Banach Circle Fractal: Find factor scale $s$ such that between small circle is not overlap in $n$-th step as $n\to\infty$ 
Source: from stackoverflow
I want to draw a Banach Fractal as picture above.

*

*First step, draw a circle with radius $r$.

*Second step, draw 9 smaller circles with radius $s\cdot r$ ($s$ is
factor scale, $0<s<1$) [one circle in center and 8 circles in circle
side].

*Third step, each 9 smaller circle from second step, is drawn by 9
smaller circles with radius $s^2\cdot r$ [one circle in center and 8 circles in circle
side].

*etc.

In picture above, from third step the smaller circle is overlap. Now, I want to calculate the factor scale $s$ such that the smaller circle is not overlap in $n$-th step as $n\to\infty$.
I try to calculate as follows.
For first step until infinity step, total radius of smaller circle is
$$sr+s^2 r+s^3 r+\ldots = \dfrac{sr}{1-s}.$$
So, for first step until infinity step total diameter of smaller circle is
$$T_d=2\dfrac{sr}{1-s}.$$
The circumference of first step circle is $2\pi r$.
Because of we draw 8 circles in first step circle side, now I divide circumference by 8.
So, total diameter per small circle in "first circle side" is
$$T_d=\dfrac{2\pi r}{8}=\dfrac{\pi r}{4}$$
Now, we have equality
$$2\dfrac{sr}{1-s}=\dfrac{\pi r}{4}.$$
Solving the equality:
\begin{align}
&2\dfrac{sr}{1-s}=\dfrac{\pi r}{4}\\
\iff &\dfrac{s}{1-s}=\dfrac{\pi }{8}\\
\iff &8s=\pi-\pi s\\
\iff &(8+\pi) s=\pi \\
\iff & \dfrac{\pi}{8+\pi}.
\end{align}
Now I have the factor scale $s=\dfrac{\pi}{8+\pi}$ such that between small circle is not overlap in $n$-th step as $n\to\infty$.
Is it correct answer?
I think it over a hour and I'm not sure with my answer.

I guess you could just draw it and see:

Close, but not quite.  The correct value of
$$
s = \frac{\sin(\pi/8)}{\sin(\pi/8) + \sin(5\pi/8)} = \frac{1}{2+\sqrt{2}}
$$
should yield the so-called octa-gasket:

You can see why this should work once you see how the following lines perfectly partition the pieces:

Once you get to that point, finding $s$ is a simple application of trigonometry.
You can find the code for the pictures on Observable.
