Extension of Descartes' "Kissing Circles" Theorem Descartes' "Kissing Circle" Theorem relates the radii, $r_1$, $r_2$, $r_3$, $r_4$, of four mutually-tangent circles thusly:
$$( k_1 + k_2 + k_3 + k_4 )^2 = 2 ( k_1^2 + k_2^2 + k_3^2 + k_4^2 ) \qquad\text{where } k_i := \pm\frac{1}{r_i}$$
with the "$\pm$" sign indicating internal or external tangency of the corresponding circle.
Suppose, given three mutually-tangent circles, we determine the external fourth kissing circle, and then we want to find out the radius of a 5th circle that touches the 4th circle and any two of the circles of the original three circles. How do we find out the radius of this 5th circle?
 A: You might want to look at this paper and the references given in it:
http://www.math.washington.edu/~julia/teaching/445_Spring2013/DescartesAndTheApollonianGasket%20%281%29.pdf
A: If circles of radius $r_1$, $r_2$, $r_3$, $r_4$ are mutually tangent, and also circles of radius (say) $r_2$, $r_3$, $r_4$, $r_5$, then we have
$$\begin{align}
\left( k_1 + k_2 + k_3 + k_4 \right)^2 &= 2 \left( k_1^2 + k_2^2 + k_3^2 + k_4^2 \right) \\
\left( k_2 + k_3 + k_4 + k_5 \right)^2 &= 2 \left( k_2^2 + k_3^2 + k_4^2 + k_5^2 \right)
\end{align}$$
with $k_i = \pm 1/r_i$ as desired.
If you need the fourth circle's radius, then, of course, you can solve the equations in stages: get $k_4$ from the first, and use that in the second to get $k_5$.
If you don't really care about the fourth radius, we can eliminate $k_4$ from the equations. (For instance, subtract one from the other to get rid of the $k_4^2$ terms, and solve for $k_4$ in the resulting linear equation. Then substitute this $k_4$ into either of the original equations.) Assuming $k_3 \neq k_5$, we arrive at this relation:
$$16 k_1^2 + 16 k_2^2 + 
   k_3^2 + k_5^2 + 16 k_1 k_2 - 8 k_1 k_3 - 8 k_1 k_5 - 8 k_2 k_3 - 8 k_2 k_5 - 2 k_3 k_5  = 0$$ 
Thus,
$$k_5 = 4 k_1 + 4 k_2 + k_3 \pm 4 \sqrt{k_1 k_2 + k_2 k_3 +k_3 k_1}$$
