MathsICSE Class 9 · Chapter 13

Circle ICSE Class 9: Chord Properties, Arcs and Solved Questions

Q: Which circle theorems must I prove in Class IX?

The line from the centre bisecting a chord is perpendicular to it; equal chords are equidistant from the centre; equal central angles give equal arcs and conversely. The perpendicular bisecting the chord, equidistant chords being equal, and the chord–arc results are used without proof.

Q: Are angle properties like 'angle in a semicircle' in Class IX?

No. Angle properties of circles, cyclic quadrilaterals and tangents are in the Class X syllabus.

Q: Why must the three points not be collinear?

If they are on one line, the perpendicular bisectors are parallel and never meet, so no centre exists.

Q: Is a diameter a chord?

Yes; it is the longest chord, passing through the centre.

Q: What does 'subtend' mean?

An arc or chord subtends an angle at a point when lines from its ends to that point form the angle.

Updated 15 October 2026 · Fact-checked

In ICSE Class 9 Circle: the line from the centre to the mid-point of a chord is perpendicular to it, and the perpendicular from the centre bisects the chord; equal chords are equidistant from the centre and conversely; one circle passes through three non-collinear points; equal arcs subtend equal central angles, and equal chords cut equal arcs.

What Circle is about

A chord is a segment joining two points of a circle; a diameter is a chord through the centre. An arc is the part of the circle between two points, and the angle it subtends at the centre is the angle between the radii to its ends.

The CISCE 2028 Class IX syllabus covers (a) chord properties: a line joining the centre to the mid-point of a chord (other than a diameter) meets it at right angles; a perpendicular dropped from the centre halves the chord (no proof needed); chords of equal length lie at equal distances from the centre; chords at equal distances are of equal length (no proof needed); exactly one circle can be drawn through three points not on a line; and (b) arc and chord properties: arcs that subtend equal angles at the centre are equal, and conversely; equal chords cut off equal arcs, and conversely (without proof).

Almost every numerical question uses one right triangle: radius as hypotenuse, half-chord and distance from the centre as the legs. School exams set 3-mark length questions, 4-mark proofs of the two theorems you must prove, and harder questions on parallel or intersecting chords.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Circle is in the 4. Geometry (iii) Circle part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Perpendicular from the centre and the mid-point of a chord

Theorem (prove): if a line joins the centre of a circle to the mid-point of a chord that is not a diameter, it is perpendicular to that chord.

Proof: let $M$ be the mid-point of chord $A B$ of a circle with centre $O$ . In $△ O A M$ and $△ O B M$ : $O A = O B$ (radii), $A M = B M$ (given), $O M$ common. So they are congruent (SSS) and $∠ O M A = ∠ O M B$ . These form a linear pair, so each is $9 0^{\circ}$ .

Converse (use without proof): the perpendicular from the centre to a chord bisects the chord.

Worked example

A chord of length 24 cm is drawn in a circle of radius 13 cm. Find its distance from the centre.

Try it first, then show the solutionSolution

The perpendicular $O M$ bisects the chord: $A M = 12$ cm.
In right $△ O M A$ : $O M = 1 3^{2} - 1 2^{2} = 25 = 5$ cm.

Answer: 5 cm

The radius–half-chord–distance triangle

For a chord of length $2 l$ at distance $d$ from the centre of a circle of radius $r$ :

$r^{2} = d^{2} + l^{2}$ .

Know any two of $r$ , $d$ , $l$ and you can find the third. The closer a chord is to the centre, the longer it is; the diameter ( $d = 0$ ) is the longest chord.

Worked example

A chord 30 cm long is 8 cm from the centre. Find the radius.

Try it first, then show the solutionSolution

Half-chord $= 15$ cm.
$r = 1 5^{2} + 8^{2} = 289 = 17$ cm.

Answer: 17 cm

Equal chords and their distances

Theorem (prove): equal chords of a circle are equidistant from the centre.

Proof: let $A B = C D$ , with $O M ⊥ A B$ and $O N ⊥ C D$ . Perpendiculars from the centre bisect the chords, so $A M = \frac{1}{2} A B = \frac{1}{2} C D = C N$ . In right triangles $O M A$ and $O N C$ : $O A = O C$ (radii, hypotenuses) and $A M = C N$ . So they are congruent (RHS), giving $O M = O N$ .

