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.

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

Marking Scheme

11 mark: recognising that the radius to the point of contact (4 cm) is perpendicular to the chord and bisects it, and setting up $5^2 = 4^2 + AP^2$ .
21 mark: solving $AP = 3$ cm and doubling to get the chord $AB = 6$ cm.
3Award 0.5 mark for a correct labelled figure; deduct 0.5 mark if the student stops at $AP = 3$ 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 $90^\circ$ and bisects it; by Pythagoras half the chord is $\sqrt{5^2 - 4^2} = 3$ cm, so the whole chord is 6 cm.

Analysis & 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 $90^\circ$ 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 $\sqrt{41}$ ) 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.

Common Mistakes

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

Previously Asked

2016Section BQ122 marks

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 $2\sqrt{R^2 - r^2}$ — here $2\sqrt{25 - 16} = 6$ cm.

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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 $90^\circ$ .

Why do we multiply the 3 cm by 2?

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

Is there a quick formula?

Yes: $\text{chord} = 2\sqrt{R^2 - r^2}$ where R is the larger radius and r the smaller. Here $2\sqrt{5^2 - 4^2} = 2\sqrt{9} = 6$ cm.