MathsICSE Class 9 · Circle

Equal Chords, Arcs and Central Angles; One Circle Through Three Points (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

In a circle, equal chords are equidistant from the centre, cut off equal arcs and subtend equal angles at the centre, and each of these implies the others. Exactly one circle passes through three points that are not on a line; its centre is where the perpendicular bisectors of two of the joining segments meet.

Equal chords AB and CD sit the same distance from O (OM = ON) and subtend equal angles at the centre, because the triangles are congruent.

Illustration drawn for QuestionBuddy.

Understand Equal Chords, Arcs, Central Angles and the Circle Through Three Points

Equal chords and distances (prove). If $A B = C D$ in a circle with centre $O$ , and $O M ⊥ A B$ , $O N ⊥ C D$ , then $A M = \frac{1}{2} A B = \frac{1}{2} C D = C N$ . In right triangles $O M A$ and $O N C$ : hypotenuses $O A = O C$ (radii) and $A M = C N$ , so they are congruent (RHS) and $O M = O N$ . The converse (chords equidistant from the centre are equal) is used without proof.

Arcs and central angles. In one circle (or equal circles), arcs that subtend equal angles at the centre are equal, and equal arcs subtend equal angles at the centre. Equal chords cut off equal arcs and conversely (without proof). Putting these together: equal chords subtend equal angles at the centre, because $△ O A B ≅ △ O C D$ (SSS) when $A B = C D$ .

Chord equal to the radius. $O A = O B = A B = r$ , so $△ O A B$ is equilateral and the chord subtends $6 0^{\circ}$ at the centre; its distance from the centre is $\frac{3}{2} r$ .

One circle through three points. A circle through $A$ , $B$ has its centre on the perpendicular bisector of $A B$ ; through $B$ , $C$ on the perpendicular bisector of $B C$ . If $A, B, C$ are not collinear, these bisectors meet at exactly one point $O$ , and $O A = O B = O C$ . So exactly one circle exists (centre $O$ , radius $O A$ ). If $A, B, C$ are collinear the bisectors are parallel and no circle exists.

Useful special cases.

Right-angled triangle: the centre is the mid-point of the hypotenuse and the radius is half the hypotenuse.
Isosceles triangle with equal sides $a$ and altitude $h$ to the base: the centre lies on the altitude and the radius is $R = \frac{a ^{2}}{2 h}$ (from $R^{2} = (h - R)^{2} + (half base)^{2}$ and $a^{2} = h^{2} + (half base)^{2}$ ).

Reading a question. "Equal chords" suggests equidistant from the centre; "equal angle at the centre" suggests equal arcs and equal chords; "three points" suggests perpendicular bisectors.

Equal Chords, Arcs, Central Angles and the Circle Through Three Points: formulas to remember

Equal chords: $A B = C D ⟺ O M = O N$; Distances from the centre.
Equal arcs: $arc A B = arc C D ⟺ ∠ A O B = ∠ C O D$; In the same circle.
Chord equal to the radius: $A B = r \Rightarrow ∠ A O B = 6 0^{\circ}$; Equilateral triangle OAB.
Circle through 3 points: $centre = intersection of perpendicular bisectors$; Points must not be collinear.
Isosceles triangle circle: $R = \frac{a ^{2}}{2 h}$; Equal sides a, altitude h to the base.

How to solve Equal Chords, Arcs, Central Angles and the Circle Through Three Points questions

Decide whether the question is about distances, angles at the centre or a circle through points.

1For equal chords, draw both perpendiculars from O and use RHS congruence.
2For angle questions, join the ends of the chords to O and look for isosceles or equilateral triangles.
3For three points, draw two perpendicular bisectors; their meeting point is the centre.
4Form equations if an unknown is shared (for example an angle written in terms of $x$ ).
5Verify by checking equal radii or equal distances at the end.

Equal Chords, Arcs, Central Angles and the Circle Through Three Points 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.

Equal cuts through a round table

Suppose two straight rulers of equal length are laid as chords on a circular tabletop of radius 13 cm, each 24 cm long.

How far is each ruler from the centre, and what can you say about the angles they subtend?

Try it first, then show the solutionSolution

$d = 1 3^{2} - 1 2^{2} = 5$ cm for each ruler.
Equal chords subtend equal angles at the centre.

Answer: 5 cm each; equal angles at the centre.

Memory hook: Equal chords, equal distance, equal angle.

Locating a fountain from three lamp posts

Imagine three lamp posts $A, B, C$ forming a triangle with $A B = A C = 20$ m and $B C = 24$ m, and a circular path passing through all three.

