MathsICSE Class 10 · Circles

The Angle at the Centre Theorem: Angle at the Centre Is Double the Angle at the Circumference (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The angle an arc subtends at the centre of a circle is double the angle it subtends at any point on the remaining part of the circle. So if $\angle AOB = 110^\circ$, a point $C$ on the major arc sees $AB$ at $55^\circ$. For a point on the minor arc, use the reflex angle at the centre.

Both angles stand on the same arc AB: the angle at the centre O (110°) is double the angle at C on the remaining circle (55°).

Illustration drawn for QuestionBuddy.

Understand Angle at the Centre Theorem

This is the first angle property in the CISCE syllabus for Circles, and its proof is to be taught (it is not marked 'without proof').

The theorem. Let $O$ be the centre and let the arc $A B$ subtend $∠ A O B$ at the centre. If $C$ is any point on the remaining part of the circle (the arc not being looked at), then $∠ A O B = 2∠ A C B$ .

Which angle at the centre? Always the one that 'faces' the same arc as $C$ does.

$C$ on the major arc: use the ordinary angle $∠ A O B$ (less than $18 0^{\circ}$ ). $∠ A C B = \frac{1}{2} ∠ A O B$ , an acute angle.
$D$ on the minor arc: use the reflex angle $A O B$ , which is $36 0^{\circ} - ∠ A O B$ . $∠ A D B = \frac{1}{2} (36 0^{\circ} - ∠ A O B)$ , an obtuse angle.
$∠ A O B = 18 0^{\circ}$ ( $A B$ a diameter): every point on the circle sees $A B$ at $9 0^{\circ}$ . This is the angle in a semicircle.

**Proof (one diagram, centre inside $∠ A C B$ ).** Join $C O$ and produce it to $D$ .

1. $O A = O C$ (radii), so $∠ O A C = ∠ O C A$ .
2. $∠ A O D$ is an exterior angle of $△ A O C$ , so $∠ A O D = ∠ O A C + ∠ O C A = 2∠ O C A$ .
3. In the same way, $∠ B O D = 2∠ O C B$ .
4. Adding, $∠ A O B = 2 (∠ O C A + ∠ O C B) = 2∠ A C B$ .

When $O$ lies outside $∠ A C B$ , subtract in step 4 instead of adding; when the arc is major, the same steps give reflex $∠ A O B = 2∠ A C B$ .

The partner fact: radii make isosceles triangles. In $△ O A B$ , $O A = O B$ , so $∠ O A B = ∠ O B A = \frac{18 0 ^{\circ} - ∠ A O B}{2}$ . Combining this with the theorem gives a useful shortcut for $C$ on the major arc, $∠ O A B + ∠ A C B = 9 0^{\circ}$ .

Where the centre sits. If all angles of $△ A B C$ are acute, $O$ is inside the triangle. If $∠ A$ is obtuse, $O$ lies outside, beyond $B C$ , and $∠ O B C = ∠ A - 9 0^{\circ}$ .

Writing reasons. Every step needs a reason in brackets, for example '(angle at centre = 2 × angle at circumference)' or '( $O A = O B$ , radii)'. The 2024 examiners' report says many candidates gave none.

Angle at the Centre Theorem: formulas to remember

Angle at the centre theorem: $∠ A O B = 2∠ A C B$; $C$ on the arc opposite the arc $A B$ being looked at.
Point on the minor arc: $∠ A D B = \frac{1}{2} (36 0^{\circ} - ∠ A O B)$; Uses the reflex angle at the centre.
Base angle of the radius triangle: $∠ O A B = ∠ O B A = \frac{18 0 ^{\circ} - ∠ A O B}{2}$; $O A = O B$ (radii).
Shortcut: $∠ O A B + ∠ A C B = 9 0^{\circ}$; $C$ on the major arc $A B$ .

How to solve Angle at the Centre Theorem questions

Find the arc, find the angle at the centre on that arc, then halve or double, giving a reason each time.

1Mark the centre $O$ and the chord or arc that both angles stand on.
2Decide which side of the chord the point on the circle is: on the major arc use $∠ A O B$ ; on the minor arc use the reflex angle $36 0^{\circ} - ∠ A O B$ .
3Apply 'centre = 2 × circumference' in the right direction: halve to go from centre to circumference, double to go back.
4Look for radii: every triangle with two radii as sides is isosceles; use equal base angles and the angle sum $18 0^{\circ}$ .
5Check with a second route (for example $∠ O A B + ∠ A C B = 9 0^{\circ}$ ) and write a reason beside every line.

