Circle Theorems · Study Notes · Edexcel IGCSE 4MA1 Higher

Seven circle theorems for Higher Tier. Memorise them by name; state the name when using.

At a glance

Angle at centre = 2 × angle at circumference (same arc).
Angle in a semicircle = $90°$ .
Angles in the same segment are equal.
Opposite angles in cyclic quadrilateral sum to $180°$ .
Tangent meets radius at $90°$ .
Tangents from external point are equal in length.
Alternate segment theorem: tangent-chord angle = angle in alternate segment.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.12 — Identify and use the seven circle theorems.
4.12 — State theorem names when applying.
4.12 — Solve problems involving tangents, chords, inscribed angles.

The seven circle theorems

▼

Memorise all seven by name. State the name when using.

1. Angle at centre is twice the angle at circumference (same arc).

If a chord subtends an angle of $\alpha$ at the centre, it subtends $\alpha/2$ at any point on the major arc.

2. Angle in a semicircle is $90°$ .

(Special case of theorem 1: if the central angle is $180°$ , the inscribed angle is $90°$ .)

Theorem 1: the central angle is double the angle at the circumference. Theorem 2 is the case where that arc is a semicircle.

3. Angles in the same segment are equal.

Two chords from the same arc subtend the same angle wherever measured on the major arc.

4. Opposite angles in a cyclic quadrilateral sum to $180°$ .

A 4-vertex inscribed shape has opposite angles supplementary.

5. Tangent is perpendicular to radius at the point of contact.

If a line touches the circle at point $P$ , and $O$ is the centre, then $OP \perp$ tangent.

6. Two tangents from an external point are equal in length.

If two tangents are drawn from outside the circle, the segments from external point to tangent points are equal.

A tangent meets the radius at 90°, and the two tangents drawn from one external point are equal in length.

7. Alternate segment theorem.

The angle between a tangent and a chord = the angle in the alternate segment (the segment on the other side of the chord).

Worked qualitative. Why is the angle in a semicircle $90°$ ?

Diameter creates a central angle of $180°$ .
Inscribed angle: $180/2 = 90°$ .
This is theorem 2 — special case of 1.

Edexcel tip. ALWAYS state the theorem name. 'Angles in the same segment' or 'angle at centre is twice angle at circumference' — exact phrasing.

Seven theorems memorised.
State name when using.
Apply to find unknown angles.
Often combined in multi-step problems.

Working with circle theorems

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Identify the configuration. Apply the right theorem.

Method:

Identify what features are present (tangent? chord? cyclic quadrilateral?).
Match to a theorem.
Apply.

Common configurations:

Feature	Theorem
Tangent at $P$	Tangent ⊥ radius (#5)
Two tangents from $P$	Equal lengths (#6)
Inscribed quadrilateral	Opposite angles supplementary (#4)
Diameter as chord	$90°$ angle elsewhere (#2)
Chord with two angles on either side	Equal in same segment (#3)
Tangent and chord meeting	Alternate segment (#7)

Worked example. Tangent at $P$ , chord $PQ$ , angle between them $40°$ . Find angle subtended by $PQ$ at the far side.

Alternate segment theorem.
Far side angle = $40°$ .

Multi-step problems. Often you'll combine 2-3 theorems in one problem.

Worked qualitative. Why are circle theorems useful?

Without them, you'd need to set up triangles and use long Pythagoras / trig calculations.
With them, many angles can be found directly.
They embody beautiful geometric symmetries.

Edexcel tip. Mark schemes for circle problems award:

M1 for identifying the configuration.
M1 for naming the theorem.
A1 for the angle.

Always state the theorem!

Match configuration to theorem.
Apply directly.
Multi-step: combine theorems.
Always name the theorem.

Quick recap

Seven theorems: state by name.
Centre = 2 × circumference (same arc).
Semicircle: $90°$ .
Same segment: equal.
Cyclic quadrilateral: opposite sum to $180°$ .
Tangent ⊥ radius.
Two tangents from external point: equal.
Alternate segment.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Angle at centre = 2 × angle at circumference (same arc).
Angle in semicircle = $90°$ .
Same segment: angles equal.
Cyclic quad: opposite angles sum to $180°$ .
Tangent ⊥ radius.
Tangents from external: equal.
Alternate segment: tangent-chord = inscribed.

