# 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?

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, $$k_5 = 4 k_1 + 4 k_2 + k_3 \pm 4 \sqrt{k_1 k_2 + k_2 k_3 +k_3 k_1}$$

• Any ideas about finding the radii of nth such circle ? – karan kapoor Jul 12 '15 at 21:51
• I mean is there some pattern ? Or we need to compute previous to calculate the further ones ? – karan kapoor Jul 12 '15 at 21:51
• There must certainly be a pattern, but I'm not sure how straightforward it is. As the other answer suggests, you should consider looking for resources on the Apollonian Gasket. See, for instance, Wikipedia's article. – Blue Jul 13 '15 at 4:52

You might want to look at this paper and the references given in it: