MathsCBSE Class 9 · Circles

Chord Theorems and Distance from the Centre: Class 9 Circles

Q: Is the foot of the perpendicular from the centre always the midpoint of the chord?

Yes, for every chord, including a diameter (where the distance is zero and the centre itself is the midpoint).

Q: Can two different chords be equal but at different distances?

No. Equal chords are always at equal distances from the centre, and chords at equal distances are equal.

Q: How do I find the centre of a circle from a drawing?

Draw two chords, construct the perpendicular bisector of each, and mark where the bisectors meet. That point is the centre.

Updated 17 October 2026 · Fact-checked

A chord and the centre of a circle form an isosceles triangle, and the perpendicular from the centre to a chord bisects it. So every chord sum reduces to one right triangle with radius $r$ as hypotenuse, half the chord $\frac{c}{2}$ and the distance $d$ from the centre: $r^2 = d^2 + \left(\frac{c}{2}\right)^2$. Equal chords are equally far from the centre and subtend equal angles at it, and a longer chord lies nearer the centre.

Circle with centre M, two chords AB and CD, and dashed perpendiculars ME and MG from the centre to each chord. — Notice the dashed perpendiculars from the centre M to each chord. Equal chords lie at equal distances from the centre.
Picture: Illuyanka · CC BY-SA 4.0 · Wikimedia Commons

Understand Chord Theorems and Distance from the Centre

Take a circle with centre $O$ and radius $r$ , and a chord $A B$ . Because $O A = O B = r$ , triangle $O A B$ is isosceles. Everything on this page comes from that one fact together with congruent triangles.

Perpendicular from the centre. If $O M ⊥ A B$ with $M$ on $A B$ , then $△ O M A ≅ △ O M B$ by RHS (right angle at $M$ , common side $O M$ , hypotenuses $O A = O B$ ). So $A M = M B$ : the perpendicular from the centre bisects the chord. The converse holds too: the line joining the centre to the midpoint of a chord is perpendicular to the chord. A further consequence is that the perpendicular bisector of any chord passes through the centre, which is how you locate the centre of a circle drawn on paper (draw two chords, bisect them perpendicularly, and see where the bisectors meet).

Equal chords. If $A B = C D$ then they subtend equal angles at the centre (SSS on $△ A O B$ and $△ C O D$ ), and the converse is true by SAS. Equal chords are also equidistant from the centre: with $O M ⊥ A B$ and $O N ⊥ C D$ , $A M = C N$ (halves of equal chords) and $O A = O C$ , so $O M = O N$ from the right triangles. Conversely, chords at equal distances from the centre are equal.

Longer means nearer. Since $d^{2} = r^{2} - (\frac{c}{2})^{2}$ , a larger $c$ gives a smaller $d$ . The diameter has $d = 0$ and is the longest chord, of length $2 r$ . A chord at distance $d \geq r$ from the centre cannot exist as a chord: $d = r$ gives a tangent line, which touches the circle at a single point.

Parallel chords. Two parallel chords have the same perpendicular through $O$ , so their distances from the centre lie along one line. If they are on opposite sides of the centre the gap between them is $d_{1} + d_{2}$ ; on the same side it is $∣ d_{1} - d_{2} ∣$ . Whenever a problem says "find the distance between parallel chords", check both positions, because the problem often does not tell you which one applies.

Typical traps. Always halve the chord before using Pythagoras. Never use the chord itself as a side of the right triangle with $r$ . And in a figure with several chords, draw one perpendicular from $O$ to each chord and label the half-lengths first; the algebra then takes care of itself.

Chord Theorems and Distance from the Centre: formulas to remember

Chord, radius and distance: $r^{2} = d^{2} + (\frac{c}{2})^{2}$; $c$ = chord length, $d$ = distance of the chord from the centre.
Chord length: $c = 2 r^{2} - d^{2}$; Longest when $d = 0$ (diameter, $2 r$ ).
Distance from the centre: $d = r^{2} - (\frac{c}{2})^{2}$; Non-negative and less than $r$ ( $d = 0$ for a diameter).
Equal chords: $A B = C D ⟺ ∠ A O B = ∠ C O D ⟺ O M = O N$; Equal chords: equal central angles and equal distances.
Parallel chords: $gap = d_{1} + d_{2} or ∣ d_{1} - d_{2} ∣$; Opposite sides of the centre or the same side.
Base angles at the centre: $∠ O A B = ∠ O B A = 9 0^{\circ} - \frac{1}{2} ∠ A O B$; From the isosceles triangle $O A B$ .