How it’s examined

Circle theorems on every Higher Tier paper (5-8 marks). Often multi-step, combining 2-3 theorems. Examiner reports flag (1) not naming the theorem, (2) wrong theorem applied, (3) missing the configuration.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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

1Angle at centre = 2 × angle at circumference
Higher• circle theorem
▼
Question
Inscribed angle subtends arc with central angle $80°$ . Find the inscribed angle.
Step-by-step solution
1. Step 1
  Theorem: angle at centre = 2 × angle at circumference (same arc).
2. Step 2
  Inscribed angle = $80 / 2 = 40°$ .
Answer
$40°$
2Angle in a semicircle
Foundation• Adapted from 4MA1/1H May/Jun 2024 Q12• semicircle
▼
Question
Triangle inscribed in a circle with one side as the diameter. The angle opposite the diameter is?
Step-by-step solution
1. Step 1
  Theorem: angle in a semicircle = $90°$ .
2. Step 2
  The angle subtended by a diameter is always $90°$ .
Answer
$90°$
Examiner tip
When you see a triangle inscribed in a circle with one side a diameter — it's right-angled at the third vertex.
3Cyclic quadrilateral
Higher• cyclic quadrilateral
▼
Question
Cyclic quadrilateral has opposite angles $80°$ and $x$ . Find $x$ .
Step-by-step solution
1. Step 1
  Theorem: opposite angles of a cyclic quadrilateral sum to $180°$ .
2. Step 2
  $x = 180 - 80 = 100°$ .
Answer
$x = 100°$
4Tangent perpendicular to radius
Higher• tangent
▼
Question
Tangent at point $P$ . The angle between the tangent and the radius $OP$ is?
Step-by-step solution
1. Step 1
  Theorem: tangent meets radius at $90°$ .
Answer
$90°$
5Alternate segment theorem
Higher• Adapted from 4MA1/2H May/Jun 2024 Q23• alternate segment
▼
Question
A tangent from $P$ meets a circle at $T$ . The angle between the tangent and chord $TQ$ is $50°$ . What's the angle in the alternate segment?
Step-by-step solution
1. Step 1
  Theorem: angle between tangent and chord = angle in the alternate segment (subtended by the same chord).
2. Step 2
  Both equal $50°$ .
Answer
$50°$
Examiner tip
Memorise the seven theorems and STATE THE NAME when applying.
6Two tangents from the same point [1 mark]
Getting started• Paper 1H/2H style — 1 mark• circle theorems, tangents
▼
Question
$P$ is a point outside a circle. $PA$ and $PB$ are tangents to the circle, touching it at $A$ and $B$ .

$PA = 9.4$ cm.

Write down the length of $PB$ . [1]
Step-by-step solution
1. Step 1
  Tangents drawn to a circle from the same external point are equal in length.
2. Step 2
  B1 So $PB = PA$ .
  $PB = 9.4 \text{ cm}$
Answer
$9.4$ cm
Examiner tip
Examiner tip: This fact often sits inside a longer question: it makes triangle $PAB$ isosceles and the quadrilateral $OAPB$ a kite.

Common mistake: Treating $PA$ and $PB$ as unrelated and trying Pythagoras with a radius that is not given.
7Reflex angle at the centre from an obtuse angle at the circumference [2 marks]
Getting started• Paper 1H/2H style — 2 marks• circle theorems, angle at the centre
▼
Question
$A$ , $B$ and $C$ are points on a circle, centre $O$ . $B$ is on the minor arc $AC$ .

Angle $ABC = 124°$ .