Angle at the Centre Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Angle at the centre and at the edge

Suppose a circular pond has a fountain at its centre $O$ and two lamp posts $A$ and $B$ on its edge. From the fountain, the lamp posts are $11 0^{\circ}$ apart.

A child stands on the far side of the pond, on the edge, away from both lamp posts. At what angle does she see the two posts?

Try it first, then show the solutionSolution

The child is at a point $C$ on the major arc $A B$ .
$∠ A C B = \frac{1}{2} ∠ A O B = \frac{1}{2} \times 11 0^{\circ}$ (angle at centre = 2 × angle at circumference).

Answer: $5 5^{\circ}$ , wherever she stands on that far arc.

Memory hook: From the edge you see half of what the centre sees.

Major and minor arcs on a clock face

Imagine the numbers 12 and 4 on a round clock face. Each hour mark is $3 0^{\circ}$ apart at the centre.

Find the angle that 12 and 4 make at the centre, at the 8 o'clock mark, and at the 2 o'clock mark.

Try it first, then show the solutionSolution

At the centre: 4 hours apart, so $4 \times 3 0^{\circ} = 12 0^{\circ}$ .
The 8 o'clock mark is on the major arc: $\frac{1}{2} \times 12 0^{\circ} = 6 0^{\circ}$ .
The 2 o'clock mark is on the minor arc between 12 and 4: use the reflex angle, $\frac{1}{2} (36 0^{\circ} - 12 0^{\circ}) = 12 0^{\circ}$ .

Answer: $12 0^{\circ}$ at the centre, $6 0^{\circ}$ at 8 o'clock, $12 0^{\circ}$ at 2 o'clock.

Memory hook: Inside the small arc, use the big (reflex) angle.

Tricks to remember Angle at the Centre Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Centre doubles, edge halves
The centre is closer, so it sees the arc twice as wide as any point on the edge does.
Use it when: Getting the direction right: $∠ A O B = 13 0^{\circ}$ gives $6 5^{\circ}$ at the edge, never $26 0^{\circ}$ .
Small arc, big angle
A point on the minor arc pairs with the reflex angle at the centre.
Use it when: Obtuse angles at the circumference.
Two radii, two equal angles
See $O$ joined to two points on the circle? Mark the base angles equal at once.
Use it when: Finding $∠ O A B$ , $∠ O B C$ and similar angles.
The 90 shortcut
$∠ O A B + ∠ A C B = 9 0^{\circ}$ when $C$ is on the major arc.
Use it when: Quick checks and one-line MCQs.

Angle at the Centre Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: From the circumference to the centre · 1 mark · MCQBoard level

$O$ is the centre of a circle and $C$ is a point on the major arc $A B$ . If $∠ A C B = 3 8^{\circ}$ , then $∠ A O B$ is:
(a) $1 9^{\circ}$
(b) $3 8^{\circ}$
(c) $7 6^{\circ}$
(d) $15 2^{\circ}$

Try it first, then show the solutionSolution

$∠ A O B = 2∠ A C B = 2 \times 3 8^{\circ} = 7 6^{\circ}$ (angle at centre = 2 × angle at circumference).

Answer: (c)

Example 2: Both arcs at once · 3 marks · Short answerBoard level

$O$ is the centre of a circle and $∠ A O B = 11 6^{\circ}$ . $C$ is a point on the major arc $A B$ and $D$ a point on the minor arc $A B$ . Find $∠ O A B$ , $∠ A C B$ and $∠ A D B$ .

Try it first, then show the solutionSolution

$O A = O B$ (radii), so $∠ O A B = \frac{18 0 ^{\circ} - 11 6 ^{\circ}}{2} = 3 2^{\circ}$ .
$∠ A C B = \frac{1}{2} \times 11 6^{\circ} = 5 8^{\circ}$ (angle at centre = 2 × angle at circumference).
Reflex $∠ A O B = 36 0^{\circ} - 11 6^{\circ} = 24 4^{\circ}$ , so $∠ A D B = \frac{1}{2} \times 24 4^{\circ} = 12 2^{\circ}$ .
Check: $∠ O A B + ∠ A C B = 3 2^{\circ} + 5 8^{\circ} = 9 0^{\circ}$ ✓

Answer: $∠ O A B = 3 2^{\circ}$ , $∠ A C B = 5 8^{\circ}$ , $∠ A D B = 12 2^{\circ}$

Example 3: Perpendicular from the centre to a chord · 3 marks · Short answerHarder board level

$O$ is the centre of a circle and $A B$ is a chord. $O M ⊥ A B$ and $∠ A O M = 5 4^{\circ}$ . $C$ is a point on the major arc $A B$ . Find (a) $∠ A O B$ , (b) $∠ A C B$ , (c) $∠ O A B$ .

Try it first, then show the solutionSolution

$△ O M A ≅ △ O M B$ (RHS: $O A = O B$ , $O M$ common, right angles at $M$ ), so $∠ B O M = ∠ A O M = 5 4^{\circ}$ .
(a) $∠ A O B = 2 \times 5 4^{\circ} = 10 8^{\circ}$ .
(b) $∠ A C B = \frac{1}{2} \times 10 8^{\circ} = 5 4^{\circ}$ (angle at centre = 2 × angle at circumference).
(c) In $△ O A M$ , $∠ O A B = 18 0^{\circ} - 9 0^{\circ} - 5 4^{\circ} = 3 6^{\circ}$ .

Answer: (a) $10 8^{\circ}$ (b) $5 4^{\circ}$ (c) $3 6^{\circ}$

Example 4: An obtuse angle: the centre outside the triangle · 4 marks · Long answerHarder board level

$A$ , $B$ and $C$ lie on a circle with centre $O$ , and $∠ B A C = 11 2^{\circ}$ . Find (a) reflex $∠ B O C$ , (b) $∠ B O C$ , (c) $∠ O B C$ . Does $O$ lie inside or outside $△ A B C$ ?

Try it first, then show the solutionSolution

$∠ B A C$ is obtuse, so $A$ is on the minor arc $B C$ and the arc $B C$ it looks at is the major arc.
(a) Reflex $∠ B O C = 2 \times 11 2^{\circ} = 22 4^{\circ}$ (angle at centre = 2 × angle at circumference).
(b) $∠ B O C = 36 0^{\circ} - 22 4^{\circ} = 13 6^{\circ}$ .
(c) $O B = O C$ (radii), so $∠ O B C = \frac{18 0 ^{\circ} - 13 6 ^{\circ}}{2} = 2 2^{\circ}$ . (Check: $11 2^{\circ} - 9 0^{\circ} = 2 2^{\circ}$ .)
$O$ and $A$ are on opposite sides of $B C$ , so $O$ lies outside the triangle.

Answer: (a) $22 4^{\circ}$ (b) $13 6^{\circ}$ (c) $2 2^{\circ}$ ; $O$ is outside $△ A B C$ .

Example 5: Proving and using the 90° shortcut · 4 marks · Long answerHOTS / competency

$O$ is the centre of a circle, $A B$ a chord and $C$ a point on the major arc $A B$ .
(a) Prove that $∠ O B A + ∠ A C B = 9 0^{\circ}$ .
(b) If $∠ O B A = (2 x - 7)^{\circ}$ and $∠ A C B = (x + 13)^{\circ}$ , find $x$ and both angles.

Try it first, then show the solutionSolution

(a) $O A = O B$ (radii), so $∠ O B A = ∠ O A B$ and $∠ A O B = 18 0^{\circ} - 2∠ O B A$ .
$∠ A O B = 2∠ A C B$ (angle at centre = 2 × angle at circumference), so $2∠ A C B = 18 0^{\circ} - 2∠ O B A$ , giving $∠ O B A + ∠ A C B = 9 0^{\circ}$ .
(b) $(2 x - 7) + (x + 13) = 90$ , so $3 x + 6 = 90$ and $x = 28$ .
$∠ O B A = 4 9^{\circ}$ , $∠ A C B = 4 1^{\circ}$ ; check $∠ A O B = 18 0^{\circ} - 9 8^{\circ} = 8 2^{\circ} = 2 \times 4 1^{\circ}$ ✓

Answer: (b) $x = 28$ ; $∠ O B A = 4 9^{\circ}$ , $∠ A C B = 4 1^{\circ}$

Example 6: Case study: gates of a circular park · 4 marks · Case-basedHOTS / competency

A circular park has radius 30 m and a fountain at its centre $O$ . Two gates $A$ and $B$ on the boundary make $∠ A O B = 12 0^{\circ}$ . A straight path joins the gates.
(a) At what angle does a visitor at $C$ , on the major arc, see the two gates?
(b) At what angle does a visitor at $D$ , on the minor arc, see them?
(c) How far is the path $A B$ from the fountain?

