What are the basic Circle Theorems covered in the Cambridge IGCSE Mathematics curriculum?

The Cambridge IGCSE Mathematics curriculum covers several key Circle Theorems, including the Angle at the Centre Theorem, the Angle in a Semicircle Theorem, the Angles in the Same Segment Theorem, the Cyclic Quadrilateral Theorem, and the Tangent-Secant Theorem. These theorems help in understanding the properties and relationships between angles and segments in and around circles.

How does the Angle at the Centre Theorem work in circle geometry?

The Angle at the Centre Theorem states that the angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the circumference. This theorem is fundamental in solving problems related to angles in circle geometry, as it establishes a clear relationship between central and inscribed angles.

Can you explain the Cyclic Quadrilateral Theorem?

The Cyclic Quadrilateral Theorem states that in a cyclic quadrilateral (a four-sided figure where all vertices lie on the circumference of a circle), the opposite angles sum up to 180 degrees. This theorem is crucial for solving problems involving cyclic quadrilaterals and is often used in conjunction with other circle theorems to find unknown angles.

What is the significance of the Tangent-Secant Theorem in circle geometry?

The Tangent-Secant Theorem, also known as the Power of a Point Theorem, states that if a tangent and a secant are drawn from a point outside a circle, then the square of the length of the tangent segment is equal to the product of the lengths of the entire secant segment and its external segment. This theorem is significant for solving problems involving lengths and distances in circle geometry.

How can the Angle in a Semicircle Theorem be applied in solving geometry problems?

The Angle in a Semicircle Theorem states that any angle subtended by a diameter on the circumference of a circle is a right angle (90 degrees). This theorem is particularly useful in solving problems where right angles need to be identified or verified, and it is often applied in conjunction with other geometric principles to solve complex problems in circle geometry.

Why is "Stating the angle without naming the theorem" a common mistake in Circle Theorems?

Why it happens: Students see the answer and write it. How to avoid it: Mark schemes credit the THEOREM name — "angle at centre", "same segment", "alternate segment" etc. Always include it. Source: 0580/42 — every series

Why is "Doubling instead of halving (or vice versa) for centre/circumference" a common mistake in Circle Theorems?

Why it happens: Students forget which is twice the other. How to avoid it: Mnemonic: angle at the CENTRE is BIGGER. So centre = 2 × circumference.

Why is "Pairing adjacent angles in a cyclic quadrilateral" a common mistake in Circle Theorems?

Why it happens: Confusing "opposite" with "adjacent". How to avoid it: Opposite angles in a cyclic quad sum to 180 — they're across from each other, not next to each other.

Why is "Picking an angle from the wrong segment" a common mistake in Circle Theorems?

Why it happens: Confusing which segment is "alternate". How to avoid it: Alternate segment is on the OTHER side of the chord from the tangent angle.

Cambridge IGCSE Mathematics (0580)

Circle Theorems

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Short Notes - Circle Theorems

Detailed Study Notes

Detailed notes on Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Circle Theorems — Cambridge IGCSE 0580 Maths Extended (2026)

The seven named theorems that turn diagrams into angle equations. Cambridge mark schemes credit the theorem name as much as the answer — memorise them precisely.

At a glance

Angle at centre = 2 × angle at circumference subtending the same arc.
Angles in the same segment are equal.
Angle in a semicircle = $90\degree$ .
Cyclic quadrilateral: opposite angles sum to $180\degree$ .
Tangent ⊥ radius at the point of contact.
Two tangents from a point are equal in length.
Alternate segment theorem: angle between tangent and chord = angle in alternate segment.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E4.7 — Use the following circle theorems: angle at centre, angles in same segment, angle in semicircle, cyclic quadrilateral, tangent-radius, equal tangents, alternate segment.

Angle at centre = 2 × angle at circumference

Same arc, same side. The centre angle is double.

If two angles in a circle subtend the same arc — one at the centre and one at the circumference (on the same side) — the angle at the centre is twice the angle at the circumference.

Worked. Centre angle = $130\degree$ . Find the circumference angle subtending the same arc.

$\theta = 130 / 2 = 65\degree$ .
Reason: angle at centre is twice angle at circumference.

Special case. If the chord is a DIAMETER, the centre angle is $180\degree$ , so the circumference angle is $90\degree$ — that's the angle in a semicircle = $90\degree$ theorem.

Chord AB subtends 130° at centre O and half of that, 65°, at circumference point P.

Same arc, same side.
Centre angle = $2 \times$ circumference angle.
Diameter case: angle in semicircle = $90\degree$ .
ALWAYS state the theorem name in the reason.

Angles in the same segment

All angles subtended by the same chord, on the same side, are equal.

