# Find the radius of the circle inscribed in the overlapping region of the $2$ semi-circles, given that the circle touches line $P_1P_2$.

$$C_1$$ 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.

Drop $$ED \perp PQ$$.
$$\triangle EC_{1}D$$ is right-angled with $$C_1 D=3r/4$$, $$ED=R$$, $$C_1E=r-R$$
• Oh completely forgot I could represent $C_1E$ as $r - R$ . – Anonymous Nov 25 '20 at 13:11