Try it first, then show the solutionSolution

(a) $∠ A C B = \frac{1}{2} \times 12 0^{\circ} = 6 0^{\circ}$ .
(b) Reflex $∠ A O B = 24 0^{\circ}$ , so $∠ A D B = 12 0^{\circ}$ .
(c) Draw $O M ⊥ A B$ . $O A = O B$ , so $O M$ bisects $∠ A O B$ : $∠ A O M = 6 0^{\circ}$ and $∠ O A M = 3 0^{\circ}$ .
In a $3 0^{\circ}$ – $6 0^{\circ}$ – $9 0^{\circ}$ triangle the side opposite $3 0^{\circ}$ is half the hypotenuse: $O M = \frac{1}{2} \times 30 = 15$ m.

Answer: (a) $6 0^{\circ}$ (b) $12 0^{\circ}$ (c) 15 m

Where students lose marks in Angle at the Centre Theorem

The slips examiners see most often, and what to write instead.

Mistake: Doubling the wrong way
The angle at the circumference is half the angle at the centre, not double it.
Wrong working
$∠ A O C = 13 0^{\circ}$ , so $∠ A B C = 13 0^{\circ}$ (or $26 0^{\circ}$ ).
Right working
$∠ A B C = \frac{1}{2} \times 13 0^{\circ} = 6 5^{\circ}$ .
Do this: Say 'centre is double' aloud and check that the circumference angle is the smaller one (for a point on the major arc).
Usually costs: Reported in 2024 Q7(i): candidates took $\angle ABC$ as $130^\circ$.
Mistake: No reasons given
Circle-theorem answers carry marks for the reason as well as the value.
Wrong working
$∠ A C B = 5 8^{\circ}$ .
Right working
$∠ A C B = 5 8^{\circ}$ (angle at centre = 2 × angle at circumference).
Do this: Write the theorem in brackets beside every angle you find.
Usually costs: Reported in 2024 Q7(i): many candidates did not give reasons.
Mistake: Using the ordinary angle for a point on the minor arc
A point on the minor arc looks at the major arc, which subtends the reflex angle at the centre.
Wrong working
$∠ A O B = 10 0^{\circ}$ , $D$ on the minor arc, so $∠ A D B = 5 0^{\circ}$ .
Right working
$∠ A D B = \frac{1}{2} (36 0^{\circ} - 10 0^{\circ}) = 13 0^{\circ}$ .
Do this: Shade the arc the point is looking at before you halve anything.
Usually costs: The whole part, and every later part that uses it.

Exam tips

2024 Q7(i) (3 marks): $O M ⊥ A C$ with $∠ C O M = 6 5^{\circ}$ gave $∠ A O C = 13 0^{\circ}$ and $∠ A B C = 6 5^{\circ}$ ; the report lists doubling the wrong way and missing reasons as the common errors.
Mark radii as equal on the diagram first; most 'find the angle' questions hide an isosceles triangle.
If a part is worth 3 or 4 marks, find the angle by two routes when you can; the examiners suggest this as a check.
The proof of this theorem is in the CISCE syllabus: learn it with the construction 'join $C O$ and produce it to $D$ '.

