Question

Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Answer 1

Let O be the center of the concentric circles and let AB be a chord of the outer circle, touching the inner circle at P. Join OA and OP.

Now, the radius through the point of contact is perpendicular to the tangent.

OP $\perp$ AB.

Since, the perpendicular from the center to a chord, bisects the chord, AP = PB. Now, in right $△$OPA, we have OA = 13 cm and OP = 5 cm.

${OP}^2$ + ${AP}^2$ = ${OA}^2$ $\Rightarrow$ ${AP}^2$ = ${OA}^2$ – ${OP}^2$ = ( ${13}^2$ – $5^2$) = (169 – 25) = 144.

$\Rightarrow$ AP =$\sqrt{144}$ = 12 cm.

AB = 2AP = (2 $\times$ 12) cm = 24 cm.

Hence, the length of chord AB = 24 cm.