Converse (use without proof): chords equidistant from the centre are equal.

Worked example

Two chords $A B$ and $C D$ of a circle are each 8 cm from the centre. If $A B = 12$ cm, find $C D$ and the radius.

Try it first, then show the solutionSolution

Equidistant chords are equal: $C D = 12$ cm.
$r = 6^{2} + 8^{2} = 10$ cm.

Answer: CD = 12 cm; r = 10 cm

One circle through three points

Result: there is one and only one circle passing through three given points that are not in a straight line.

Reason: the centre must be equidistant from $A$ and $B$ , so it lies on the perpendicular bisector of $A B$ ; it must also lie on the perpendicular bisector of $B C$ . Since $A$ , $B$ , $C$ are not collinear, these two bisectors are not parallel and meet at exactly one point $O$ . With centre $O$ and radius $O A$ , the circle passes through all three points, and no other centre is possible.

If the three points are collinear, the bisectors are parallel and no circle passes through them.

Worked example

Describe how to draw the circle through three non-collinear points $P$ , $Q$ , $R$ .

Try it first, then show the solutionSolution

Join $P Q$ and $QR$ .
Construct the perpendicular bisectors of $P Q$ and $QR$ ; let them meet at $O$ .
With centre $O$ and radius $O P$ draw the circle; it passes through $Q$ and $R$ as well.

Answer: Centre at the meeting point of the perpendicular bisectors.

Arcs, central angles and chords

In the same circle (or in equal circles):

Arcs and angles: if two arcs subtend equal angles at the centre, the arcs are equal; conversely, equal arcs subtend equal angles at the centre.
Chords and arcs (use without proof): equal chords cut off equal arcs; conversely, equal arcs have equal chords.

So equal chords subtend equal angles at the centre. A chord equal to the radius forms an equilateral triangle with the centre, so it subtends $6 0^{\circ}$ .

Worked example

In a circle with centre $O$ , chords $A B$ and $C D$ are equal and $∠ A O B = 8 0^{\circ}$ . Find $∠ C O D$ and $∠ O C D$ .

Try it first, then show the solutionSolution

Equal chords cut off equal arcs, which subtend equal angles at the centre: $∠ C O D = 8 0^{\circ}$ .
$O C = O D$ (radii), so $∠ O C D = \frac{18 0 ^{\circ} - 8 0 ^{\circ}}{2} = 5 0^{\circ}$ .

Answer: ∠COD = 80°; ∠OCD = 50°

Two chords: parallel or on a common line

Parallel chords: the perpendicular from the centre to one is also perpendicular to the other, so their distances from the centre lie along the same line.

Same side of the centre: distance between chords $= d_{1} - d_{2}$ .
Opposite sides: distance $= d_{1} + d_{2}$ .

Concentric circles: a line cutting both circles meets the outer one at $A$ , $B$ and the inner one at $C$ , $D$ . The perpendicular from the centre bisects both chords, so $A C = B D$ .

Worked example

In a circle of radius 5 cm, $A B = 6$ cm and $C D = 8$ cm are parallel chords. Find the distance between them if they lie (i) on the same side (ii) on opposite sides of the centre.

Try it first, then show the solutionSolution

Distance of $A B$ : $25 - 9 = 4$ cm; of $C D$ : $25 - 16 = 3$ cm.
(i) $4 - 3 = 1$ cm.
(ii) $4 + 3 = 7$ cm.

Answer: (i) 1 cm (ii) 7 cm

Maths in real life: Circle

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.

Chord length from radius and distance

Suppose a circular pond has radius 10 m and a straight footbridge crosses it 6 m from the centre.

How long is the footbridge over the water?

Try it first, then show the solutionSolution

Half-length $= 1 0^{2} - 6^{2} = 8$ m.
Length $= 16$ m.

Answer: 16 m.

Memory hook: Radius, distance, half-chord: a right triangle every time.

Circle through three points

Imagine three villages that are not in a straight line want a water tower the same distance from all three.

Where should the tower go, and is there more than one choice?

Try it first, then show the solutionSolution

The tower must lie on the perpendicular bisector of each pair of villages.
Two such bisectors meet at exactly one point, which is the unique choice.

Answer: At the meeting point of the perpendicular bisectors; only one such place.

Memory hook: Three points, one circle, one centre.