Work out the size of the reflex angle $AOC$ . [2]
Step-by-step solution
1. Step 1
  M1 Angle $ABC$ stands on the major arc $AC$ , and the angle that this arc makes at the centre is the reflex angle $AOC$ . The angle at the centre is twice the angle at the circumference.
  $\text{reflex } \angle AOC = 2 \times 124^\circ$
2. Step 2
  A1 Evaluate.
  $\text{reflex } \angle AOC = 248^\circ$
Answer
$248°$
Examiner tip
Examiner tip: The theorem still works when the angle at the circumference is obtuse; the angle at the centre is then reflex. Check: the other angle $AOC$ is $360° - 248° = 112°$ , which is twice the $56°$ you would find at any point on the major arc.

Common mistake: Deciding that an angle at the centre cannot be more than $180°$ , and giving $124° \div 2 = 62°$ or $180° - 124° = 56°$ .
8Length of a chord from its distance to the centre [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• chords, pythagoras
▼
Question
A circle has centre $O$ and radius $9$ cm.

The chord $PQ$ is $5$ cm from $O$ .

Calculate the length of $PQ$ . Give your answer correct to $3$ significant figures. [3]
Step-by-step solution
1. Step 1
  M1 The perpendicular from $O$ to $PQ$ meets it at the midpoint $M$ , and $OMP$ is right-angled at $M$ with hypotenuse $OP = 9$ cm.
  $PM^{2} = 9^{2} - 5^{2} = 81 - 25 = 56$
2. Step 2
  M1 $PM$ is half of the chord, so double it. Keep the exact value until the end.
  $PQ = 2\sqrt{56} = 14.966\ldots$
3. Step 3
  A1 Round to $3$ significant figures.
  $PQ = 15.0 \text{ cm}$
Answer
$15.0$ cm
Examiner tip
Examiner tip: The perpendicular from the centre bisects the chord, so the triangle gives only half of $PQ$ . Write " $\times 2$ " as a separate line so you do not forget it.

Common mistake: Stopping at $PM = 7.48$ cm, which is only half the chord. The other loss is rounding early: truncating $PM$ to $7.4$ and doubling gives $14.8$ cm, which is not accurate to $3$ significant figures.
9Angles where two chords cross [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• circle theorems, same segment, angles
▼
Question
$A$ , $B$ , $C$ and $D$ are points, in that order, on a circle.

The chords $AC$ and $BD$ intersect at $X$ .

Angle $BAC = 35°$ and angle $AXB = 100°$ .

Work out the size of angle $ACD$ . Give reasons for your working. [4]
Step-by-step solution
1. Step 1
  M1 In triangle $ABX$ , angle $BAX = 35°$ and angle $AXB = 100°$ . Use the angle sum.
  $\angle ABX = 180^\circ - 35^\circ - 100^\circ$
2. Step 2
  A1 Angle $ABX$ is the same angle as angle $ABD$ .
  $\angle ABD = 45^\circ$
3. Step 3
  B1 Angles $ABD$ and $ACD$ both stand on the chord $AD$ from the same side, so they are equal.
  $\angle ACD = 45^\circ$
4. Step 4
  B1 Reasons: angles in a triangle add up to $180°$ ; angles in the same segment are equal.
Answer
$45°$ , with reasons
Examiner tip
Examiner tip: Sketch the four points in order round the circle before you start. The two chords then cross inside, and you can see which angles stand on the same chord.

Common mistake: Pairing angle $ACD$ with angle $BAC$ because they look alike on the sketch. Angles in the same segment must stand on the same chord; $BAC$ stands on $BC$ , not on $AD$ .
10Radius and angle from a tangent and a line through the centre [5 marks]
Stretch• Paper 1H/2H style — 5 marks• tangents, intersecting chords, trigonometry
▼
Question
$P$ is a point outside a circle with centre $O$ .

$PT$ is a tangent to the circle at $T$ , and $PT = 12$ cm.

The straight line from $P$ through $O$ meets the circle at $A$ and then at $B$ , with $PA = 8$ cm.

