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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?

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

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  • $\begingroup$ Any ideas about finding the radii of nth such circle ? $\endgroup$ – karan kapoor Jul 12 '15 at 21:51
  • $\begingroup$ I mean is there some pattern ? Or we need to compute previous to calculate the further ones ? $\endgroup$ – karan kapoor Jul 12 '15 at 21:51
  • $\begingroup$ 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. $\endgroup$ – Blue Jul 13 '15 at 4:52
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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

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