Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.
I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.
Best Answer
Both the Soddy-Gosset Theorem and Descartes' Theorem say $$ \left(\frac12+\frac13-\frac15+\frac1r\right)^2=2\left(\frac1{2^2}+\frac1{3^2}+\frac1{5^2}+\frac1{r^2}\right) $$ which means that $r=\frac{30}{19}\gt\frac32$.