How to solve Chord Theorems and Distance from the Centre questions

Almost every chord question is one right triangle. Build it first, then calculate.

1Mark the centre $O$ and drop a perpendicular $O M$ to each chord.
2Write the half-chord $A M = \frac{c}{2}$ on the figure; the perpendicular bisects the chord.
3Write $r^{2} = d^{2} + (\frac{c}{2})^{2}$ in the right triangle $O M A$ and substitute the known values.
4Solve for the unknown (take the positive root) and re-double the half-chord if the chord length is asked.
5For two parallel chords, compute each distance separately and combine them as sum or difference; give both cases unless the figure fixes one.
6For proofs, name the congruent triangles (RHS, SAS or SSS) and state the corresponding parts equal.

Chord Theorems 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.

Distance of a chord from the centre

Suppose a round pizza of radius 15 cm is cut by one straight cut that passes 9 cm away from the middle of the pizza.

Find the length of the cut across the pizza, and say whether a cut 12 cm from the middle would be longer or shorter.

Try it first, then show the solutionSolution

The cut is a chord at distance $d = 9$ cm, with $r = 15$ cm.
Half the chord $= 1 5^{2} - 9^{2} = 225 - 81 = 144 = 12$ cm.
The full chord is $2 \times 12 = 24$ cm.
At $d = 12$ : half $= 225 - 144 = 81 = 9$ , so the chord is $18$ cm: shorter, because it is farther from the middle.

Answer: The cut is 24 cm long; a cut 12 cm from the middle would be 18 cm long.

Memory hook: The farther from the middle, the shorter the cut.

Tricks to remember Chord Theorems and Distance from the Centre

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

Half, square, subtract
Half the chord, square it, subtract from $r^{2}$ , take the root: that is $d$ . Reverse the order to get the chord.
Use it when: Any chord-and-distance numerical.
Perpendicular means bisect
A line from the centre that is square to a chord cuts it in two equal pieces. Say it as one phrase when writing a proof.
Use it when: Proof questions needing $A M = M B$ .
Fat chord, short distance
Chord length and distance from the centre move in opposite directions.
Use it when: Comparing two chords without calculating.

Chord Theorems and Distance from the Centre: solved examples

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

Example 1: Round table cut-out · 3 marks · Short answerBoard level

A circular dining table has radius 65 cm. A straight edge is cut from it so that the edge lies 25 cm from the middle of the table. How long is the straight edge?

Try it first, then show the solutionSolution

The straight edge is a chord; let $O M ⊥$ chord, so $O M = 25$ cm and $O A = 65$ cm. (1)
$A M^{2} = O A^{2} - O M^{2} = 6 5^{2} - 2 5^{2} = 4225 - 625 = 3600$ , so $A M = 60$ cm. (1)
The perpendicular bisects the chord, so the edge is $2 \times 60 = 120$ cm long. (1)

Answer: The straight edge is 120 cm long.

Example 2: Chords at equal distances · 3 marks · Short answerBoard level

Prove that two chords of a circle that are at equal distances from the centre are equal in length.

Try it first, then show the solutionSolution

Let $O M ⊥ A B$ and $O N ⊥ C D$ with $O M = O N$ . (½)
In right triangles $O M A$ and $O N C$ : $O M = O N$ , hypotenuse $O A = O C$ (radii), right angles at $M$ and $N$ ; so $△ O M A ≅ △ O N C$ by RHS. (1½)
By CPCT $A M = C N$ . The perpendicular from the centre bisects each chord, so $A B = 2 A M = 2 C N = C D$ . (1)

Answer: Chords equidistant from the centre are equal: $A B = C D$ .

Example 3: Length and distance related · 5 marks · Long answerHarder board level

In a circle of radius 13 cm, the length of a chord exceeds its distance from the centre by 19 cm. Find the length of the chord.

