Question

If $\Omega_{1}$ be a circle with centre $O$ and diameter $AB\&P$ be a point on the segment $OB$ . Suppose another circle $\Omega_{2}$ with centre $P$ lies in the interior of $\Omega_{1}.$ Tangents are drawn from $A$ and $B$ to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$ and $B_{1}$ are on the opposite sides of $AB$ . Given that $A_{1}B=5, \, AB_{1}=15$ and $OP=10,$ If $r$ is the radius of $\Omega_{1}$ Then $\left[\frac{r}{10}\right]$ equals

Answer 1

$\frac{r_{1}}{r + 10}=\frac{5}{2 r}$ ......(i)
$\frac{r_{1}}{r - 10}=\frac{15}{2 r}$ ......(ii)
Dividing equation (i) by equation (ii), we get
$\frac{r - 10}{r + 10}=\frac{1}{3} \, \, \Rightarrow 3r-30=r+10 \, \Rightarrow r=20$ .