(a) Work out the radius of the circle. [3]

(b) Work out the size of angle $TPO$ . Give your answer correct to $1$ decimal place. [2]
Step-by-step solution
1. Step 1
  M1 (a) For a tangent and a secant from the same point, $PT^{2} = PA \times PB$ .
  $12^{2} = 8 \times PB$
2. Step 2
  A1 Solve for $PB$ .
  $PB = 144 \div 8 = 18 \text{ cm}$
3. Step 3
  A1 $AB$ passes through $O$ , so it is a diameter.
  $AB = 18 - 8 = 10 \text{ cm}, \quad r = 5 \text{ cm}$
4. Step 4
  M1 (b) Angle $OTP = 90°$ because the tangent is perpendicular to the radius. In triangle $OTP$ , $OT = 5$ is opposite angle $TPO$ and $PT = 12$ is adjacent to it.
  $\tan TPO = \dfrac{5}{12}$
5. Step 5
  A1 Use the inverse tangent.
  $\angle TPO = \tan^{-1}\left(\dfrac{5}{12}\right) = 22.619\ldots = 22.6^\circ$
Answer
(a) $5$ cm (b) $22.6°$
Examiner tip
Examiner tip: Check with Pythagoras: $OP = 8 + 5 = 13$ and $5^{2} + 12^{2} = 169 = 13^{2}$ , so the right angle at $T$ is consistent.

Common mistake: Giving $PB - PA = 10$ as the radius. $AB$ is the diameter, so the radius is half of it.

Question 1
Paper 1H/2H style — 1 mark1 mark
$ABCD$ is a cyclic quadrilateral. The side $DC$ is extended to the point $E$ , so that $DCE$ is a straight line.

Angle $BCE = 71°$ .

Write down the size of angle $BAD$ . [1]
Model answer
Angle $BCD = 180° - 71° = 109°$ , and angle $BAD = 180° - 109°$ .

Answer: $71°$
Why this scores
Mark scheme:
- B1 for $71$ cao
Why this scores full marks: The script links the exterior angle to the interior angle at $C$ (angles on a straight line), then to the opposite angle at $A$ (opposite angles of a cyclic quadrilateral add up to $180°$ ). The exterior angle of a cyclic quadrilateral always equals the interior angle at the opposite vertex.
Question 2
Paper 1H/2H style — 2 marks2 marks
$PQ$ is a chord of a circle with centre $O$ .

$M$ is the point on $PQ$ such that $OM$ is perpendicular to $PQ$ .

$PM = 6.2$ cm.

(a) Write down the length of $QM$ . [1]

(b) Give a reason for your answer. [1]
Model answer
(a) Answer: $6.2$ cm

(b) The perpendicular from the centre of a circle to a chord bisects the chord.
Why this scores
Mark scheme:
- B1 for $6.2$
- B1 for the perpendicular from the centre to a chord bisects the chord oe
Why this scores full marks: The reason names all three parts of the property: the line comes from the centre, it is perpendicular to the chord, and it bisects the chord. " $M$ is the middle" restates the answer without giving a reason.
Question 3
Paper 1H/2H style — 2 marks2 marks
$A$ , $B$ and $C$ are points on a circle and $AB$ is a diameter.

Angle $CAB = 2x°$ and angle $CBA = 3x°$ .

Find the value of $x$ . [2]
Model answer
Angle $ACB = 90°$ (angle in a semicircle)

$2x + 3x + 90 = 180$

$5x = 90$

Answer: $x = 18$
Why this scores
Mark scheme:
- M1 for $2x + 3x + 90 = 180$ or $2x + 3x = 90$
- A1 for $18$ cao
Why this scores full marks: The script puts the right angle at $C$ , the point opposite the diameter, and then uses the angle sum of the triangle. Candidates who forget the right angle write $5x = 180$ and get $36$ , which scores nothing.
Question 4
Paper 1H/2H style — 3 marks3 marks
$PA$ and $PB$ are tangents to a circle with centre $O$ , touching the circle at $A$ and $B$ .

$OA = 6$ cm and $PA = 8$ cm.