Theorem. If multiple angles at the circumference are subtended by the SAME chord, on the SAME side of that chord, they are all EQUAL.

Worked. Two angles at the circumference subtended by the same chord, on the same side. One is $48\degree$ . The other:

$48\degree$ .
Reason: angles in the same segment are equal.

Tip. Look for two angles whose vertices are on the same arc and whose sides reach the same two endpoints (the chord).

P and Q lie on the same arc above chord AB, so both subtend equal angles.

Same chord + same side → equal angles.
Cite the theorem by name.

Cyclic quadrilateral

Opposite angles in a cyclic quadrilateral sum to $180\degree$ .

A cyclic quadrilateral has all four vertices on the circle.

Theorem. Opposite angles of a cyclic quadrilateral sum to $180\degree$ .

Worked. $ABCD$ is cyclic. $\angle A = 78\degree$ . Find $\angle C$ .

$\angle A + \angle C = 180\degree \Rightarrow \angle C = 102\degree$ .
Reason: opposite angles in a cyclic quadrilateral.

Don't confuse with adjacent. Opposite means ACROSS, not next to. Adjacent angles in a cyclic quad don't generally satisfy the rule.

Opposite vertices A and C lie across the quadrilateral; their angles add to 180°.

All four vertices on the circle.
Opposite angles sum to $180\degree$ .
Two pairs of opposite angles each sum to $180\degree$ .

Tangent properties

Tangent ⊥ radius at the point of contact. Two tangents from an external point are equal.

Tangent perpendicular to radius. A tangent meets the radius drawn to the point of contact at $90\degree$ .

Worked. Tangent $TA$ touches circle at $A$ ; centre $O$ . The angle $\angle OAT = 90\degree$ .

Reason: tangent perpendicular to radius at point of contact.

Two tangents from an external point. If two tangents are drawn to a circle from an external point $P$ , the two tangent lengths are EQUAL.

Worked. $PA$ and $PB$ are tangents to a circle from $P$ , touching at $A$ and $B$ . Then $PA = PB$ .

Reason: tangents from an external point are equal.

Each radius meets its tangent at 90°, and the two tangents from P are equal in length.

Alternate segment theorem. The angle between a tangent and a chord at the point of contact equals the angle in the alternate (opposite) segment subtended by that chord.

Worked. Tangent $TA$ at $A$ . Chord $AB$ makes angle $42\degree$ with the tangent. Find the angle in the alternate segment subtended by $AB$ .

$42\degree$ .
Reason: alternate segment theorem.

The tangent-chord angle equals the angle subtended by the same chord in the alternate segment.

Tangent ⊥ radius at the contact point.
Two tangents from external point: equal lengths.
Alternate segment: angle between tangent and chord = angle in opposite segment.
Cite each theorem by NAME for full marks.

Quick recap

Angle at centre = 2 × angle at circumference (same arc).
Angles in the same segment: equal.
Angle in a semicircle: $90\degree$ .
Cyclic quadrilateral: opposite angles sum to $180\degree$ .
Tangent ⊥ radius at contact.
Two tangents from a point: equal.
Alternate segment: tangent-chord angle = angle in alternate segment.
ALWAYS state the theorem name as your reason.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Angle at centre theorem — angle at centre = $2 \times$ angle at circumference subtending the same arc.
Angles in the same segment — angles subtended by the same arc/chord, on the same side, are equal.
Angle in a semicircle — $90\degree$ .
Cyclic quadrilateral — opposite angles sum to $180\degree$ .
Tangent perpendicular to radius — at the point of contact.
Two tangents from an external point — equal in length.
Alternate segment theorem — angle between tangent and chord = angle in alternate segment.

How it’s examined

Circle theorems are guaranteed on Paper 4 (5-8 marks). Cambridge stacks 2-3 theorems in one diagram and asks for an angle, with reasons. Paper 2 has simpler 2-3 mark single-theorem items. Examiner reports flag failure to STATE the theorem name as the recurring lost-mark scenario — even when the angle calculation is correct.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E4.7); 0580/42 Oct/Nov 2024 — Q9 (cyclic quadrilateral); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Circle Theorems

Step-by-step solutions to past-paper-style questions on circle theorems, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Circle Theorems

Almost every circle theorems exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A circle diagram with one angle to find — using a single named theorem such as angle at the centre, same segment, cyclic quadrilateral or alternate segment.

How to approach it

Identify which arc or chord the angle stands on, apply the one matching theorem (halve/double for centre, equate for same segment, $180\degree$ for opposite cyclic angles), and name the theorem beside the value.

