Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

## Circle – PRMO 2017, Problem 27

Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. 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 $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

- $9$
- $40$
- $34$
- $20$

**Key Concepts**

Geometry

Circle

## Check the Answer

But try the problem first…

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

## Try with Hints

First hint

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$

From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$

[

\begin{array}{l}

\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \

\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}

\end{array}

]

Can you now finish the problem ……….

Second Hint

Again, $\Delta B P E \sim \Delta B A B_{1}$

Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$

$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Final Step

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

## Other useful links

- https://www.cheenta.com/ordered-pairs-prmo-2019-problem-18/
- https://www.youtube.com/watch?v=VLyrlx2DWdA&t=20s

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