How far from $A$ is the centre of the path?

Try it first, then show the solutionSolution

Altitude from $A$ : $h = 2 0^{2} - 1 2^{2} = 16$ m.
$R = \frac{2 0 ^{2}}{2 \times 16} = 12.5$ m.

Answer: The centre is 12.5 m from $A$ , on the altitude.

Memory hook: Perpendicular bisectors meet at the centre.

Tricks to remember Equal Chords, Arcs, Central Angles and the Circle Through Three Points

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

Equal chord trio
Equal chords, equal arcs and equal central angles always go together.
Use it when: Switching between the three facts.
Radius chord = 60
A chord of length r makes an equilateral triangle with the centre.
Use it when: Angle and distance problems.
Two bisectors
Two perpendicular bisectors pin the centre; the third is automatic.
Use it when: Circle through three points.

Equal Chords, Arcs, Central Angles and the Circle Through Three Points: solved examples

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

Example 1: Chord equal to the radius · 1 mark · MCQBoard level

A chord of a circle is equal in length to the radius. The angle it subtends at the centre is:
(a) $3 0^{\circ}$
(b) $4 5^{\circ}$
(c) $6 0^{\circ}$
(d) $9 0^{\circ}$

Try it first, then show the solutionSolution

$O A = O B = A B$ , so $△ O A B$ is equilateral.
Each angle is $6 0^{\circ}$ .

Answer: (c) $6 0^{\circ}$

Example 2: Arcs from central angles · 3 marks · Short answerBoard level

In a circle, arc $A B$ subtends $4 0^{\circ}$ at the centre and arc $C D$ subtends $12 0^{\circ}$ . If arc $A B$ is $5$ cm long, find the length of arc $C D$ and the angle arc $P Q$ subtends if it equals arc $A B$ .

Try it first, then show the solutionSolution

Arc length in one circle is proportional to the central angle: arc $C D = 5 \times \frac{120}{40} = 15$ cm.
Equal arcs subtend equal angles, so arc $P Q$ subtends $4 0^{\circ}$ .

Answer: Arc $C D = 15$ cm; $∠ P O Q = 4 0^{\circ}$ .

Example 3: Equal parallel chords · 3 marks · Short answerHarder board level

Two equal parallel chords of length $24$ cm are drawn in a circle of radius $13$ cm. Find the distance between them.

Try it first, then show the solutionSolution

Equal chords are equidistant from the centre: each is $d = 1 3^{2} - 1 2^{2} = 5$ cm away.
Being parallel and distinct, they lie on opposite sides of the centre.
Distance between them $= 5 + 5 = 10$ cm.

Answer: 10 cm

Example 4: Chords and central angles with surds · 4 marks · Short answerHarder board level

A circle has centre $O$ and radius $12$ cm. Chord $P Q$ subtends $6 0^{\circ}$ at the centre and chord $R S$ subtends $9 0^{\circ}$ . Find $P Q$ , the distance of $P Q$ from $O$ , and $R S$ .

Try it first, then show the solutionSolution

$∠ P O Q = 6 0^{\circ}$ and $O P = O Q$ , so $△ O P Q$ is equilateral: $P Q = 12$ cm.
Distance $= 1 2^{2} - 6^{2} = 108 = 63$ cm $\approx 10.39$ cm.
$∠ R O S = 9 0^{\circ}$ with $O R = O S = 12$ : $R S = 144 + 144 = 122$ cm.

Answer: $P Q = 12$ cm, distance $63$ cm, $R S = 122$ cm.

Example 5: The circle through three points · 5 marks · Long answerHOTS / competency

$A$ , $B$ , $C$ are three points with $A B = A C = 10$ cm and $B C = 12$ cm. Show that exactly one circle passes through them and find its radius.

Try it first, then show the solutionSolution

The points are not collinear (they form a triangle), so the perpendicular bisectors of $A B$ and $B C$ meet at one point $O$ ; the circle with centre $O$ is unique.
Let $M$ be the mid-point of $B C$ ; $A M ⊥ B C$ and $A M = 1 0^{2} - 6^{2} = 8$ cm. The centre lies on $A M$ (the perpendicular bisector of $B C$ ).
Let $O A = O B = R$ . Then $O M = 8 - R$ and in right triangle $O M B$ : $R^{2} = (8 - R)^{2} + 6^{2}$ .
$R^{2} = 64 - 16 R + R^{2} + 36$ , so $16 R = 100$ and $R = 6.25$ cm.
Check: $O M = 1.75$ ; $1.7 5^{2} + 36 = 3.0625 + 36 = 39.0625 = 6.2 5^{2}$ .