Calculate the size of angle $APB$ . Give your answer correct to $1$ decimal place. [3]
Model answer
Angle $OAP = 90°$ (tangent perpendicular to radius)

$\tan APO = \dfrac{6}{8}$

$\angle APO = 36.869\ldots^\circ$

$OP$ bisects angle $APB$ (triangles $OAP$ and $OBP$ are congruent), so

$\angle APB = 2 \times 36.869\ldots = 73.739\ldots$

Answer: $73.7°$
Why this scores
Mark scheme:
- M1 for $\tan APO = \dfrac{6}{8}$ oe (or finding $OP = 10$ and a correct trig ratio)
- M1 for $2 \times$ their angle $APO$
- A1 for awrt $73.7$
Why this scores full marks: The script uses the right angle between the tangent and the radius to make triangle $OAP$ right-angled, and carries the unrounded $36.869\ldots$ into the doubling. Rounding to $36.9$ first gives $73.8$ , which loses the accuracy mark.
Question 5
Paper 1H/2H style — 3 marks3 marks
$A$ , $B$ and $C$ are points on a circle.

$SAT$ is the tangent to the circle at $A$ . $S$ and $B$ are on opposite sides of the line $AC$ .

Angle $SAC = 57°$ and angle $BAC = 64°$ .

Work out the size of angle $ACB$ . [3]
Model answer
Angle $ABC = 57°$ (alternate segment theorem)

$\angle ACB = 180^\circ - 57^\circ - 64^\circ$

Answer: $59°$
Why this scores
Mark scheme:
- B1 for angle $ABC = 57$
- M1 for $180 - 57 - 64$ oe
- A1 for $59$ cao
Why this scores full marks: The script matches the tangent angle to the correct chord. Angle $SAC$ is between the tangent and chord $AC$ , so it equals the angle at $B$ , on the far side of $AC$ . A quick check: angle $TAB = 180° - 57° - 64° = 59°$ , and by the same theorem it equals angle $ACB$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
$A$ , $B$ , $C$ and $D$ are points on a circle. $PAB$ and $PCD$ are straight lines, with $P$ outside the circle.

$PA = 3.5$ cm, $AB = 8.5$ cm and $PC = 4.2$ cm.

Calculate the length of $CD$ . [3]
Model answer
$PB = 3.5 + 8.5 = 12$

$PA \times PB = PC \times PD$

$3.5 \times 12 = 4.2 \times PD$

$PD = 42 \div 4.2 = 10$

$CD = 10 - 4.2$

Answer: $5.8$ cm
Why this scores
Mark scheme:
- M1 for $3.5 \times 12 = 4.2 \times PD$ oe
- A1 for $PD = 10$
- A1 for $5.8$ cao
Why this scores full marks: Every length in the product is measured from $P$ . The script finds $PD$ first and then subtracts $PC$ , because the question asks for the part of the line inside the circle.
Question 7
Paper 1H/2H style — 4 marks4 marks
$A$ , $B$ , $C$ and $D$ are points, in that order, on a circle. $AB$ is parallel to $DC$ .

The chords $AC$ and $BD$ intersect at $X$ .

Angle $BAC = 36°$ .

Work out the size of angle $AXB$ . Give reasons for your working. [4]
Model answer
Angle $ACD = 36°$ (alternate angles are equal, $AB$ parallel to $DC$ )

Angle $ABD = 36°$ (angles in the same segment are equal, both on chord $AD$ )

$\angle AXB = 180^\circ - 36^\circ - 36^\circ$

(angles in a triangle add up to $180°$ )

Answer: $108°$
Why this scores
Mark scheme:
- B1 for angle $ACD = 36$ with reason alternate angles, or angle $BDC = 36$ with reason angles in the same segment
- B1 for angle $ABD = 36$ with a correct reason
- M1 for $180 - 36 - 36$
- A1 for $108$ cao
Why this scores full marks: Each angle is paired with its reason as it is found, using the words "alternate angles" and "angles in the same segment". The script then sees that triangle $ABX$ has two equal angles, $36°$ at $A$ and at $B$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
$A$ , $B$ , $C$ and $D$ are points on a circle, centre $O$ . $C$ and $D$ are both on the major arc $AB$ .

Angle $AOB = (5y + 20)°$ and angle $ACB = (2y + 16)°$ .

