Geometry problem about two externally touching circles Two circles of radius $~12~$ and $~3~$ touch externally. A line intersecting both of them intersects first circle at points $P$ and $Q$, second circle - at points $R$ and $S$. Three resulting line segments, two inside the circles and the one between them, are equal: $PQ=QR=RS$. Find their common length.
I have prepared a picture with Geogebra to illustrate

I was trying to solve this with no luck. 
After formulating it as a system of equations based on coordinates, with the origin being circles' common point and $X$-axis on the line connecting their centers, Wolfram Alpha helped me to find that the answer should be $\frac{3}{2}\sqrt{13}~$. 
Can you give any hints on how to solve this?

Let $PQ=QR=RS=2x$, $AM$ and $O_2N$ be perpendiculars to $PS$ 
and $O_2K$ be a perpendicular to $AM$. 
Thus, since $$KO_2=MN=x+2x+x=4x,$$ $$AK=\sqrt{12^2-x^2}-\sqrt{3^2-x^2}$$ and $$AO_2=12+3=15,$$ by the Pythagoras's theorem for $\Delta AO_2K$ we obtain:
$$(4x)^2+\left(\sqrt{12^2-x^2}-\sqrt{3^2-x^2}\right)^2=15^2.$$
Can you end it now?
I got $PQ=QR=RS=\frac{3\sqrt{13}}{2}.$