Common trap

Writing the angle with no theorem named, doubling instead of halving for centre/circumference, or pairing adjacent rather than opposite cyclic angles. Examiner reports flag all three every series.

Multi-step problem

Recognise it by

A tangent diagram combining several facts — tangent perpendicular to radius plus Pythagoras, or two tangents from an external point forming a kite.

How to approach it

Use the tangent-radius right angle to set up a right triangle (for Pythagoras) or a kite $OAPB$ with two right angles (angle sum $360\degree$ ), then solve for the unknown.

Common trap

Placing $OP$ as a leg instead of the hypotenuse in Pythagoras, or forgetting that both radii meet the tangents at $90\degree$ . Examiner reports flag both.

1Use "angle at the centre is twice angle at the circumference"
ExtendedDirect calculation• centre theorem
Question
$O$ is the centre. Angle at the centre subtended by arc $AB$ is $130\degree$ . Find the angle at the circumference subtended by the same arc.
Step-by-step solution
1. Step 1
  Angle at centre = $2\times$ angle at circumference (same arc, same side).
  $\theta = \dfrac{130}{2} = 65\degree$
Answer
$65\degree$
2Cyclic quadrilateral
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q9• cyclic quadrilateral
Question
$ABCD$ is a cyclic quadrilateral. $\angle A = 78\degree$ and $\angle B = 102\degree$ . Find $\angle C$ and $\angle D$ .
Step-by-step solution
1. Step 1
  Opposite angles in a cyclic quadrilateral sum to $180\degree$ .
  $\angle C = 180 - 78 = 102\degree$
2. Step 2
  Same rule for $\angle D$ .
  $\angle D = 180 - 102 = 78\degree$
Answer
$\angle C = 102\degree,\ \angle D = 78\degree$
3Tangent perpendicular to radius
ExtendedDirect calculation• tangent
Question
A tangent touches a circle at $T$ . $OT$ is the radius. The angle $\angle OTQ$ (between tangent and radius) is given. State its value.
Step-by-step solution
1. Step 1
  The tangent at any point is perpendicular to the radius drawn to that point.
Answer
$90\degree$
4Use the alternate segment theorem
ExtendedDirect calculation• alternate segment
Question
Line $TA$ is tangent at $A$ . Chord $AB$ creates an angle of $42\degree$ with the tangent. Find the angle in the alternate segment subtended by $AB$ .
Step-by-step solution
1. Step 1
  Alternate segment: angle between tangent and chord = angle in the alternate segment.
Answer
$42\degree$
Examiner tip
Always cite "alternate segment theorem" by name — mark schemes credit it explicitly.
5Angles in the same segment
ExtendedDirect calculation• same segment
Question
Two angles are subtended at the circumference by the same chord, on the same side. One is $48\degree$ . Find the other.
Step-by-step solution
1. Step 1
  Angles in the same segment are equal.
Answer
$48\degree$
6Angle in a semicircle
CoreDirect calculation• semicircle, diameter
Question
$AB$ is the diameter of a circle. $C$ is a point on the circumference (not on $AB$ ). State the value of $\angle ACB$ and give a reason.
Step-by-step solution
1. Step 1
  The angle subtended by a diameter at the circumference is a right angle (the angle at the centre is $180\degree$ because $AB$ is a straight line, and the angle at the circumference is half of this).
Answer
$\angle ACB = 90\degree$ (angle in a semicircle)
Examiner tip
The examiner report flags candidates who write $90\degree$ without stating the reason. Always cite "angle in a semicircle" — the reason carries an independent mark.
7Combine angle at centre and same segment
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q12• centre theorem, same segment
Question
Points $A$ , $B$ , $C$ , $D$ lie on a circle with centre $O$ . The angle $\angle AOC = 140\degree$ (reflex angle ignored — take the smaller angle). $D$ lies on the major arc $AC$ . Find $\angle ABC$ and $\angle ADC$ .
Step-by-step solution
1. Step 1
  Both $\angle ABC$ and $\angle ADC$ are angles at the circumference subtended by the same arc $AC$ (the minor arc).
2. Step 2
  Angle at centre = $2\times$ angle at circumference.
  $\angle ABC = \angle ADC = \dfrac{140}{2} = 70\degree$
Answer
$\angle ABC = 70\degree,\ \angle ADC = 70\degree$ (angles in the same segment are equal).
Examiner tip
The examiner report flags candidates who use the angle at the centre but apply the doubling in the wrong direction. Centre is BIGGER, so divide by $2$ when going centre → circumference.
8Cyclic quadrilateral with parallel chord
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q9• cyclic quadrilateral
Question
$PQRS$ is a cyclic quadrilateral. $\angle PQR = 88\degree$ and $\angle QRS = 105\degree$ . Find $\angle QPS$ and $\angle PSR$ .
Step-by-step solution
1. Step 1
  Opposite angles of a cyclic quadrilateral sum to $180\degree$ .
2. Step 2
  $\angle QPS$ is opposite $\angle QRS$ .
  $\angle QPS = 180 - 105 = 75\degree$
3. Step 3
  $\angle PSR$ is opposite $\angle PQR$ .
  $\angle PSR = 180 - 88 = 92\degree$
Answer
$\angle QPS = 75\degree,\ \angle PSR = 92\degree$
9Use tangent perpendicular to radius (Pythagoras)
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q13• tangent, radius, Pythagoras
Question
From an external point $P$ , a tangent of length $24$ cm touches a circle of radius $7$ cm at $T$ . Find the distance $OP$ from the centre to $P$ .
Step-by-step solution
1. Step 1
  The tangent $PT$ is perpendicular to the radius $OT$ at the point of contact, so triangle $OTP$ is right-angled at $T$ .
2. Step 2
  Apply Pythagoras with hypotenuse $OP$ .
  $OP^{2} = OT^{2} + PT^{2} = 7^{2} + 24^{2} = 49 + 576 = 625$
3. Step 3
  Square root both sides.
  $OP = \sqrt{625} = 25\,\text{cm}$
Answer
$OP = 25\,\text{cm}$
Examiner tip
The examiner report flags candidates who set up Pythagoras with $OP$ as a leg instead of the hypotenuse. The radius and tangent are the two perpendicular legs; the line to the external point is the hypotenuse.
10Two tangents from an external point
ChallengeMulti-step problem• tangent, external point
Question
From an external point $P$ , two tangents $PA$ and $PB$ touch a circle with centre $O$ at $A$ and $B$ . Given $\angle APB = 54\degree$ , find $\angle AOB$ .
Step-by-step solution
1. Step 1
  Tangents from an external point are equal in length, and the line $OP$ bisects $\angle APB$ . Both radii $OA$ and $OB$ are perpendicular to the tangents at the points of contact.
2. Step 2
  The quadrilateral $OAPB$ has angles summing to $360\degree$ , with right angles at $A$ and $B$ .
  $\angle AOB + \angle APB + 90 + 90 = 360$
3. Step 3
  Solve for $\angle AOB$ .
  $\angle AOB = 360 - 90 - 90 - 54 = 126\degree$
Answer
$\angle AOB = 126\degree$
Examiner tip
The examiner report flags candidates who forget that BOTH radii are perpendicular to the tangents. The kite $OAPB$ has two right angles, which closes the problem in one step.