Practice questions on Angle at the Centre Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$O$ is the centre and $∠ A O B = 8 4^{\circ}$ . For a point $D$ on the minor arc $A B$ , $∠ A D B$ is:
(a) $4 2^{\circ}$
(b) $8 4^{\circ}$
(c) $9 6^{\circ}$
(d) $13 8^{\circ}$
Hint
Use the reflex angle $27 6^{\circ}$ .
Answer
(d)
Q2 · 2 marks · Short answerBoard level
$O$ is the centre of a circle and $∠ O B A = 2 7^{\circ}$ . Find $∠ A O B$ and $∠ A C B$ , where $C$ is on the major arc $A B$ .
Hint
$O A = O B$ .
Answer
$∠ A O B = 18 0^{\circ} - 5 4^{\circ} = 12 6^{\circ}$ ; $∠ A C B = 6 3^{\circ}$ .
Q3 · 4 marks · Long answerHarder board level
$A$ , $B$ , $C$ lie on a circle with centre $O$ . $∠ A B C = 7 2^{\circ}$ and $∠ B A C = 4 6^{\circ}$ . Find $∠ A O C$ , $∠ O A C$ , $∠ A O B$ and $∠ O A B$ , and check your answers using $∠ B A C$ .
Hint
First find $∠ A C B$ .
Answer
$∠ A O C = 14 4^{\circ}$ , $∠ O A C = 1 8^{\circ}$ ; $∠ A C B = 6 2^{\circ}$ , so $∠ A O B = 12 4^{\circ}$ and $∠ O A B = 2 8^{\circ}$ . Check: $1 8^{\circ} + 2 8^{\circ} = 4 6^{\circ} = ∠ B A C$ ✓
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $∠ A O B = 10 0^{\circ}$ and $C$ lies on the minor arc $A B$ , then $∠ A C B = 5 0^{\circ}$ .
Reason (R): The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Which angle at the centre does $C$ on the minor arc pair with?
Answer
(b): $∠ A C B = \frac{1}{2} (36 0^{\circ} - 10 0^{\circ}) = 13 0^{\circ}$ .
Q5 · 3 marks · Short answerHarder board level
$A B$ is a chord of a circle with centre $O$ . $C$ is on the major arc and $D$ on the minor arc. $∠ A C B = (3 x + 5)^{\circ}$ and $∠ A D B = (5 x - 1)^{\circ}$ . Find $x$ , both angles and $∠ A O B$ .
Hint
$∠ A C B = \frac{1}{2} ∠ A O B$ and $∠ A D B = \frac{1}{2} (36 0^{\circ} - ∠ A O B)$ ; add them.
Answer
$∠ A C B + ∠ A D B = 18 0^{\circ}$ : $8 x + 4 = 180$ , $x = 22$ ; $∠ A C B = 7 1^{\circ}$ , $∠ A D B = 10 9^{\circ}$ , $∠ A O B = 14 2^{\circ}$ .
Q6 · 3 marks · Short answerHarder board level
$A B$ is a diameter of a circle with centre $O$ , and $C$ is a point on the circle with $∠ B O C = 6 4^{\circ}$ . Find $∠ B A C$ , $∠ A C O$ and $∠ A B C$ , giving reasons.
Hint
Arc $B C$ subtends $∠ B O C$ at the centre and $∠ B A C$ at $A$ .
Answer
$∠ B A C = 3 2^{\circ}$ (angle at centre = 2 × angle at circumference); $∠ A C O = ∠ O A C = 3 2^{\circ}$ ( $O A = O C$ ); $∠ A B C = 9 0^{\circ} - 3 2^{\circ} = 5 8^{\circ}$ (angle in a semicircle).
Q7 · 4 marks · Long answerHOTS / competency
$A$ , $B$ and $C$ lie on a circle with centre $O$ , and $O A B C$ is a parallelogram. $D$ is a point on the major arc $A C$ . Find $∠ A O C$ and $∠ A D C$ .
Hint
$O A = O C$ (radii) and opposite sides of a parallelogram are equal: what kind of triangle is $O A B$ ?
Answer
$A B = O C = O A = O B$ , so $△ O A B$ and $△ O B C$ are equilateral and $∠ A O C = 6 0^{\circ} + 6 0^{\circ} = 12 0^{\circ}$ . Then $∠ A D C = 6 0^{\circ}$ . (Check: $∠ A B C = 12 0^{\circ} = ∠ A O C$ , opposite angles of a parallelogram.)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the angle at the centre theorem.Show answer
The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle.
2. $∠ A O B = 15 0^{\circ}$ . Angle at a point on the major arc?Show answer
$7 5^{\circ}$
3. $∠ A O B = 15 0^{\circ}$ . Angle at a point on the minor arc?Show answer
$10 5^{\circ}$
4. $∠ O A B$ when $∠ A O B = 7 0^{\circ}$ ?Show answer
$5 5^{\circ}$
5.Is the proof of this theorem required?Show answer
Yes; it is not one of the 'without proof' theorems.

Angle at the Centre Theorem: frequently asked questions

What does the angle at the centre theorem say?

The angle subtended by an arc at the centre is twice the angle it subtends at any point on the rest of the circle. For example, if the centre angle is 100°, the angle at the circumference is 50°.

What happens when the point is on the minor arc?

Use the reflex angle at the centre (360° minus the ordinary angle) and halve it. The answer is obtuse; with a centre angle of 100° it is 130°.

Do I need to prove this theorem in the ICSE exam?

It can be asked. The CISCE syllabus says proofs of the circle theorems are to be taught unless marked 'without proof', and this one is not so marked.