \Rightarrow 5x = 150 \Rightarrow x = 30$

④$\angle A = 100^\circ$ and $\angle C = 80^\circ$. Check: $100 + 80 = 180$ ✓

Worked Example 6 — Multi-step chase

$AB$ is a diameter of a circle with centre $O$. $C$ is a point on the circle with $\angle CAB = 34^\circ$. Find $\angle CBA$ and $\angle COB$.

①$AB$ is a diameter, so $\angle ACB = 90^\circ$ (angle in a semicircle).

②Angle sum of triangle $ABC$: $\angle CBA = 180 - 90 - 34 = 56^\circ$.

③For $\angle COB$: it is the angle at the centre standing on arc $CB$, and $\angle CAB = 34^\circ$ is the angle at the circumference on the same arc.

④$\angle COB = 2 \times 34 = 68^\circ$

Check: triangle $OCB$ is isosceles ($OC = OB$, radii), so its base angles are $(180-68)\div 2 = 56^\circ$ — matching $\angle CBA$ ✓

Worked Example 7 — Alternate segment

$TA$ is a tangent to a circle at $A$, and $AB$ is a chord. The angle between the tangent and the chord, $\angle TAB$, is $58^\circ$. $C$ is a point in the alternate segment. Find $\angle ACB$.

①The alternate segment theorem says the tangent–chord angle equals the angle in the alternate segment.

②$\angle ACB = \angle TAB = 58^\circ$

5 Quick Reference

Semicircle

Angle in a semicircle is $90^\circ$.

Centre

Angle at the centre is twice the angle at the circumference.

Same segment

Angles in the same segment are equal.

Cyclic quad

Opposite angles add to $180^\circ$.

Tangent–radius

They meet at $90^\circ$.

Two tangents

Equal in length; the shape formed is a kite.

Chord

The perpendicular from the centre bisects it.

Alternate segment

Tangent–chord angle equals the angle in the alternate segment.

Reflex trap

Doubling an obtuse circumference angle gives the reflex centre angle.

Always

Quote the theorem by name as your reason.

6 Practice Questions

Question 1

$AB$ is a diameter of a circle and $C$ lies on the circle. Angle $ABC = 27^\circ$. Find angle $BAC$, giving reasons.

▶ Show solution

$\angle ACB = 90^\circ$ — the angle in a semicircle is $90^\circ$.

$\angle BAC = 180 - 90 - 27 = 63^\circ$ — angle sum of a triangle.

Question 2

$O$ is the centre of a circle. $A$, $B$ and $C$ are on the circumference with $\angle ACB = 41^\circ$. Find $\angle AOB$.

▶ Show solution

$\angle AOB = 2 \times 41 = 82^\circ$

Reason: the angle at the centre is twice the angle at the circumference.

Question 3

In cyclic quadrilateral $PQRS$, $\angle P = 96^\circ$ and $\angle Q = 71^\circ$. Find $\angle R$ and $\angle S$.

▶ Show solution

$\angle R = 180 - 96 = 84^\circ$ — opposite angles of a cyclic quadrilateral add to $180^\circ$.

$\angle S = 180 - 71 = 109^\circ$ — same reason.

Check: $96 + 71 + 84 + 109 = 360^\circ$ ✓

Question 4

$A$, $B$, $C$ and $D$ lie on a circle. $\angle ACB = 38^\circ$. Find $\angle ADB$, giving a reason.

▶ Show solution

Both angles stand on the same chord $AB$ and are on the same side of it.

$\angle ADB = 38^\circ$ — angles in the same segment are equal.

Question 5

A tangent touches a circle at $T$. $O$ is the centre and $P$ is a point on the tangent with $\angle TOP = 62^\circ$. Find $\angle TPO$.

▶ Show solution

$\angle OTP = 90^\circ$ — the angle between a tangent and a radius is $90^\circ$.

$\angle TPO = 180 - 90 - 62 = 28^\circ$ — angle sum of a triangle.

Question 6

$A$, $B$ and $C$ lie on a circle with centre $O$, and $\angle ACB = 128^\circ$ where $C$ is on the minor arc. Find the obtuse angle $AOB$.

▶ Show solution

Angle at the centre $= 2 \times 128 = 256^\circ$ — but this is the reflex angle.

Obtuse $\angle AOB = 360 - 256 = 104^\circ$.

Question 7

$TA$ and $TB$ are tangents to a circle with centre $O$, touching at $A$ and $B$. $\angle ATB = 46^\circ$. Find (a) $\angle AOB$, (b) $\angle OAB$.

▶ Show solution

(a) $\angle OAT = \angle OBT = 90^\circ$ (tangent–radius).

Angles of quadrilateral $OATB$: $\angle AOB = 360 - 90 - 90 - 46 = 134^\circ$.

(b) $OA = OB$ (radii), so triangle $OAB$ is isosceles.

$\angle OAB = (180 - 134) \div 2 = 46 \div 2 = 23^\circ$.

Question 8

A tangent at $A$ makes an angle of $73^\circ$ with the chord $AB$. $C$ is a point in the alternate segment. Find $\angle ACB$, and find $\angle ABC$ given that $\angle BAC = 55^\circ$.

▶ Show solution

$\angle ACB = 73^\circ$ — alternate segment theorem.

Angle sum of triangle $ABC$: $\angle ABC = 180 - 73 - 55 = 52^\circ$.

Question 9

$AB$ is a diameter of a circle with centre $O$. $C$ and $D$ lie on the circle, on the same side of $AB$. $\angle CAB = 25^\circ$ and $\angle DAB = 61^\circ$.

(a) Find $\angle ACB$. (b) Find $\angle ADB$. (c) Find $\angle CBD$.

▶ Show solution

(a) $\angle ACB = 90^\circ$ — angle in a semicircle.

(b) $\angle ADB = 90^\circ$ — same reason ($AB$ is still the diameter).

(c) In triangle $ABC$: $\angle ABC = 180 - 90 - 25 = 65^\circ$.

In triangle $ABD$: $\angle ABD = 180 - 90 - 61 = 29^\circ$.

$C$ and $D$ are on the same side, so $\angle CBD = 65 - 29 = 36^\circ$.

Question 10

$A$, $B$, $C$ and $D$ lie on a circle with centre $O$. $TA$ is a tangent at $A$. $\angle TAB = 64^\circ$ and $\angle ADC = 118^\circ$.

(a) Find $\angle ACB$, with a reason. (b) Find $\angle ABC$. (c) Find the reflex angle $AOC$. (d) Find $\angle BAC$ given that $\angle ABC$ is the angle you found in (b) and $\angle ACB$ is from (a).

▶ Show solution

(a) $\angle ACB = \angle TAB = 64^\circ$ — alternate segment theorem.

(b) $ABCD$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$:

$\angle ABC = 180 - 118 = 62^\circ$.

(c) $\angle ABC = 62^\circ$ is the angle at the circumference standing on arc $ADC$.

The angle at the centre on the same arc is $2 \times 62 = 124^\circ$ — this is the non-reflex $\angle AOC$ measured through $D$'s side.

So the reflex angle $AOC = 360 - 124 = 236^\circ$.

(d) In triangle $ABC$: $\angle BAC = 180 - 64 - 62 = 54^\circ$ — angle sum of a triangle.

Circle Theorems (G10) · GCSE Maths Revision · Created with MathJax