Key Formulae — Circle Theorems

The formulae you need to memorise for circle theorems on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Angle at centre = 2 × angle at circumference
$\theta_{\text{centre}} = 2 \times \theta_{\text{circumference}}$
When to use
When two angles subtend the same arc, one at the centre and one on the circumference (same side).
Angles in the same segment are equal
$\theta_{1} = \theta_{2}$
When to use
Two angles at the circumference subtending the same chord, on the same side.
Cyclic quadrilateral
$\angle A + \angle C = 180\degree,\quad \angle B + \angle D = 180\degree$
When to use
Opposite angles in a cyclic quadrilateral sum to $180\degree$ .
Tangent ⊥ radius
$\angle (OT, \text{tangent}) = 90\degree$
When to use
Tangent meets the radius drawn to the point of contact at $90\degree$ .
Alternate segment
$\angle (\text{tangent, chord}) = \angle (\text{in alternate segment})$
When to use
Angle between tangent and chord at the point of contact equals the inscribed angle in the alternate segment.

Key Definitions and Keywords — Circle Theorems

Definitions to memorise and the exact keywords mark schemes credit for circle theorems answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Tangent
Examiner keyword
A line that touches a circle at exactly one point.
Chord
Examiner keyword
A straight line joining two points on a circle.
Cyclic quadrilateral
Examiner keyword
A quadrilateral whose four vertices lie on a single circle.
Segment of a circle
The region bounded by a chord and an arc.
Inscribed (circumference) angle
An angle whose vertex is on the circumference and whose two sides are chords.