Finding a radius from a broken piece

Suppose an archaeologist finds part of a circular plate. A straight edge across it measures 24 cm between two points of the rim, and the rim is 8 cm from the mid-point of this edge, measured at right angles.

Find the radius of the original plate.

Try it first, then show the solutionSolution

Let the radius be $r$ ; the chord is $r - 8$ from the centre and half-chord is 12.
$r^{2} = (r - 8)^{2} + 144 \Rightarrow 16 r = 208 \Rightarrow r = 13$ cm.

Answer: 13 cm.

Memory hook: Distance from centre = radius minus the bulge.

Topics in this chapter

Circle formula sheet

Radius, distance, half-chord: $r^{2} = d^{2} + (\frac{c}{2})^{2}$; c is the chord length.
Chord from r and d: $c = 2 r^{2} - d^{2}$; d < r.
Parallel chords: $d_{1} \pm d_{2}$; − same side, + opposite sides.
Equal chords: $A B = C D ⟺ O M = O N$; OM, ON perpendicular distances.
Equal arcs: $A B ⌢ = C D ⌢ ⟺ ∠ A O B = ∠ C O D$; Same or equal circles.

Circle: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Chord, perpendicular and radius

Any chord-length question or the proof that the line from the centre bisecting a chord is perpendicular to it.

Draw

Draw a circle with centre O.
Draw a chord AB that is not a diameter.
Draw OM from O to the mid-point M of AB.
Join OA (and OB for the proof).

Label

O (centre)
A, B (ends of the chord)
M (mid-point)
right-angle mark at M
r on OA, d on OM, half-chord on AM

Watch out: For numerical questions label r, d and the half-chord before writing r² = d² + l².

How the school exam asks Circle

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find a length	Two of radius, chord, distance from the centre are given; find the third.	Right triangle with the radius as hypotenuse and half the chord as one leg.
Prove a chord theorem	Prove the line from the centre bisecting a chord is perpendicular to it, or that equal chords are equidistant.	Congruent triangles (SSS or RHS) with radii as equal sides.
Parallel chords	Find the distance between parallel chords, or the radius given the chords and their distance apart.	Distances from the centre along one line; same side subtract, opposite sides add.
Equal chords and arcs	Given equal chords, find angles at the centre or prove segments equal.	Equal chords ↔ equal arcs ↔ equal central angles ↔ equal distances.
Circle through three points	Explain why only one circle passes through three non-collinear points, or construct it.	Perpendicular bisectors meet at one point.

Tricks to remember Circle

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

Radius is the roof
In the chord triangle, the radius is always the hypotenuse (the roof); half-chord and distance are the walls.
Use it when: r² = d² + l².
Closer is longer
The nearer a chord is to the centre, the longer it is; the diameter is closest and longest.
Use it when: Comparing chords from distances.
Equal chain
Equal chords ⇔ equal arcs ⇔ equal central angles ⇔ equal distances from the centre.
Use it when: Arc and chord problems.
Same side subtract
Parallel chords on the same side: subtract distances; opposite sides: add.
Use it when: Distance between parallel chords.

Circle: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Longest chord · 1 mark · MCQExam level

The longest chord of a circle of radius 7 cm is:
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 3.5 cm

Try it first, then show the solutionSolution

The longest chord is a diameter $= 2 r$ .

Answer: (b)

Example 2: Chord from radius and distance · 2 marks · Short answerExam level

Find the length of a chord which is 6 cm from the centre of a circle of radius 10 cm.

Try it first, then show the solutionSolution

Half-chord $= 100 - 36 = 8$ cm.
Chord $= 16$ cm.

Answer: 16 cm

Example 3: Concentric circles · 3 marks · Short answerExam level

Two concentric circles have centre $O$ . A line cuts the outer circle at $A$ and $B$ and the inner circle at $C$ and $D$ (in the order $A$ , $C$ , $D$ , $B$ ). Prove that $A C = B D$ .

Try it first, then show the solutionSolution

Draw $O M ⊥ A B$ .
In the outer circle, $O M$ bisects chord $A B$ : $A M = B M$ . In the inner circle, $O M$ bisects chord $C D$ : $C M = D M$ .
Subtract: $A M - C M = B M - D M$ , so $A C = B D$ .

Answer: Proved.

Example 4: Radius from parallel chords · 4 marks · Long answerHarder exam level

$A B$ and $C D$ are parallel chords on opposite sides of the centre, with $A B = 10$ cm, $C D = 24$ cm, and the distance between them 17 cm. Find the radius.

