# Prove that radius of circle $r$ exceeds $3/2$

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.

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 $$r=\frac{30}{19}\gt\frac32$$.
Hint. Let $$A=(-5,0)$$ and let $$B=(5,0)$$. Then $$P=(1,0)$$ and if $$(x,y)$$ is the center of the fourth circle and $$k$$ is its radius, it follows that $$\begin{cases} (x+2)^2+y^2=(k+3)^2\\(x-3)^2+y^2=(k+2)^2\\x^2+y^2=(5-k)^2 \end{cases}.$$ Now you should be able to find the precise value of $$k$$.
• See my edit. Note that by solving the system we get $k=30/19$. – Robert Z Jan 10 '19 at 8:24
• Oh wait, what do you mean by $k$ is its center? – Yellow Jan 10 '19 at 8:37
• Sorry $k$ is the radius. The same notation you used in your question. – Robert Z Jan 10 '19 at 8:40