Common Mistakes and Misconceptions — Circle Theorems

The traps other students keep falling into on circle theorems questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Stating the angle without naming the theorem
0580/42 — every series
Why it happens
Students see the answer and write it.
How to avoid it
Mark schemes credit the THEOREM name — "angle at centre", "same segment", "alternate segment" etc. Always include it.
✕Doubling instead of halving (or vice versa) for centre/circumference
Why it happens
Students forget which is twice the other.
How to avoid it
Mnemonic: angle at the CENTRE is BIGGER. So centre = 2 × circumference.
✕Pairing adjacent angles in a cyclic quadrilateral
Why it happens
Confusing "opposite" with "adjacent".
How to avoid it
Opposite angles in a cyclic quad sum to $180\degree$ — they're across from each other, not next to each other.
✕Picking an angle from the wrong segment
Why it happens
Confusing which segment is "alternate".
How to avoid it
Alternate segment is on the OTHER side of the chord from the tangent angle.

Circle Theorems — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Circle Theorems

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Short Notes - Circle Theorems

Detailed Study Notes

Detailed notes on Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Circle Theorems — Cambridge IGCSE 0580 Maths Extended (2026)

The seven named theorems that turn diagrams into angle equations. Cambridge mark schemes credit the theorem name as much as the answer — memorise them precisely.

At a glance

Angle at centre = 2 × angle at circumference subtending the same arc.
Angles in the same segment are equal.
Angle in a semicircle = $90\degree$ .
Cyclic quadrilateral: opposite angles sum to $180\degree$ .
Tangent ⊥ radius at the point of contact.
Two tangents from a point are equal in length.
Alternate segment theorem: angle between tangent and chord = angle in alternate segment.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E4.7 — Use the following circle theorems: angle at centre, angles in same segment, angle in semicircle, cyclic quadrilateral, tangent-radius, equal tangents, alternate segment.

Angle at centre = 2 × angle at circumference

Same arc, same side. The centre angle is double.

If two angles in a circle subtend the same arc — one at the centre and one at the circumference (on the same side) — the angle at the centre is twice the angle at the circumference.

Worked. Centre angle = $130\degree$ . Find the circumference angle subtending the same arc.

$\theta = 130 / 2 = 65\degree$ .
Reason: angle at centre is twice angle at circumference.

Special case. If the chord is a DIAMETER, the centre angle is $180\degree$ , so the circumference angle is $90\degree$ — that's the angle in a semicircle = $90\degree$ theorem.

Chord AB subtends 130° at centre O and half of that, 65°, at circumference point P.

Same arc, same side.
Centre angle = $2 \times$ circumference angle.
Diameter case: angle in semicircle = $90\degree$ .
ALWAYS state the theorem name in the reason.

Angles in the same segment

All angles subtended by the same chord, on the same side, are equal.

Theorem. If multiple angles at the circumference are subtended by the SAME chord, on the SAME side of that chord, they are all EQUAL.

Worked. Two angles at the circumference subtended by the same chord, on the same side. One is $48\degree$ . The other:

$48\degree$ .
Reason: angles in the same segment are equal.

Tip. Look for two angles whose vertices are on the same arc and whose sides reach the same two endpoints (the chord).

P and Q lie on the same arc above chord AB, so both subtend equal angles.

Same chord + same side → equal angles.
Cite the theorem by name.

Cyclic quadrilateral

Opposite angles in a cyclic quadrilateral sum to $180\degree$ .

A cyclic quadrilateral has all four vertices on the circle.

Theorem. Opposite angles of a cyclic quadrilateral sum to $180\degree$ .

Worked. $ABCD$ is cyclic. $\angle A = 78\degree$ . Find $\angle C$ .

$\angle A + \angle C = 180\degree \Rightarrow \angle C = 102\degree$ .
Reason: opposite angles in a cyclic quadrilateral.

Don't confuse with adjacent. Opposite means ACROSS, not next to. Adjacent angles in a cyclic quad don't generally satisfy the rule.

Opposite vertices A and C lie across the quadrilateral; their angles add to 180°.

All four vertices on the circle.
Opposite angles sum to $180\degree$ .
Two pairs of opposite angles each sum to $180\degree$ .

Tangent properties

Tangent ⊥ radius at the point of contact. Two tangents from an external point are equal.

Tangent perpendicular to radius. A tangent meets the radius drawn to the point of contact at $90\degree$ .

Worked. Tangent $TA$ touches circle at $A$ ; centre $O$ . The angle $\angle OAT = 90\degree$ .

Reason: tangent perpendicular to radius at point of contact.

Two tangents from an external point. If two tangents are drawn to a circle from an external point $P$ , the two tangent lengths are EQUAL.

Worked. $PA$ and $PB$ are tangents to a circle from $P$ , touching at $A$ and $B$ . Then $PA = PB$ .

Reason: tangents from an external point are equal.

Each radius meets its tangent at 90°, and the two tangents from P are equal in length.

Alternate segment theorem. The angle between a tangent and a chord at the point of contact equals the angle in the alternate (opposite) segment subtended by that chord.