Try it first, then show the solutionSolution

Let the distance be $d$ cm, so the chord is $(d + 19)$ cm and half the chord is $\frac{d + 19}{2}$ . (1)
Right triangle: $d^{2} + (\frac{d + 19}{2})^{2} = 1 3^{2}$ . (1)
Multiply by 4: $4 d^{2} + d^{2} + 38 d + 361 = 676$ , so $5 d^{2} + 38 d - 315 = 0$ . (1)
Split the middle term (product $5 \times (- 315) = - 1575$ , factors $63$ and $- 25$ ): $5 d^{2} + 63 d - 25 d - 315 = d (5 d + 63) - 5 (5 d + 63) = (5 d + 63) (d - 5) = 0$ . (1)
Distance cannot be negative, so $d = 5$ (the root $- \frac{63}{5}$ is rejected); chord $= 5 + 19 = 24$ cm. (½)
Check: half $= 12$ , $5^{2} + 1 2^{2} = 169 = 1 3^{2}$ . (½)

Answer: The chord is 24 cm long (its distance from the centre is 5 cm).

Example 4: Common chord, one degenerate case · 3 marks · Short answerHarder board level

Two circles of radii 5 cm and 3 cm intersect at $A$ and $B$ , and $A B = 6$ cm. Find the distance between their centres.

Try it first, then show the solutionSolution

$A B$ is a common chord; the line joining the centres is the perpendicular bisector of $A B$ and meets it at its midpoint $M$ with $A M = 3$ . (½)
Larger circle (centre $C_{1}$ ): $C_{1} M = 5^{2} - 3^{2} = 4$ cm. (1)
Smaller circle (centre $C_{2}$ ): $C_{2} M = 3^{2} - 3^{2} = 0$ , so $C_{2} = M$ and $A B$ is a diameter of the smaller circle. (1)
So $C_{1} C_{2} = 4 + 0 = 4$ cm, whichever side $C_{1}$ lies on. (½)

Answer: The centres are 4 cm apart.

Example 5: Three points, one circle · 4 marks · Case-basedHOTS / competency

Case: three paths meet a park at the points $A (0, 0)$ , $B (8, 0)$ and $C (0, 6)$ (coordinates in metres). A round fountain is to be built so that its boundary passes through all three points. (i) Why does exactly one such circle exist? (ii) Find its centre using chord bisectors. (iii) Find its radius. (iv) Is a lamp at $D (8, 6)$ on the same circle?

Try it first, then show the solutionSolution

(i) $A$ , $B$ , $C$ are not on one line, and exactly one circle passes through three non-collinear points. (1)
(ii) The centre lies on the perpendicular bisector of chord $A B$ , that is $x = 4$ , and on that of chord $A C$ , that is $y = 3$ . So the centre is $(4, 3)$ . (1)
(iii) Radius $=$ distance to $A$ $= 4^{2} + 3^{2} = 5$ m; check $B$ : $4^{2} + 3^{2} = 5$ and $C$ : $4^{2} + 3^{2} = 5$ . (1)
(iv) Distance from $(4, 3)$ to $D (8, 6)$ is $16 + 9 = 5$ m, equal to the radius, so $D$ is on the circle. (1)

Answer: (i) Three non-collinear points fix one circle; (ii) centre $(4, 3)$ ; (iii) radius 5 m; (iv) yes, $D$ lies on it.

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

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

Mistake: Using the whole chord in Pythagoras
The right triangle uses the half-chord, but students write the full length.
Wrong working
$d = 1 0^{2} - 1 6^{2}$ for a 16 cm chord in a circle of radius 10 cm.
Right working
$d = 1 0^{2} - 8^{2} = 6$ cm.
Do this: Write "half chord = ..." as the first line.
Usually costs: Usually 2 of 3 marks, and the root of a negative number is a red flag.
Mistake: Giving only one case for parallel chords
Assuming both chords are on opposite sides of the centre.
Wrong working
Distance $= 5 + 12 = 17$ cm only.
Right working
$17$ cm (opposite sides) or $7$ cm (same side).
Do this: Always write both cases unless the question fixes the side.
Usually costs: 1 to 2 marks.
Mistake: Proof without naming the congruence rule
Writing "triangles are congruent" with no reason.
Wrong working
$△ O M A ≅ △ O N C$ , so $O M = O N$ .
Right working
$△ O M A ≅ △ O N C$ (RHS: right angles, $O A = O C$ , $A M = C N$ ), so $O M = O N$ (CPCT).
Do this: List three equalities and the rule before using CPCT.
Usually costs: 1 mark in 3-mark proofs.

Exam tips

Draw the perpendicular from the centre first; it turns every chord problem into a right triangle.
In proofs, state "perpendicular from the centre bisects the chord" as a reason; it earns a mark of its own.
For two chords, find both distances before combining them.
Check that $d < r$ and that the half-chord is at most $r$ ; if not, a number is wrong.

