If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$. The radius of circle $S_3$ which touches externally with $S_1$ and $S_2$ and internally with $S$ is?
I tried making a diagram and figuring out, but cannot bring a relation.
Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that $$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$ (See, Loney, Plane Trigonometry, Page 187, Ex 27 , Q-29).
Applying this to the triangle $CC_1C_2$, we have $$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$ from which we get $r = 30/19$.