Work out the size of angle $ADB$ . Give reasons for your working. [4]
Model answer
The angle at the centre is twice the angle at the circumference:

$5y + 20 = 2(2y + 16)$

$5y + 20 = 4y + 32$

$y = 12$

Angle $ACB = 2(12) + 16 = 40°$

Angle $ADB$ = angle $ACB$ (angles in the same segment are equal)

Answer: $40°$
Why this scores
Mark scheme:
- M1 for $5y + 20 = 2(2y + 16)$ oe
- A1 for $y = 12$
- B1 for $40$ (ft their $y$ )
- B1 for both reasons: angle at centre twice angle at circumference; angles in the same segment
Why this scores full marks: The script doubles the angle at the circumference, not the angle at the centre, and brackets the whole expression: $2(2y + 16)$ . It then substitutes back, because the question asks for an angle, not for $y$ . Check: angle $AOB = 5(12) + 20 = 80°$ , which is twice $40°$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
$A$ , $B$ , $C$ and $D$ are points on a circle and $ABCD$ is a cyclic quadrilateral.

$TA$ is the tangent to the circle at $A$ . $T$ and $C$ are on opposite sides of the line $AB$ .

Angle $TAB = 62°$ and angle $ADC = 100°$ .

Work out the size of angle $BAC$ . Give a reason for each stage of your working. [5]
Model answer
Angle $ACB = 62°$ (alternate segment theorem)

Angle $ABC = 180° - 100° = 80°$ (opposite angles of a cyclic quadrilateral add up to $180°$ )

$\angle BAC = 180^\circ - 62^\circ - 80^\circ$

(angles in a triangle add up to $180°$ )

Answer: $38°$
Why this scores
Mark scheme:
- B1 for angle $ACB = 62$
- B1 for angle $ABC = 80$
- M1 for $180 - 62 - 80$ oe
- A1 for $38$ cao
- B1 for all three reasons
Why this scores full marks: The script finds the alternate segment correctly: angle $TAB$ is between the tangent and chord $AB$ , so it equals the angle at $C$ , on the far side of $AB$ . It then uses the cyclic quadrilateral for angle $ABC$ (opposite $D$ ), not for angle $BAC$ . Every stage has its standard reason, which the final B1 requires.
Question 10
Paper 1H/2H style — 5 marks5 marks
A circle has centre $O$ and radius $7.5$ cm.

$AB$ and $CD$ are parallel chords of the circle, on opposite sides of $O$ .

$AB = 12$ cm and $CD = 9$ cm.

Work out the distance between the two chords. [5]
Model answer
The perpendicular from $O$ to each chord bisects it, so the half-chords are $6$ cm and $4.5$ cm.

Distance from $O$ to $AB$ :

$\sqrt{7.5^{2} - 6^{2}} = \sqrt{20.25} = 4.5$

Distance from $O$ to $CD$ :

$\sqrt{7.5^{2} - 4.5^{2}} = \sqrt{36} = 6$

The chords are on opposite sides of $O$ , so add:

$4.5 + 6 = 10.5$

Answer: $10.5$ cm
Why this scores
Mark scheme:
- M1 for $7.5^{2} - 6^{2}$ oe
- A1 for $4.5$
- M1 for $7.5^{2} - 4.5^{2}$ oe
- A1 for $6$
- A1 for $10.5$ cao
Why this scores full marks: The script halves each chord before using Pythagoras, subtracts because the radius is the hypotenuse, and adds the two distances because the chords are on opposite sides of the centre. If they were on the same side, the distance would be $6 - 4.5 = 1.5$ cm; reading that detail in the question is worth the last mark.

4MA1 Circle Theorems

At a glance

What you’ll learn

The seven circle theorems

Working with circle theorems

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Angle at centre = 2 × angle at circumference

2Angle in a semicircle

3Cyclic quadrilateral

4Tangent perpendicular to radius

5Alternate segment theorem

6Two tangents from the same point [1 mark]

7Reflex angle at the centre from an obtuse angle at the circumference [2 marks]

8Length of a chord from its distance to the centre [3 marks]

9Angles where two chords cross [4 marks]

10Radius and angle from a tangent and a line through the centre [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10