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# Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017, Question 27. You may use sequential hints to solve the problem.

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

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

## Try with Hints

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 ……….

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……..

Dividing (1) by (2)

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

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