Worked. Tangent $TA$ at $A$ . Chord $AB$ makes angle $42\degree$ with the tangent. Find the angle in the alternate segment subtended by $AB$ .

$42\degree$ .
Reason: alternate segment theorem.

The tangent-chord angle equals the angle subtended by the same chord in the alternate segment.

Tangent ⊥ radius at the contact point.
Two tangents from external point: equal lengths.
Alternate segment: angle between tangent and chord = angle in opposite segment.
Cite each theorem by NAME for full marks.

Quick recap

Angle at centre = 2 × angle at circumference (same arc).
Angles in the same segment: equal.
Angle in a semicircle: $90\degree$ .
Cyclic quadrilateral: opposite angles sum to $180\degree$ .
Tangent ⊥ radius at contact.
Two tangents from a point: equal.
Alternate segment: tangent-chord angle = angle in alternate segment.
ALWAYS state the theorem name as your reason.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Angle at centre theorem — angle at centre = $2 \times$ angle at circumference subtending the same arc.
Angles in the same segment — angles subtended by the same arc/chord, on the same side, are equal.
Angle in a semicircle — $90\degree$ .
Cyclic quadrilateral — opposite angles sum to $180\degree$ .
Tangent perpendicular to radius — at the point of contact.
Two tangents from an external point — equal in length.
Alternate segment theorem — angle between tangent and chord = angle in alternate segment.

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E4.7); 0580/42 Oct/Nov 2024 — Q9 (cyclic quadrilateral); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Circle Theorems

Step-by-step solutions to past-paper-style questions on circle theorems, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Circle Theorems

Almost every circle theorems exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A circle diagram with one angle to find — using a single named theorem such as angle at the centre, same segment, cyclic quadrilateral or alternate segment.

How to approach it

Common trap

Writing the angle with no theorem named, doubling instead of halving for centre/circumference, or pairing adjacent rather than opposite cyclic angles. Examiner reports flag all three every series.

Multi-step problem

Recognise it by

A tangent diagram combining several facts — tangent perpendicular to radius plus Pythagoras, or two tangents from an external point forming a kite.

How to approach it

Use the tangent-radius right angle to set up a right triangle (for Pythagoras) or a kite $OAPB$ with two right angles (angle sum $360\degree$ ), then solve for the unknown.

Common trap

Placing $OP$ as a leg instead of the hypotenuse in Pythagoras, or forgetting that both radii meet the tangents at $90\degree$ . Examiner reports flag both.