Try it first, then show the solutionSolution

Let $A B$ be $x$ cm from the centre, so $C D$ is $(17 - x)$ cm away.
$r^{2} = 5^{2} + x^{2} = 1 2^{2} + (17 - x)^{2}$ .
$25 + x^{2} = 144 + 289 - 34 x + x^{2} \Rightarrow 34 x = 408 \Rightarrow x = 12$ .
$r = 25 + 144 = 13$ cm.

Answer: 13 cm

Example 5: Equal chords meeting at a point · 4 marks · Long answerHarder exam level

Equal chords $A B$ and $C D$ of a circle with centre $O$ intersect at $P$ . Prove that $O P$ bisects the angle between the chords.

Try it first, then show the solutionSolution

Draw $O M ⊥ A B$ and $O N ⊥ C D$ .
Equal chords are equidistant: $O M = O N$ .
In right triangles $O M P$ and $O N P$ : $O P$ common (hypotenuse), $O M = O N$ ; congruent (RHS).
So $∠ O P M = ∠ O P N$ : $O P$ bisects the angle between the chords.

Answer: Proved.

Example 6: Chord equal to the radius · 3 marks · Short answerHarder exam level

A chord of a circle is equal to its radius. Find the angle it subtends at the centre, and the distance of the chord from the centre if the radius is 6 cm.

Try it first, then show the solutionSolution

$O A = O B = A B$ : $△ O A B$ is equilateral, so the angle is $6 0^{\circ}$ .
Distance $= 6^{2} - 3^{2} = 27 = 33 \approx 5.20$ cm.

Answer: 60°; 3√3 ≈ 5.20 cm

Example 7: Intersecting equal chords · 5 marks · Long answerHOTS / competency

Equal chords $A B$ and $C D$ of a circle intersect at $P$ inside the circle. Prove that $P B = P D$ and $P A = P C$ .

Try it first, then show the solutionSolution

Draw $O M ⊥ A B$ and $O N ⊥ C D$ ; then $M$ , $N$ are mid-points and $O M = O N$ (equal chords).
$△ O M P ≅ △ O N P$ (RHS), so $M P = N P$ .
$M B = \frac{1}{2} A B = \frac{1}{2} C D = N D$ .
Taking $P$ between $M$ and $B$ , and between $N$ and $D$ : $P B = M B - M P = N D - N P = P D$ .
Then $P A = A B - P B = C D - P D = P C$ .

Answer: Proved.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

A circular park with paths

A circular park has radius 25 m and centre $O$ . Two straight parallel paths cross it: path $P_{1}$ is 48 m long and path $P_{2}$ is 40 m long. A third straight path $P_{3}$ is the same length as $P_{1}$ but runs in a different direction.

(i) · 3 marks
Find the distance of each of $P_{1}$ and $P_{2}$ from the centre.
Answer
Answer: 7 m and 15 m
$P_{1}$ : $625 - 576 = 7$ m. $P_{2}$ : $625 - 400 = 15$ m.
(ii) · 3 marks
Find the distance between $P_{1}$ and $P_{2}$ if they are on opposite sides of the centre, and if they are on the same side.
Answer
Answer: 22 m; 8 m
Opposite: $7 + 15 = 22$ m. Same side: $15 - 7 = 8$ m.
(iii) · 4 marks
How far is $P_{3}$ from the centre, and which property did you use? Which of the three paths is the shortest, and why?
Answer
Answer: 7 m; P₂ (40 m) is the shortest
$P_{3} = P_{1}$ , and equal chords are equidistant from the centre, so 7 m. The farther a chord is from the centre, the shorter it is; $P_{2}$ is farthest (15 m), so it is the shortest path, 40 m.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find its distance from the centre. (3 marks)

Step	Marks
Perpendicular from the centre bisects the chord: half-chord 8 cm	1
$d^{2} = 1 0^{2} - 8^{2} = 36$	1
$d = 6$ cm	1

Prove that the straight line drawn from the centre of a circle to bisect a chord (not a diameter) is perpendicular to the chord. (4 marks)