Answer: Unique circle; radius $6.25$ cm.

Where students lose marks in Equal Chords, Arcs, Central Angles and the Circle Through Three Points

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

Mistake: Assuming equal angles means equal chords in different circles
The equalities hold in the same circle or equal circles.
Wrong working
Chord in a small circle equals chord in a big circle because the angles at the centre are equal.
Right working
Equal angles give equal chords only when the radii are equal.
Do this: Check the circles first.
Usually costs: Reasoning mark.
Mistake: Wrong centre for a right triangle
The centre is on the hypotenuse mid-point, not the right-angle vertex.
Wrong working
Radius $=$ the longer leg.
Right working
Radius $=$ half the hypotenuse.
Do this: Right angle in a semicircle idea.
Usually costs: Final answer.
Mistake: Omitting that points must be non-collinear
The existence statement needs it.
Wrong working
A circle passes through any three points.
Right working
A unique circle passes through three points not on a line.
Do this: State the condition.
Usually costs: One mark.

Exam tips

For the equal chord proof, write the congruence as RHS and name the right angles.
Write "equal chords of the same circle" before claiming equal central angles.
In circle-through-three-points problems, use the altitude of an isosceles triangle; the centre lies on it.
Class IX is a school-exam class; these items are set at the top of that range and some at foundation-olympiad level.

Practice questions on Equal Chords, Arcs, Central Angles and the Circle Through Three Points

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
The number of circles that can be drawn through three collinear points is:
(a) 0
(b) 1
(c) 2
(d) infinitely many
Hint
The perpendicular bisectors are parallel.
Answer
(a) 0
Q2 · 3 marks · Short answerHarder board level
$A B$ and $C D$ are equal chords of a circle with centre $O$ , with $∠ A O B = (5 x - 15)^{\circ}$ and $∠ C O D = (3 x + 15)^{\circ}$ . Find $x$ . If $A B = 9$ cm, find the radius.
Hint
Equal chords subtend equal angles at the centre.
Answer
$5 x - 15 = 3 x + 15$ , $x = 15$ , so each angle is $6 0^{\circ}$ . $△ O A B$ is equilateral, so the radius is $9$ cm.
Q3 · 3 marks · Short answerHarder board level
The sides of a right-angled triangle $A B C$ are $A B = 6$ cm, $B C = 8$ cm, with the right angle at $B$ . Find the radius of the circle through $A$ , $B$ , $C$ .
Hint
The mid-point of the hypotenuse is equidistant from all three vertices.
Answer
$A C = 36 + 64 = 10$ cm. The centre is the mid-point of $A C$ , so the radius is $5$ cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Exactly one circle passes through three non-collinear points.
Reason (R): The perpendicular bisectors of two of the joining segments meet at exactly one point.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
The centre must be on both bisectors.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Three sensors $A$ , $B$ , $C$ in a park form a triangle with $A B = A C = 20$ m and $B C = 24$ m. A circular fence is to pass through all three.
(a) Find the height of $A$ above $B C$ .
(b) Find the radius of the fence.
(c) Find the distance of the centre from $B C$ and say whether the centre is inside the triangle.
Hint
$R^{2} = (h - R)^{2} + 1 2^{2}$ .
Answer
(a) $400 - 144 = 16$ m (b) $R^{2} = (16 - R)^{2} + 144$ gives $32 R = 256 + 144 = 400$ , so $R = 12.5$ m (c) $16 - 12.5 = 3.5$ m from $B C$ , on the same side as $A$ , so the centre is inside the triangle.

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.What is true of equal chords and the centre?Show answer
They are at equal distances from the centre.
2.A chord is exactly as long as the radius of its circle. How large is the angle it makes at the centre?Show answer
60 degrees.
3.Where is the centre of the circle through the vertices of a right triangle?Show answer
At the mid-point of the hypotenuse.
4.Three points are not on one straight line. How many circles go through all of them?Show answer
Exactly one.

Equal Chords, Arcs, Central Angles and the Circle Through Three Points: frequently asked questions

Is the equal chords theorem proved or used?

The ICSE syllabus asks you to prove that equal chords are equidistant from the centre; the converse is used without proof.

What if the three points are on a line?

No circle passes through them.

Do equal arcs always have equal chords?

Yes, in the same circle or equal circles.