Question

In a circle with centre P, a chord AB is parallel to a tangent and intersects the radius drawn from a point of contact at the mid point of the radius. If AB = 12, find the radius of the circle.

Answer 1

Let the radius of the circle be r.
Then AO = r and OP =$\frac{r}{2}$
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{AB}\parallel \mathrm{CD}\mathrm{and}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{tangent}. \\ \mathrm{i}.\mathrm{e}.,\angle \mathrm{POB}=\angle \mathrm{PQD}=90°(\mathrm{Corresponding}\mathrm{angle}) \\ \mathrm{A}\hspace{0.17em}\mathrm{perpendicular}\mathrm{line}\mathrm{drawn}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{to}\mathrm{the}\mathrm{chord}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \therefore \mathrm{OB}=6\mathrm{units} \\ \mathrm{In}\mathrm{right}∆\mathrm{POB},\mathrm{we}\mathrm{have}: \\ {\mathrm{PB}}^2={\mathrm{PO}}^2+{\mathrm{OB}}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow r^2=\frac{r^2}{4}+36 \\ \Rightarrow r^2-\frac{r^2}{4}=36 \\ \Rightarrow 3r^2=144 \\ \Rightarrow r^2=48 \\ \Rightarrow r=4\sqrt{3}\mathrm{units} \end{gathered}$$