Step	Marks
Figure: centre $O$ , chord $A B$ , mid-point $M$ ; join $O A$ , $O B$ , $O M$	1
$O A = O B$ (radii), $A M = B M$ (given), $O M$ common	1
$△ O A M ≅ △ O B M$ (SSS), so $∠ O M A = ∠ O M B$	1
They form a linear pair, so each is $9 0^{\circ}$ : $O M ⊥ A B$	1

Two parallel chords of a circle are 6 cm and 8 cm long and lie on the same side of the centre, 1 cm apart. Find the radius. (5 marks)

Step	Marks
Let the 8 cm chord be $x$ cm from the centre; the 6 cm chord is $(x + 1)$ cm away (shorter chord is farther)	1
$r^{2} = 4^{2} + x^{2} = 3^{2} + (x + 1)^{2}$	1
$16 + x^{2} = 9 + x^{2} + 2 x + 1 \Rightarrow 2 x = 6$	1
$x = 3$	1
$r = 16 + 9 = 5$ cm	1

Where students lose marks in Circle

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

Mistake: Using the full chord instead of half
The right triangle has half the chord as a leg.
Wrong working
$d = 1 3^{2} - 2 4^{2}$
Right working
$d = 1 3^{2} - 1 2^{2} = 5$
Do this: Write 'perpendicular from the centre bisects the chord' and halve first.
Usually costs: The answer.
Mistake: Not stating the property used
Steps like 'AM = BM' need the reason 'perpendicular from the centre bisects the chord'.
Do this: Bracket the property after each step.
Usually costs: Method marks.
Mistake: Adding distances for chords on the same side
Same-side chords are separated by the difference of their distances.
Do this: Sketch the circle and both chords before deciding.
Usually costs: 1 to 2 marks.
Mistake: Proving results marked 'without proof'
The syllabus states several results without proof; proving them wastes time.
Do this: Prove only: line from centre bisecting a chord is perpendicular; equal chords are equidistant; arcs with equal central angles are equal and conversely.
Usually costs: Time.
Mistake: Applying chord properties to a diameter case
The first theorem excludes diameters, whose mid-point is the centre itself.
Do this: Check the chord is not a diameter.
Usually costs: 1 mark.

Practice questions on Circle

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 · MCQExam level
A chord of a circle of radius 17 cm is 8 cm from the centre. Its length is:
(a) 15 cm
(b) 30 cm
(c) 18 cm
(d) 25 cm
Hint
Half-chord √(289 − 64).
Answer
(b)
Q2 · 2 marks · Short answerExam level
Find the radius of a circle in which a chord of 40 cm is 15 cm from the centre.
Hint
Half-chord 20.
Answer
$400 + 225 = 25$ cm.
Q3 · 2 marks · Short answerExam level
Two chords $A B$ and $C D$ of a circle are equal, and $A B$ is 5 cm from the centre. How far is $C D$ from the centre? State the result used.
Hint
Equal chords.
Answer
5 cm; equal chords are equidistant from the centre.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): The perpendicular bisectors of any two chords of a circle meet at its centre.
Reason (R): The perpendicular from the centre to a chord bisects it.
(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
What if the two chords are parallel?
Answer
(b). Parallel chords have the same perpendicular bisector line, so it does not fix a single point; (A) fails for them. (R) is true.
Q5 · 3 marks · Short answerExam level
In a circle of radius 13 cm, two parallel chords of lengths 10 cm and 24 cm lie on the same side of the centre. Find the distance between them.
Hint
Distances 12 and 5.
Answer
$12 - 5 = 7$ cm.
Q6 · 3 marks · Short answerHarder exam level
$A B$ is a diameter of a circle with centre $O$ , and chord $C D$ is parallel to $A B$ . If $A B = 26$ cm and $C D = 10$ cm, find the distance between $A B$ and $C D$ .
Hint
The diameter passes through O.
Answer
Distance of $C D$ from $O$ $= 169 - 25 = 12$ cm, which is the distance between the chords.
Q7 · 3 marks · Short answerHarder exam level
$M$ and $N$ are the mid-points of equal chords $A B$ and $C D$ of a circle with centre $O$ . Prove that $∠ O M N = ∠ O N M$ .
Hint
OM = ON.
Answer
Lines from the centre to the mid-points are perpendicular to the chords, so $O M$ and $O N$ are the distances; equal chords are equidistant, so $O M = O N$ and $△ O M N$ is isosceles: $∠ O M N = ∠ O N M$ .
Q8 · 3 marks · Short answerHarder exam level
A chord $P Q$ of a circle of radius 10 cm subtends $9 0^{\circ}$ at the centre. Find $P Q$ and its distance from the centre.
Hint
Right isosceles triangle OPQ.
Answer
$P Q = 100 + 100 = 102 \approx 14.14$ cm. Distance $= 100 - 50 = 52 \approx 7.07$ cm.
Q9 · 4 marks · Long answerHarder exam level
Two circles of radii 10 cm and 17 cm intersect at $A$ and $B$ , and $A B = 16$ cm. Find the distance between their centres if they lie on opposite sides of $A B$ .
Hint
The line of centres is the perpendicular bisector of AB.
Answer
Half of $A B$ is 8 cm. Distances from $A B$ : $100 - 64 = 6$ cm and $289 - 64 = 15$ cm. Centres are $6 + 15 = 21$ cm apart.
Q10 · 3 marks · Short answerHarder exam level
Prove that if two arcs of a circle subtend equal angles at the centre, their chords are equal.
Hint
SAS with radii.
Answer
If $∠ A O B = ∠ C O D$ : $O A = O C$ , $O B = O D$ (radii) and the included angles are equal, so $△ A O B ≅ △ C O D$ (SAS) and $A B = C D$ .
Q11 · 3 marks · Short answerHarder exam level
A chord of length 8 cm is at a distance of 3 cm from the centre. Find the length of a chord of the same circle at a distance of 4 cm from the centre, and say why it is shorter.
Hint
Find r first.
Answer
$r = 16 + 9 = 5$ cm. New chord $= 2 25 - 16 = 6$ cm. It is farther from the centre, so shorter.
Q12 · 4 marks · Long answerHOTS / competency
$A B$ and $A C$ are two equal chords of a circle with centre $O$ . Prove that the bisector of $∠ B A C$ passes through $O$ .
Hint
Show OA bisects ∠BAC using equal distances or congruent triangles.
Answer
In $△ O A B$ and $△ O A C$ : $O B = O C$ (radii), $A B = A C$ (given), $O A$ common, so they are congruent (SSS) and $∠ O A B = ∠ O A C$ . Hence $A O$ is the bisector of $∠ B A C$ , i.e. the bisector passes through $O$ .

