Question 272 marksVery Short AnswerSection B

  1. Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Circles
Chord of larger circle touching smaller concentric circle

Official answer

The chord of the larger circle is 6 cm long. Half the chord (3 cm), the smaller radius (4 cm), and the larger radius (5 cm) form a right triangle, so the chord is twice the half-length found via Pythagoras.

Answer keywords:concentric circlestangent perpendicular to radiuschord bisected by perpendicular from centrePythagorashalf chord 3 cmchord 6 cmradius 5radius 4

Marking scheme

  1. 1.1 mark: recognising that the radius to the point of contact (4 cm) is perpendicular to the chord and bisects it, and setting up 52=42+AP25^2 = 4^2 + AP^2.
  2. 2.1 mark: solving AP=3AP = 3 cm and doubling to get the chord AB=6AB = 6 cm.
  3. 3.Award 0.5 mark for a correct labelled figure; deduct 0.5 mark if the student stops at AP=3AP = 3 cm without doubling.

Hint

The chord is a tangent to the inner circle, so the inner radius (4 cm) is perpendicular to it and bisects it; use Pythagoras with hypotenuse 5 cm, then double the half-chord.

Quick oral answer

The chord touches the inner circle, so the 4 cm radius meets it at 90∘90^\circ and bisects it; by Pythagoras half the chord is 52−42=3\sqrt{5^2 - 4^2} = 3 cm, so the whole chord is 6 cm.

Common mistakes

  1. 1.Reporting the answer as 3 cm (the half-chord AP) and forgetting to double it to get the full chord AB=6AB = 6 cm.
  2. 2.Treating 5 cm and 4 cm as the two perpendicular sides and computing 25+16=41\sqrt{25+16} = \sqrt{41}, instead of realising the 5 cm radius is the hypotenuse of the right triangle.
  3. 3.Not drawing/labelling the perpendicular from the centre, then failing to justify why the chord is bisected at the point of tangency.

Analysis and explanation

Combine two circle theorems — tangent ⊥\perp radius, and perpendicular from centre bisects a chord — to find the chord length.


Concept

  • The chord of the larger circle touching the smaller circle meets the common radius at 90∘90^\circ at the point of contact.
  • That perpendicular equals the smaller radius (4 cm) and bisects the chord, forming a right triangle with the larger radius (5 cm) as hypotenuse.
  • Pythagoras gives half-chord = 3 cm, so full chord = 6 cm.

Common mistakes

  • Reporting 3 cm and forgetting to double it for the full chord.
  • Using 5 and 4 as the two legs (giving 41\sqrt{41}) instead of recognising 5 cm is the hypotenuse.

Real-world

  • This is exactly how one finds the width of a straight road that just grazes the inner edge of a circular park.

Previously asked

2016Section B2 marksQ12

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Interesting facts

Concentric circles share a centre but have different radii — the region between them is called an annulus, the same shape as a washer or a CD/DVD.

The result generalises neatly: for concentric circles of radii R and r, the length of a chord of the larger that touches the smaller is always 2R2−r22\sqrt{R^2 - r^2} — here 225−16=62\sqrt{25 - 16} = 6 cm.

Question details

Q27 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Circles
Topic
Chord of larger circle touching smaller concentric circle
Bloom's level
apply
Syllabus topic
Unit 4: Geometry — Circles: Tangent to a circle, tangent perpendicular to radius at point of contact, concentric circles (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the inner radius perpendicular to the chord?

The chord of the larger circle is a tangent to the smaller circle. By the tangent-radius theorem, a tangent is perpendicular to the radius drawn to the point of contact, so the 4 cm radius meets the chord at 90∘90^\circ.

Why do we multiply the 3 cm by 2?

The perpendicular from the centre bisects the chord, so it only gives half the chord (AP=3AP = 3 cm). The full chord AB is twice this, i.e. 6 cm.

Is there a quick formula?

Yes: chord=2R2−r2\text{chord} = 2\sqrt{R^2 - r^2} where R is the larger radius and r the smaller. Here 252−42=29=62\sqrt{5^2 - 4^2} = 2\sqrt{9} = 6 cm.