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$C_1$ and $C_2$ are centres of $2$ identical semi-circles with radius $r$ units. We have $(C_1P_1):(PQ) = 1:7$. Find the radius of the circle inscribed in the overlapping region of the $2$ semi-circles, given that the circle touches line $P_1P_2$.

What I Tried: Here is a picture in Geogebra :-

What I did is assume $C_1P_1 = C_2P_2 = x$, and we have $PQ = 7x$. Now :- $$PC_1 + C_1P_2 + P_2C_2 + C_2Q = PQ$$ We have :- $PC_1 = C_1P_2 = C_2Q = r$, and we get :- $$3r + x = 7x$$ $$\rightarrow r = 2x$$ This implies that :- $$C_1P_1 = P_1P_2 = P_2C_2$$

I am not sure how to start finding the radius of the inscribed circle, one thing I did is drop a perpendicular from $E$ to $PQ$ and then try to use Pythagorean Theorem, but that did not help as I expected.

Can anyone help me? Thank You.

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Drop $ED \perp PQ$.

$\triangle EC_{1}D $ is right-angled with $C_1 D=3r/4$, $ED=R$, $C_1E=r-R$

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  • $\begingroup$ Oh completely forgot I could represent $C_1E$ as $r - R$ . $\endgroup$ – Anonymous Nov 25 '20 at 13:11

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