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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.

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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 $r=\frac{30}{19}\gt\frac32$.

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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 $k$.

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  • $\begingroup$ Ah, coordinate bashing (if I am not wrong) $\endgroup$ – Yellow Jan 10 '19 at 8:08
  • $\begingroup$ See my edit. Note that by solving the system we get $k=30/19$. $\endgroup$ – Robert Z Jan 10 '19 at 8:24
  • $\begingroup$ Oh wait, what do you mean by $k$ is its center? $\endgroup$ – Yellow Jan 10 '19 at 8:37
  • $\begingroup$ Sorry $k$ is the radius. The same notation you used in your question. $\endgroup$ – Robert Z Jan 10 '19 at 8:40

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