Practice questions on Chord Theorems 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
Two chords of the same circle are at distances 3 cm and 5 cm from the centre. Which statement is correct?
(a) The chord at 5 cm is longer
(b) The chord at 3 cm is longer
(c) The chords are equal
(d) The comparison depends on the radius
Hint
A chord is longer when it is nearer to the centre.
Answer
(b). Chord $= 2 r^{2} - d^{2}$ decreases as $d$ increases, for any radius, so the chord at 3 cm is longer.
Q2 · 3 marks · Short answerBoard level
$A B$ and $C D$ are equal chords of a circle with centre $O$ , and $∠ A O B = 6 4^{\circ}$ . Find $∠ C O D$ and $∠ O A B$ , giving a reason for each.
Hint
Equal chords subtend equal angles at the centre; $△ O A B$ is isosceles.
Answer
$∠ C O D = 6 4^{\circ}$ because equal chords subtend equal angles at the centre (SSS). In isosceles $△ O A B$ , $∠ O A B = \frac{18 0 ^{\circ} - 6 4 ^{\circ}}{2} = 5 8^{\circ}$ .
Q3 · 3 marks · Short answerHarder board level
A chord $A B$ of a circle of radius 12 cm subtends a right angle at the centre. Find the length of $A B$ and its distance from the centre, leaving surds.
Hint
$△ O A B$ is a right isosceles triangle; the foot of the perpendicular is the midpoint of $A B$ .
Answer
$A B^{2} = 1 2^{2} + 1 2^{2} = 288$ , so $A B = 122$ cm. Half of $A B$ is $62$ , and the distance from $O$ is $144 - 72 = 72 = 62$ cm (the median to the hypotenuse of a right isosceles triangle equals half the hypotenuse).
Q4 · 5 marks · Long answerHarder board level
In a circle of radius 17 cm, two parallel chords have lengths 30 cm and 16 cm. Find the distance between them in each possible position, and say which positions give a valid figure.
Hint
Find each distance from the centre; both positions of the chords are valid.
Answer
Chord 30: half 15, $d_{1} = 289 - 225 = 8$ cm. Chord 16: half 8, $d_{2} = 289 - 64 = 15$ cm. Opposite sides of the centre: $8 + 15 = 23$ cm. Same side: $15 - 8 = 7$ cm. Both positions are valid (each chord fits inside the circle), so the distance is 23 cm or 7 cm.
Q5 · 4 marks · Case-basedHOTS / competency
Case: two circular ponds of radii 13 m and 15 m overlap. Their boundaries meet at $A$ and $B$ with $A B = 24$ m. (i) Find the distance from each centre to $A B$ . (ii) Find the distance between the centres if they lie on opposite sides of $A B$ . (iii) Find it if they lie on the same side of $A B$ . (iv) Why must the line joining the centres be perpendicular to $A B$ ?
Hint
The common chord is bisected by the line of centres; use half of 24 for both circles.
Answer
(i) Half of $A B$ is 12. For radius 13: $169 - 144 = 5$ m. For radius 15: $225 - 144 = 9$ m. (ii) Opposite sides: $5 + 9 = 14$ m. (iii) Same side: $9 - 5 = 4$ m. (iv) Each centre lies on the perpendicular bisector of the chord $A B$ (it is equidistant from $A$ and $B$ ), so both lie on one line, which is that perpendicular bisector.

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.Formula linking $r$ , $d$ and chord $c$ ?Show answer
$r^{2} = d^{2} + (\frac{c}{2})^{2}$ .
3.Which chord is the longest?Show answer
The diameter, of length $2 r$ .
4.How are equal chords placed relative to the centre?Show answer
Equidistant from it.
5.Number of circles through three non-collinear points?Show answer
Exactly one.

Chord Theorems and Distance from the Centre: frequently asked questions

Is the foot of the perpendicular from the centre always the midpoint of the chord?

Yes, for every chord, including a diameter (where the distance is zero and the centre itself is the midpoint).

Can two different chords be equal but at different distances?

No. Equal chords are always at equal distances from the centre, and chords at equal distances are equal.

How do I find the centre of a circle from a drawing?

Draw two chords, construct the perpendicular bisector of each, and mark where the bisectors meet. That point is the centre.