1Use "angle at the centre is twice angle at the circumference"
ExtendedDirect calculation• centre theorem
Question
$O$ is the centre. Angle at the centre subtended by arc $AB$ is $130\degree$ . Find the angle at the circumference subtended by the same arc.
Step-by-step solution
1. Step 1
  Angle at centre = $2\times$ angle at circumference (same arc, same side).
  $\theta = \dfrac{130}{2} = 65\degree$
Answer
$65\degree$
2Cyclic quadrilateral
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q9• cyclic quadrilateral
Question
$ABCD$ is a cyclic quadrilateral. $\angle A = 78\degree$ and $\angle B = 102\degree$ . Find $\angle C$ and $\angle D$ .
Step-by-step solution
1. Step 1
  Opposite angles in a cyclic quadrilateral sum to $180\degree$ .
  $\angle C = 180 - 78 = 102\degree$
2. Step 2
  Same rule for $\angle D$ .
  $\angle D = 180 - 102 = 78\degree$
Answer
$\angle C = 102\degree,\ \angle D = 78\degree$
3Tangent perpendicular to radius
ExtendedDirect calculation• tangent
Question
A tangent touches a circle at $T$ . $OT$ is the radius. The angle $\angle OTQ$ (between tangent and radius) is given. State its value.
Step-by-step solution
1. Step 1
  The tangent at any point is perpendicular to the radius drawn to that point.
Answer
$90\degree$
4Use the alternate segment theorem
ExtendedDirect calculation• alternate segment
Question
Line $TA$ is tangent at $A$ . Chord $AB$ creates an angle of $42\degree$ with the tangent. Find the angle in the alternate segment subtended by $AB$ .
Step-by-step solution
1. Step 1
  Alternate segment: angle between tangent and chord = angle in the alternate segment.
Answer
$42\degree$
Examiner tip
Always cite "alternate segment theorem" by name — mark schemes credit it explicitly.
5Angles in the same segment
ExtendedDirect calculation• same segment
Question
Two angles are subtended at the circumference by the same chord, on the same side. One is $48\degree$ . Find the other.
Step-by-step solution
1. Step 1
  Angles in the same segment are equal.
Answer
$48\degree$
6Angle in a semicircle
CoreDirect calculation• semicircle, diameter
Question
$AB$ is the diameter of a circle. $C$ is a point on the circumference (not on $AB$ ). State the value of $\angle ACB$ and give a reason.
Step-by-step solution
1. Step 1
  The angle subtended by a diameter at the circumference is a right angle (the angle at the centre is $180\degree$ because $AB$ is a straight line, and the angle at the circumference is half of this).
Answer
$\angle ACB = 90\degree$ (angle in a semicircle)
Examiner tip
The examiner report flags candidates who write $90\degree$ without stating the reason. Always cite "angle in a semicircle" — the reason carries an independent mark.
7Combine angle at centre and same segment
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q12• centre theorem, same segment
Question
Points $A$ , $B$ , $C$ , $D$ lie on a circle with centre $O$ . The angle $\angle AOC = 140\degree$ (reflex angle ignored — take the smaller angle). $D$ lies on the major arc $AC$ . Find $\angle ABC$ and $\angle ADC$ .
Step-by-step solution
1. Step 1
  Both $\angle ABC$ and $\angle ADC$ are angles at the circumference subtended by the same arc $AC$ (the minor arc).
2. Step 2
  Angle at centre = $2\times$ angle at circumference.
  $\angle ABC = \angle ADC = \dfrac{140}{2} = 70\degree$
Answer
$\angle ABC = 70\degree,\ \angle ADC = 70\degree$ (angles in the same segment are equal).
Examiner tip
The examiner report flags candidates who use the angle at the centre but apply the doubling in the wrong direction. Centre is BIGGER, so divide by $2$ when going centre → circumference.
8Cyclic quadrilateral with parallel chord
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q9• cyclic quadrilateral
Question
$PQRS$ is a cyclic quadrilateral. $\angle PQR = 88\degree$ and $\angle QRS = 105\degree$ . Find $\angle QPS$ and $\angle PSR$ .
Step-by-step solution
1. Step 1
  Opposite angles of a cyclic quadrilateral sum to $180\degree$ .
2. Step 2
  $\angle QPS$ is opposite $\angle QRS$ .
  $\angle QPS = 180 - 105 = 75\degree$
3. Step 3
  $\angle PSR$ is opposite $\angle PQR$ .
  $\angle PSR = 180 - 88 = 92\degree$
Answer
$\angle QPS = 75\degree,\ \angle PSR = 92\degree$
9Use tangent perpendicular to radius (Pythagoras)
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q13• tangent, radius, Pythagoras
Question
From an external point $P$ , a tangent of length $24$ cm touches a circle of radius $7$ cm at $T$ . Find the distance $OP$ from the centre to $P$ .
Step-by-step solution
1. Step 1
  The tangent $PT$ is perpendicular to the radius $OT$ at the point of contact, so triangle $OTP$ is right-angled at $T$ .
2. Step 2
  Apply Pythagoras with hypotenuse $OP$ .
  $OP^{2} = OT^{2} + PT^{2} = 7^{2} + 24^{2} = 49 + 576 = 625$
3. Step 3
  Square root both sides.
  $OP = \sqrt{625} = 25\,\text{cm}$
Answer
$OP = 25\,\text{cm}$
Examiner tip
The examiner report flags candidates who set up Pythagoras with $OP$ as a leg instead of the hypotenuse. The radius and tangent are the two perpendicular legs; the line to the external point is the hypotenuse.
10Two tangents from an external point
ChallengeMulti-step problem• tangent, external point
Question
From an external point $P$ , two tangents $PA$ and $PB$ touch a circle with centre $O$ at $A$ and $B$ . Given $\angle APB = 54\degree$ , find $\angle AOB$ .
Step-by-step solution
1. Step 1
  Tangents from an external point are equal in length, and the line $OP$ bisects $\angle APB$ . Both radii $OA$ and $OB$ are perpendicular to the tangents at the points of contact.
2. Step 2
  The quadrilateral $OAPB$ has angles summing to $360\degree$ , with right angles at $A$ and $B$ .
  $\angle AOB + \angle APB + 90 + 90 = 360$
3. Step 3
  Solve for $\angle AOB$ .
  $\angle AOB = 360 - 90 - 90 - 54 = 126\degree$
Answer
$\angle AOB = 126\degree$
Examiner tip
The examiner report flags candidates who forget that BOTH radii are perpendicular to the tangents. The kite $OAPB$ has two right angles, which closes the problem in one step.

