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.