MathsICSE Class 9 · Circle

Chord Length and Distance from the Centre: r² = d² + l² (ICSE Class 9 Circle)

Updated 17 October 2026 · Fact-checked

The line from the centre to the mid-point of a chord meets it at a right angle, and the perpendicular from the centre bisects the chord. So the radius, half the chord and the distance from the centre form a right-angled triangle with r² = d² + l², where the chord has length 2l.

Notice that OM meets the chord at a right angle at its mid-point, so the right triangle OMB gives r² = d² + l².

Illustration drawn for QuestionBuddy.

Understand Chord Length and Distance from the Centre

Take a circle with centre $O$ and radius $r$ , and a chord $A B$ that is not a diameter. Let $M$ be its mid-point.

Theorem (prove). $O M ⊥ A B$ . In $△ O A M$ and $△ O B M$ : $O A = O B$ (radii), $A M = B M$ (given), $O M$ is common, so they are congruent (SSS). Hence $∠ O M A = ∠ O M B$ , and being a linear pair, each is $9 0^{\circ}$ .

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

The working triangle. In right-angled $△ O M A$ , with $O M = d$ and $A M = l$ (so $A B = 2 l$ ):
$ $r^{2} = d^{2} + l^{2} .$ $
Given two of $r, d, l$ the third follows. As $d$ decreases the chord grows; at $d = 0$ the chord is a diameter, the longest chord, $2 r$ .

Parallel chords. If $A B ∥ C D$ , the perpendicular from $O$ to $A B$ is also perpendicular to $C D$ , so the two distances $d_{1}, d_{2}$ lie along one line. Distance between the chords: $∣ d_{1} - d_{2} ∣$ if they are on the same side of the centre, $d_{1} + d_{2}$ if on opposite sides. Questions often give the distance between the chords and ask for the radius: use $r^{2} = d_{1}^{2} + l_{1}^{2} = d_{2}^{2} + l_{2}^{2}$ with $d_{2} = d_{1} \pm gap$ .

Concentric circles. For two circles with the same centre, a line meeting the outer circle at $A, B$ and the inner at $C, D$ has one perpendicular from $O$ that bisects both chords, so $A C = B D$ .

Neat triples. $(5, 12, 13)$ , $(8, 15, 17)$ , $(7, 24, 25)$ , $(3, 4, 5)$ and their multiples turn up as $(d, l, r)$ . Recognise them, but still show $r^{2} = d^{2} + l^{2}$ in your working.

Level. In school papers expect a 3 mark length question; the harder items here (parallel chords with unknown radius, concentric circles) go beyond routine and are labelled advanced or HOTS.

Chord Length and Distance from the Centre: formulas to remember

Working triangle: $r^{2} = d^{2} + l^{2}$; Chord length is 2l; d is the distance from the centre.
Chord length: $A B = 2 r^{2} - d^{2}$; From the same triangle.
Distance of a chord: $d = r^{2} - l^{2}$; l is half the chord.
Parallel chords, same side: $gap = d_{1} - d_{2}$; The nearer chord is longer.
Parallel chords, opposite sides: $gap = d_{1} + d_{2}$; Centre lies between them.

How to solve Chord Length and Distance from the Centre questions

Always draw the perpendicular from the centre and mark half the chord.

1Draw $O M ⊥ A B$ and mark $A M = M B$ (half the chord).
2Write $r^{2} = d^{2} + l^{2}$ and substitute the two known lengths to find the third.
3Solve; remember the chord is $2 l$ , not $l$ .
4For two chords, form one triangle for each and use the same $r$ .
5For parallel chords, decide same side or opposite sides from the figure and add or subtract the distances.
6State units and give the final answer to what was asked (length, distance or radius).

Chord Length and Distance from the Centre 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.

Chord of a circular pond

Suppose a circular pond of radius 10 m has a straight wooden walkway across it, 6 m from the centre.

Find the length of the walkway.

Try it first, then show the solutionSolution

Half the walkway: $l = 1 0^{2} - 6^{2} = 64 = 8$ m.
Length $= 2 \times 8 = 16$ m.

Answer: 16 m.

Memory hook: Perpendicular from the centre halves the chord.

Slicing a round cheese

Imagine a circular cheese of radius 26 cm cut by a straight wire; the cut is 24 cm from the centre.

How wide is the cut?

Try it first, then show the solutionSolution

$l = 2 6^{2} - 2 4^{2} = 676 - 576 = 100 = 10$ cm.
Width $= 20$ cm.

Answer: 20 cm.

Memory hook: A (10, 24, 26) triangle: a multiple of 5-12-13.

Tricks to remember Chord Length and Distance from the Centre

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

Half first
The right triangle uses half the chord; double at the end.
Use it when: Avoiding the commonest slip.
Near is long
Chords closer to the centre are longer; the diameter is longest.
Use it when: Checking which chord is bigger.
Same side subtract
Same side: subtract distances. Opposite sides: add.
Use it when: Parallel chord questions.

Chord Length and Distance from the Centre: solved examples

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

Example 1: Chord from radius and distance · 1 mark · MCQBoard level

A chord lies $21$ cm from the centre of a circle of radius $29$ cm. Its length is:
(a) 20 cm
(b) 40 cm
(c) 42 cm
(d) 58 cm

Try it first, then show the solutionSolution

Half-chord $= 2 9^{2} - 2 1^{2} = 400 = 20$ cm.
Chord $= 40$ cm.

Answer: (b) 40 cm

Example 2: Radius from a chord · 3 marks · Short answerBoard level

A chord of length $16$ cm is at a distance of $6$ cm from the centre. Find the radius.

Try it first, then show the solutionSolution

$l = 8$ , $d = 6$ .
$r^{2} = 6^{2} + 8^{2} = 100$ , so $r = 10$ cm.

Answer: 10 cm

Example 3: Parallel chords on opposite and same sides · 4 marks · Short answerHarder board level

In a circle of radius $13$ cm two parallel chords have lengths $24$ cm and $10$ cm. Find the distance between them if they are (i) on the same side of the centre, (ii) on opposite sides.

Try it first, then show the solutionSolution

For the 24 cm chord: $d_{1} = 1 3^{2} - 1 2^{2} = 5$ cm.
For the 10 cm chord: $d_{2} = 1 3^{2} - 5^{2} = 12$ cm.
(i) Same side: $12 - 5 = 7$ cm.
(ii) Opposite sides: $12 + 5 = 17$ cm.

Answer: (i) 7 cm (ii) 17 cm

Example 4: Two chords, one circle · 4 marks · Short answerHarder board level

A chord of length $30$ cm is drawn in a circle of radius $17$ cm. Find its distance from the centre. A second chord is $15$ cm from the centre; find its length and say which chord is longer.

Try it first, then show the solutionSolution

First chord: $d = 1 7^{2} - 1 5^{2} = 64 = 8$ cm.
Second chord: $l = 1 7^{2} - 1 5^{2} = 8$ , so its length is $16$ cm.
The 30 cm chord (at 8 cm) is longer, as it is nearer the centre.

Answer: Distance 8 cm; second chord 16 cm; the 30 cm chord is longer.

Example 5: Concentric circles · 5 marks · Long answerHOTS / competency

Two concentric circles have radii $17$ cm and $10$ cm. A line at a distance of $8$ cm from the centre meets the larger circle at $A$ , $B$ and the smaller at $C$ , $D$ (in the order $A, C, D, B$ ). Find $A B$ , $C D$ and $A C$ , and prove $A C = B D$ .

Try it first, then show the solutionSolution

$A B = 2 1 7^{2} - 8^{2} = 2 \times 15 = 30$ cm.
$C D = 2 1 0^{2} - 8^{2} = 2 \times 6 = 12$ cm.
The foot $M$ of the perpendicular from the centre is the mid-point of both $A B$ and $C D$ , so $A M = M B = 15$ and $C M = M D = 6$ .
$A C = A M - C M = 15 - 6 = 9$ cm and $B D = M B - M D = 15 - 6 = 9$ cm, so $A C = B D$ .

Answer: $A B = 30$ cm, $C D = 12$ cm, $A C = B D = 9$ cm.

Where students lose marks in Chord Length and Distance from the Centre

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

Mistake: Using the whole chord in the triangle
The right triangle has half the chord as a leg.
Wrong working
$r^{2} = 5^{2} + 2 4^{2}$ for a chord of 24 cm and $d = 5$ .
Right working
$r^{2} = 5^{2} + 1 2^{2} = 169$ , $r = 13$ .
Do this: Write "half-chord =" before substituting.
Usually costs: Whole answer.
Mistake: Forgetting to double at the end
You find $l$ , but the chord is $2 l$ .
Wrong working
Chord $= r^{2} - d^{2} = 12$ .
Right working
Chord $= 2 \times 12 = 24$ .
Do this: End with "length of chord = 2l".
Usually costs: Final answer.
Mistake: Adding when the chords are on the same side
The gap is the difference when both chords are on one side of the centre.
Wrong working
Gap $= 6 + 8 = 14$ for chords on the same side.
Right working
Same side: $8 - 6 = 2$ .
Do this: Sketch the centre and both chords.
Usually costs: Final answer.

Exam tips

Always begin by drawing the perpendicular from O; it earns a mark in a proof and sets up every length question.
If a question gives a gap between parallel chords and no radius, set up two equations in the same r.
Check the final radius by testing both chords: both triangles must give the same hypotenuse.
Parallel-chord and concentric-circle questions are standard ICSE Class 9 items. Practise them, because they often carry 4 to 5 marks.

Practice questions on Chord Length and Distance from the Centre

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
A chord of length $10$ cm lies in a circle of radius $13$ cm. Its distance from the centre is:
(a) 5 cm
(b) 8 cm
(c) 12 cm
(d) 24 cm
Hint
$l = 5$ .
Answer
(c). $d = 169 - 25 = 12$ cm.
Q2 · 3 marks · Short answerHarder board level
A chord $24$ cm long is drawn in a circle of radius $15$ cm. Find its distance from the centre. At what distance would a chord of length $18$ cm lie?
Hint
Two right triangles with the same hypotenuse.
Answer
$d = 225 - 144 = 9$ cm. For 18 cm: $l = 9$ , $d = 225 - 81 = 12$ cm.
Q3 · 4 marks · Short answerHarder board level
Two parallel chords of lengths $48$ cm and $40$ cm lie on the same side of the centre and are $8$ cm apart. Find the radius.
Hint
Let the distances be $d$ and $d + 8$ ; $d^{2} + 2 4^{2} = (d + 8)^{2} + 2 0^{2}$ .
Answer
$d^{2} + 576 = d^{2} + 16 d + 64 + 400$ , so $16 d = 112$ , $d = 7$ . $r^{2} = 49 + 576 = 625$ , $r = 25$ cm (check: $1 5^{2} + 2 0^{2} = 625$ ).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The diameter is the longest chord of a circle.
Reason (R): For a diameter, the distance from the centre is $0$ , so $2 l = 2 r^{2} - 0 = 2 r$ .
(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
Chord length decreases as d increases.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A circular pond has radius $20$ m. A bridge is built along a chord $32$ m long. A second bridge is built parallel to it on the other side of the centre, $16$ m from the centre.
(a) How far is the first bridge from the centre?
(b) How long is the second bridge?
(c) How far apart are the two bridges?
Hint
Opposite sides: add the distances.
Answer
(a) $400 - 256 = 12$ m (b) $2 400 - 256 = 24$ m (c) $12 + 16 = 28$ m.

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.State the working relation for a chord of length 2l at distance d in a circle of radius r.Show answer
r² = d² + l².
3.Which chord is longest?Show answer
The diameter.
4.Two parallel chords are on opposite sides of the centre. How do you find the gap?Show answer
Add their distances from the centre.

Chord Length and Distance from the Centre: frequently asked questions

Do I have to prove that the perpendicular from the centre bisects the chord?

You prove that a line from the centre to the mid-point is perpendicular to the chord; the converse is used without proof.

Why is the chord 2l and not l?

l is the distance from the foot of the perpendicular to one end; the foot is the mid-point.

Can the same circle have two chords of equal length at different distances?

No; equal chords are at equal distances from the centre (see the next topic).