Key Formulae — Circle Theorems

The formulae you need to memorise for circle theorems on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Angle at centre = 2 × angle at circumference
$\theta_{\text{centre}} = 2 \times \theta_{\text{circumference}}$
When to use
When two angles subtend the same arc, one at the centre and one on the circumference (same side).
Angles in the same segment are equal
$\theta_{1} = \theta_{2}$
When to use
Two angles at the circumference subtending the same chord, on the same side.
Cyclic quadrilateral
$\angle A + \angle C = 180\degree,\quad \angle B + \angle D = 180\degree$
When to use
Opposite angles in a cyclic quadrilateral sum to $180\degree$ .
Tangent ⊥ radius
$\angle (OT, \text{tangent}) = 90\degree$
When to use
Tangent meets the radius drawn to the point of contact at $90\degree$ .
Alternate segment
$\angle (\text{tangent, chord}) = \angle (\text{in alternate segment})$
When to use
Angle between tangent and chord at the point of contact equals the inscribed angle in the alternate segment.

Key Definitions and Keywords — Circle Theorems

Definitions to memorise and the exact keywords mark schemes credit for circle theorems answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Tangent
Examiner keyword
A line that touches a circle at exactly one point.
Chord
Examiner keyword
A straight line joining two points on a circle.
Cyclic quadrilateral
Examiner keyword
A quadrilateral whose four vertices lie on a single circle.
Segment of a circle
The region bounded by a chord and an arc.
Inscribed (circumference) angle
An angle whose vertex is on the circumference and whose two sides are chords.

Common Mistakes and Misconceptions — Circle Theorems

The traps other students keep falling into on circle theorems questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Stating the angle without naming the theorem
0580/42 — every series
Why it happens
Students see the answer and write it.
How to avoid it
Mark schemes credit the THEOREM name — "angle at centre", "same segment", "alternate segment" etc. Always include it.
✕Doubling instead of halving (or vice versa) for centre/circumference
Why it happens
Students forget which is twice the other.
How to avoid it
Mnemonic: angle at the CENTRE is BIGGER. So centre = 2 × circumference.
✕Pairing adjacent angles in a cyclic quadrilateral
Why it happens
Confusing "opposite" with "adjacent".
How to avoid it
Opposite angles in a cyclic quad sum to $180\degree$ — they're across from each other, not next to each other.
✕Picking an angle from the wrong segment
Why it happens
Confusing which segment is "alternate".
How to avoid it
Alternate segment is on the OTHER side of the chord from the tangent angle.

Circle Theorems — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Circle Theorems

Short Notes - Circle Theorems

Detailed Study Notes

At a glance

What you’ll learn

Angle at centre = 2 × angle at circumference

Angles in the same segment

Cyclic quadrilateral

Tangent properties

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Circle Theorems

1Use "angle at the centre is twice angle at the circumference"

2Cyclic quadrilateral

3Tangent perpendicular to radius

4Use the alternate segment theorem

5Angles in the same segment

6Angle in a semicircle

7Combine angle at centre and same segment

8Cyclic quadrilateral with parallel chord

9Use tangent perpendicular to radius (Pythagoras)

10Two tangents from an external point

Angle at centre = 2 × angle at circumference

Angles in the same segment are equal

Cyclic quadrilateral

Tangent ⊥ radius

Alternate segment

Tangent

Chord

Cyclic quadrilateral

Segment of a circle

Inscribed (circumference) angle

✕Stating the angle without naming the theorem

✕Doubling instead of halving (or vice versa) for centre/circumference

✕Pairing adjacent angles in a cyclic quadrilateral

✕Picking an angle from the wrong segment

Circle Theorems — frequently asked questions

Circle Theorems

Short Notes - Circle Theorems

Detailed Study Notes

At a glance

What you’ll learn

Angle at centre = 2 × angle at circumference

Angles in the same segment

Cyclic quadrilateral

Tangent properties

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Circle Theorems

1Use "angle at the centre is twice angle at the circumference"

2Cyclic quadrilateral

3Tangent perpendicular to radius

4Use the alternate segment theorem

5Angles in the same segment

6Angle in a semicircle

7Combine angle at centre and same segment

8Cyclic quadrilateral with parallel chord

9Use tangent perpendicular to radius (Pythagoras)

10Two tangents from an external point

Angle at centre = 2 × angle at circumference

Angles in the same segment are equal

Cyclic quadrilateral

Tangent ⊥ radius

Alternate segment

Tangent

Chord

Cyclic quadrilateral

Segment of a circle

Inscribed (circumference) angle

✕Stating the angle without naming the theorem

✕Doubling instead of halving (or vice versa) for centre/circumference

✕Pairing adjacent angles in a cyclic quadrilateral

✕Picking an angle from the wrong segment

Circle Theorems — frequently asked questions