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 $A_{1}B=5,A{B}_1=15$ and $OP=10,$ If $r$ is the radius of ${\Omega }_1$ Then $\left[\frac{r}{10}\right]$ equals

Answer 1

Answer: 2

$\frac{r_1}{r+10}=\frac{5}{2r}$ ......(i)
$\frac{r_1}{r-10}=\frac{15}{2r}$ ......(ii)
Dividing equation (i) by equation (ii), we get
$\frac{r-10}{r+10}=\frac{1}{3}\Rightarrow 3r-30=r+10\Rightarrow r=20$ .