Circles in Geometric Progression? 3

$S_{n-1} = \left (\sum_{k=1}^{n-1} 2 \cdot r_k \right ) - 1 \Rightarrow {\color{#D61F06}S_n = S_{n-1} + 2 \cdot r_n}\\ \left ( 2 - r_n \right )^2 = 1^2 + \left (S_{n-1} + r_n \right )^2 \Rightarrow {\color{#D61F06}r_n = \frac{1}{2} \cdot \frac{3-(S_{n-1})^2}{2+S_{n-1}}}\\ \begin{aligned} n & | & r_n & | & S_n \\ 1 & | & 1 & | & 1 \\ 2 & | & \frac{1}{3} & | & \frac{5}{3} \\ 3 & | & \frac{1}{33} & | & \frac{19}{11} \\ 4 & | & \frac{1}{451} & | & \frac{71}{41} \\ 5 & | & \frac{1}{6273} & | & \frac{265}{153} \\ 6 & | & \frac{1}{87363} & | \\ \end{aligned}$

Let the centers of $C_A$ , $C_B$ , and $C_n$ be $A$ , $B$ , and $O_n$ respectively. We note that $AB= 2r_1 = 2$ all $O_n$ 's are on a straight line, which is the perpendicular bisector of $AB$ . The height $O_nO_1$ is $r_1 + 2r_2 + 2r_3 + \cdots + 2r_{n-1} + r_n = 2s_n - r_1 - r_n$ , where $s_n = \sum_{k=1}^n r_k$ . We also note that $AO_n$ is on a radius of $C_A$ and that

$\begin{aligned} \sqrt{(AO_1)^2+(O_1O_n)^2} + r_n & = r_A \\ \sqrt{r_1^2 + (2s_n-r_1-r_n)^2} + r_n & = 2 & \small \blue{\text{Note that }r_1 = 1} \\ \sqrt{1 + (2s_{n-1} + r_n-1)^2} & = 2 - r_n & \small \blue{\text{and }s_{n-1} = s_n - r_n} \\ 1 + (2s_{n-1} + r_n-1)^2 & = (1- (r_n-1))^2 & \small \blue{\text{Squaring both sides}} \\ 1 + 4s_{n-1}^2 + 4s_{n-1}(r_n-1) + (r_n-1)^2 & = 1 - 2(r_n-1) + (r_n-1)^2 \\ \implies r_n & = 1 - \frac {2s_{n-1}^2}{2s_{n-1}+1} \\ s_{n-1} + r_n & = s_{n-1} + 1 - \frac {2s_{n-1}^2}{2s_{n-1}+1} \\ \implies s_n & = \frac {3s_{n-1}+1}{2s_{n-1}+1} \end{aligned}$

Then we have:

$\begin{aligned} s_1 & = 1 \\ s_2 & = \frac {3+1}{2+1} = \frac 43 \\ s_3 & = \frac {12+3}{8+3} = \frac {15}{11} \\ s_4 & = \frac {45+11}{30+11} = \frac {56}{41} \\ s_5 & = \frac {168+41}{112+41} = \frac {209}{153} \\ s_6 & = \frac {627+153}{418+153} = \frac {780}{571} \end{aligned}$

Therefore, $r_6 = s_6 - s_5 = \dfrac {780}{571} - \dfrac {209}{153} = \dfrac 1{87363}$ $\implies a = \boxed{87363}$ .