There's Circles Everywhere!

$\overline{Ow_{0}} = \sqrt{4 + 2\sqrt{2}}R_{1}$

$\triangle{OA_{1}w_{0}} \sim \triangle{w_{1}A_{2}w_{0}} \implies \dfrac{\sqrt{4 + 2\sqrt{2}}R_{1}}{R_{1}} = \dfrac{R_{1} + R_{2}}{R_{1} - R_{2}} \implies$

$(\sqrt{4 + 2\sqrt{2}} - 1)R_{1} = (\sqrt{4 + 2\sqrt{2}} + 1)R_{2} \implies$ $R_{2} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{1}$

$R_{3} = \dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1}R_{2} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2R_{1}$

In General: $R_{n} = (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1}R_{1}$

$\implies A_{n} = \pi R_{n}^2 = \pi R_{1}^2((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$\implies A = \sum_{n = 1}^{\infty} A_{n} = \pi R_{1}^2\sum_{n = 1}^{\infty} ((\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^2)^{n - 1}$

$= \dfrac{(\sqrt{4 + 2\sqrt{2}} + 1)^2}{4\sqrt{4 + 2\sqrt{2}}}\pi R_{1}^2$

and

$C_{n} = 2\pi R_{n} = 2\pi R_{1}(\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1}$

$\implies C = \sum_{n = 1}^{\infty} C_{n} = 2\pi R_{1}\sum_{n = 1}^{\infty} (\dfrac{\sqrt{4 + 2\sqrt{2}} - 1}{\sqrt{4 + 2\sqrt{2}} + 1})^{n - 1} =$

$(\sqrt{4 + 2\sqrt{2}}) \pi R_{1} \implies C^2 = (\sqrt{4 + 2\sqrt{2}})^2 \pi^2 R_{1}^2$

$\implies \dfrac{C^2}{A} = 4\pi\sqrt{4 + 2\sqrt{2}} \approx \boxed{32.8375088952649884}$ .

For any measure of $\angle O$ , as the pattern repeats, the ratio of the shortest distance of any circle (for example: $OA= l_1$ ) to the apex $O$ and the diameter of the circle ( $AB=d_1$ ) is a constant. Let the constant be $k$ then:

$\frac {l_1}{d_1} = \frac {l_2}{d_2} = \frac {l_3}{d_3} = \cdots = k$

We note that:

$\begin{aligned} l_{n-1} & = d_n + l_n \\ kd_{n-1} & = d_n + kd_n \\ d_n & = \frac k{1+k} d_{n-1} \\ \implies r_n & = \frac k{1+k} r_{n-1} = \left(\frac k{1+k} \right)^{n-1}r_1 & \small \blue{\text{where }r_n = \frac {d_n}2 \text{, the radius of }n\text{th circle.}} \end{aligned}$

Then we have:

$\begin{aligned} \sum_{n=1}^\infty C_n & = \sum_{n=1}^\infty 2\pi r_n = \sum_{n=0}^\infty 2\pi \left(\frac k{1+k} \right)^nr_1 = \frac {2\pi r_1}{1-\frac k{1+k}} = 2\pi r_1(1+k) \\ \sum_{n=1}^\infty A_n & = \sum_{n=1}^\infty \pi r_n^2 = \frac {\pi r_1^2}{1-\left(\frac k{1+k}\right)^2} = \frac {\pi r_1^2(1+k)^2}{1+2k} \\ \implies \frac {\left(\sum_{n=1}^\infty C_n\right)^2}{\sum_{n=1}^\infty A_n} & = \frac {(2\pi r_1(1+k))^2}{\frac {\pi r_1^2(1+k)^2}{1+2k}} = 4\pi (1+2k) \end{aligned}$

Let us find $k$ for $\angle O = \dfrac \pi 4$ . Then $k= \dfrac {l_1}{d_1} = \dfrac {l_1}{2r_1} = \dfrac {r_1\csc \frac \pi 8 - r_1}{2r_1} = \dfrac {\csc \frac \pi 8 - 1}2 = \dfrac {\sqrt{4+2\sqrt 2}-1}2$ . Then

$\frac {\left(\sum_{n=1}^\infty C_n\right)^2}{\sum_{n=1}^\infty A_n} = 4\pi (1+2k) = 2\pi \sqrt{4+2\sqrt 2} \approx \boxed{32.8}$