Revival of Sangaku 2

$\Large~ABC~~is~~the~~enveloping~~\Delta.$

Based on the fact that the three given circles are on respective angle bisectors of ABC and its incircle is also where these three bisectors meet. So the centers of given circles are also on respective angle bisectors of ABC.
Thus from two centers $O_6~ and~ O_9$ of the given circles we draw the angle bisectors to meet at incenter I of triangle ABC.
This two bisectors and the line joining their centers forms a ASA $\Delta~IO_6O_9$ .
From this triangle the distance of I from $O_6$ is calculated and projected perpendicular to BC as t=7.70.
Add to t the distance of $O_6$ from BC=6, we get the inradius required= $\Huge \color{#D61F06}{13.70}.$

r=2√(xyz)/(√x+√y+√z -√(x+y+z)) where x=6, y=9 & z=12 Formula for in-radius as given by Fukagawa & Pedoe about which you can refer: http://mathworld.wolfram.com/MalfattiCircles.html