My motivation for this is the fact that the standard algorithm for rendering an Appollonian Gasket takes in the centers and radii of the three generating circles, but these circles need not be tangent even though an Apollonian Gasket is determined by the three points of tangency of the circles.
Before I explain my attempted solution I want to say that I want to solve the problem primarily on my own so if you have a full answer it would be appreciated if you prefaced it with a hint, "spoiler alert," or an explanation of why my solution is incorrect.
My solution is fairly straightforward, but its based on intuition and I haven't tryed to prove it or test it computationally yet. Call the three points of tangency $T_1,T_2,T_3$. Find the three angle bisectors of the vertices of $\Delta T_1 T_2 T_3$ and the idea that's based purely on my intuition is that these lines will be the tangent lines to each pair of circles. If I'm right about this, we can find three lines perpendicular to each of the tangents that pass through each point of tangency. Finding the intersection of each pair of these lines will necessarily yield the center of each circle, and calculating the radii from there is straightforward.
I'm pretty certain about that last part, my question is really whether or not the angle bisectors are the tangent lines for each pair of circles...and of course whether or not there is a simpler procedure for this entire thing.
The angle bisectors are not generally tangent lines to each pair of circles.
What you do instead is to find the perpendicular bisectors of each side of the $\triangle T_1T_2T_3$. These lines will meet at the triangle's circumcenter. Then draw lines through the circumcenter and each vertice. These lines will be tangent lines to each pair of circles. Lastly, through each vertice draw a line which is perpendicular to the tangent line through that vertice. These final lines will intersect with the perpendicular bisectors at the centers of the three circles.
It gets messy to illustrate the process but I've made an attempt in the figure above.