Question

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to


(a) $5\sqrt{2}\mathrm{cm}$
(b) $3\sqrt{5}\mathrm{cm}$
(c) $4\sqrt{10}\mathrm{cm}$
(d) $5\sqrt{10}\mathrm{cm}$

Answer 1

(c) $4\sqrt{10}\mathrm{cm}$

$$\begin{gathered} \mathrm{Given},OP=5\mathrm{}\mathrm{cm},PA=12\mathrm{cm} \\ \mathrm{Now},\text{j}\mathrm{oin}O\text{and}B. \\ \mathrm{Then},OB=3\mathrm{cm}. \\ \mathrm{Now},\angle OAP={90}^0\left(\text{T}\mathrm{angents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular} \\ \mathrm{}\mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}\right)\mathrm{} \\ \mathrm{Now},\text{in}∆OAP: \\ O{P}^2=O{A}^2+P{A}^2 \\ \Rightarrow O{P}^2=5^2+{12}^2 \\ \Rightarrow O{P}^2=25+144 \\ \Rightarrow O{P}^2=169 \\ \Rightarrow OP=\sqrt{169} \\ \Rightarrow OP=13 \\ \mathrm{Now},\mathrm{}\text{in}∆OBP: \\ P{B}^2=O{P}^2-O{B}^2 \\ \Rightarrow P{B}^2={13}^2-3^2 \\ \Rightarrow P{B}^2=169-9 \\ \Rightarrow P{B}^2=160 \\ \Rightarrow PB=\sqrt{160} \\ \Rightarrow PB=4\sqrt{10}\mathrm{cm} \end{gathered}$$