Question

In the given figure points P,B and Q are points of contact of the respective tangents.line QA is parallel to line PC.If QA=7.2 cm , PC = 5 cm, Find the radius of the circle.

Answer 1

Construction :- Join OA and OC

$$\begin{gathered} \mathrm{In}△\mathrm{AQO}\mathrm{and}∆\mathrm{ABO} \\ \mathrm{AQ}=\mathrm{AB}(\mathrm{tangents}\mathrm{are}\mathrm{equal}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{point}) \\ \mathrm{OQ}=\mathrm{OB}\left(\mathrm{radius}\right) \\ \mathrm{AO}=\mathrm{AO}\left(\mathrm{common}\right) \\ \mathrm{By}\mathrm{SSS}\mathrm{congruency}△\mathrm{AQO}\cong ∆\mathrm{ABO} \\ \angle \mathrm{QOA}=\angle \mathrm{AOB}(\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{So},\angle \mathrm{AOB}=\frac{1}{2}\angle \mathrm{BOQ}.................\left(1\right) \\ \mathrm{Similarly},\angle \mathrm{POC}=\angle \mathrm{BOC} \\ \mathrm{So},\angle \mathrm{BOC}=\frac{1}{2}\angle \mathrm{BOP}................\left(2\right) \\ \mathrm{Adding}\mathrm{equation}\left(1\right)\mathrm{and}\left(2\right) \\ \mathrm{So},\angle \mathrm{AOB}+\angle \mathrm{BOC}=\frac{1}{2}\angle \mathrm{BOQ}+\frac{1}{2}\angle \mathrm{BOP} \\ \Rightarrow \angle \mathrm{AOC}=\frac{1}{2}\left(\angle \mathrm{BOQ}+\angle \mathrm{BOP}\right)=\frac{1}{2}\times 180°=90° \\ \mathrm{So},△\mathrm{AOC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{triangle}. \\ {\mathrm{AO}}^2+{\mathrm{OC}}^2={\mathrm{AC}}^2\left(\mathrm{By}\mathrm{Pythagoras}\mathrm{theorem}\right)...\left(3\right) \\ \mathrm{Since}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{radius}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{tangent}. \\ \mathrm{Hence},△\mathrm{AQO}\mathrm{and}△\mathrm{OPC}\mathrm{are}\mathrm{right}\mathrm{triangle}. \\ \Rightarrow {\mathrm{OQ}}^2+{\mathrm{AQ}}^2+{\mathrm{OP}}^2+{\mathrm{PC}}^2={\mathrm{OA}}^2+{\mathrm{OC}}^2 \\ \Rightarrow {\mathrm{OQ}}^2+{\mathrm{AQ}}^2+{\mathrm{OP}}^2+{\mathrm{PC}}^2={\mathrm{AC}}^2\left(\mathrm{From}\mathrm{eq}3\right) \end{gathered}$$

We know that the tangents from the external point are equal.
So, AQ = AB = 7.2 cm
CP = CB = 5 cm
AC = AB + BC = 7.2+5 = 12.2 cm

Let OQ = x cm
Then ,
$$\begin{gathered} x^2+{\left(7.2\right)}^2+x^2+5^2={\left(12.2\right)}^2 \\ \Rightarrow 2x^2+51.84+25=148.84 \\ \Rightarrow 2x^2+76.84=148.84 \\ \Rightarrow 2x^2=72 \\ \Rightarrow x^2=36 \\ \Rightarrow x=6 \end{gathered}$$

So, the radius of the circle is 6 cm.