Question

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]

Answer 1

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.

Join OA and OC. Then,

OC = $\frac{18}{2}$ cm = 9 cm and OA = $\frac{30}{2}$ cm = 15 cm

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Also, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ OC ⊥ AB and C is the mid-point of AB.

In right ∆OCA,

$$\begin{gathered} {\mathrm{OA}}^2={\mathrm{OC}}^2+{\mathrm{AC}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{OA}}^2-{\mathrm{OC}}^2 \\ \Rightarrow \mathrm{AC}=\sqrt{{15}^2-9^2} \\ \Rightarrow \mathrm{AC}=\sqrt{225-81}=\sqrt{144}=12\mathrm{cm} \end{gathered}$$

∴ AB = 2AC = 2 × 12 cm = 24 cm

Thus, the required length of the chord is 24 cm.