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 does the perpendicular from the centre do to a chord?Show answer
It bisects the chord.
2.How are equal chords placed relative to the centre?Show answer
They are equidistant from it.
3.How many circles pass through three non-collinear points?Show answer
Exactly one.
4.Formula linking radius r, distance d and half-chord l?Show answer
r² = d² + l².
5.Which is the longest chord?Show answer
The diameter.
6.Angle subtended at the centre by a chord equal to the radius?Show answer
60°.
7.Equal arcs subtend … at the centre.Show answer
Equal angles.
8.Distance between parallel chords on opposite sides of the centre?Show answer
The sum of their distances from the centre.

Best order to study it

Learn the vocabulary: chord, diameter, arc, central angle.
Prove the perpendicular-bisector theorem and practise r² = d² + l² questions.
Prove 'equal chords are equidistant'; learn its converse as a result.
Learn the circle-through-three-points reasoning and construction.
Finish with arc and chord properties and parallel-chord problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Do you halve the chord before using Pythagoras?
Can you prove the line from the centre to a chord's mid-point is perpendicular?
Can you prove equal chords are equidistant?
Can you find the distance between parallel chords on either side?
Can you find a radius from two parallel chords and their gap?
Can you explain why only one circle passes through three non-collinear points?
Do you know which results are 'without proof'?

Circle: frequently asked questions

Which circle theorems must I prove in Class IX?

The line from the centre bisecting a chord is perpendicular to it; equal chords are equidistant from the centre; equal central angles give equal arcs and conversely. The perpendicular bisecting the chord, equidistant chords being equal, and the chord–arc results are used without proof.

Are angle properties like 'angle in a semicircle' in Class IX?

No. Angle properties of circles, cyclic quadrilaterals and tangents are in the Class X syllabus.

Why must the three points not be collinear?

If they are on one line, the perpendicular bisectors are parallel and never meet, so no centre exists.

Is a diameter a chord?

Yes; it is the longest chord, passing through the centre.

What does 'subtend' mean?

An arc or chord subtends an angle at a point when lines from its ends to that point form the angle.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028