Two concentric circles are of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle.

The correct answer is 16 cm.

We are given two concentric circles (circles sharing the same centre) with radii:

Radius of larger circle, R = 10 cm
Radius of smaller circle, r = 6 cm

We need to find the length of a chord of the larger circle that is tangent to (touches) the smaller circle.

Step 1 — Set up the diagram.

Let O be the common centre of both circles. Let AB be the chord of the larger circle that touches the smaller circle at point P.

Step 2 — Identify the key relationship.

Since the chord AB is a tangent to the smaller circle, the radius of the smaller circle drawn to the point of tangency is perpendicular to the chord. Therefore:

OP ⊥ AB, which means OP = 6 cm (radius of smaller circle) and the angle OPB = 90°.

Step 3 — Use the perpendicular bisector property.

A key theorem states: The perpendicular from the centre of a circle to a chord bisects the chord.

Since OP ⊥ AB, the point P is the midpoint of chord AB. Therefore:

AP = PB

Step 4 — Apply the Pythagorean Theorem in triangle OPB.

In right-angled triangle OPB:

OB = R = 10 cm (radius of larger circle)
OP = r = 6 cm (radius of smaller circle)
Angle OPB = 90°

By the Pythagorean theorem:

{\mathrm{OB}}^2={\mathrm{OP}}^2+{\mathrm{PB}}^2

{10}^2=6^2+{\mathrm{PB}}^2

100=36+{\mathrm{PB}}^2

{\mathrm{PB}}^2=100-36=64

\mathrm{PB}=\sqrt{64}=8\text{cm}

Step 5 — Find the full length of chord AB.

Since P is the midpoint of AB:

\mathrm{AB}=2\times \mathrm{PB}=2\times 8=16\text{cm}

Therefore, the length of the chord of the larger circle which touches the smaller circle is 16 cm.