TITLE: How to avoid overlap in circle fractals?
QUESTION [8 upvotes]: I had asked this on reddit and someone suggested that I try here:  
Assuming that the pattern in the image below continues infinitely, how much would each generation of circles have to decrease to avoid any overlap/collision.  In my example the diameter of each circle is 1/3 of its parent circle.  As you can see that's not small enough and overlap occurs quickly.

REPLY [4 votes]: Define the sequence of $\theta_n$ to be,
\begin{cases}
0 \ ,  & \text{if $n=0$} \\
\cfrac{3\pi}{4}\ , & \text{if $n=1$} \\
\pi \ , & \text{$n \gt 1$}
\end{cases}
Visually speaking, this is the sequence that tries to cross back on the original circle the "hardest". The exceptions are $\theta_0$ and $\theta_1$. The initial angle has to be $0$. The second angle can't be $\pi$ or it automatically hits the circle. The other angles are $\pi$ because we want the angle that takes the most direct route to the circle, but ends up just touching the edge of the first circle.
Then, as noted by Mark McClure, we have a representation with complex numbers,
$$(1) \quad \left|\sum_{k=0}^{\infty} (1+r) \cdot r^k e^{\theta_k i}\right| > 1$$
Because $\theta_n=\pi$ for $n \gt 1$, we should analyze the sum starting with $k=2$.
$e^{\pi i}=-1$, so we immediately have,
$$(2) \quad \sum_{k=2}^{\infty} (1+r) \cdot r^k e^{\theta_k i}=-\sum_{k=2}^{\infty} (1+r)r^k=\cfrac{r^2 \cdot (r+1)}{r-1}$$
So, if we substitute the value found in $(2)$ into $(1)$, we get,
$$(3) \quad \left|(1+r)+(1+r) \cdot r \cdot \cfrac{-\sqrt{2}+\sqrt{2}i}{2}+\cfrac{r^2 \cdot (r+1)}{r-1}\right| > 1$$
We can expand this using algebra, I prefer equalities,
$$(3.9) \quad \sqrt{(2-\sqrt{2}) \cdot r^4+2\sqrt{2} \cdot r^2-(2+\sqrt{2}) \cdot r+1} \cdot \cfrac{r+1}{1-r} = 1$$
More conducive for manipulations, we have,
$$(4) \quad \left((2-\sqrt{2}) \cdot r^4+2\sqrt{2} \cdot r^2-(2+\sqrt{2}) \cdot r+1\right) \cdot \cfrac{(r+1)^2}{(1-r)^2} = 1$$
This is quite an odd equation. First of all, here's a plot of $(3.9)$ against $1$.

Numerically, there are three solutions, which is unexpected. We have,
$$r=0$$
$$r=\phi=\cfrac{\sqrt{5}-1}{2}=0.618...$$
and then, this...
$$r=\cfrac{-1}{3}-\cfrac{2\cdot(4+3\sqrt{2})}{3 \cdot \left(26+9\sqrt{2}+3\sqrt{3 \cdot (114+76\sqrt{2})} \right)^{1/3}}+\cfrac{1}{3} \cdot \left(26+9\sqrt{2}+3\sqrt{3 \cdot (114+76\sqrt{2})} \right)^{1/3}=0.1660561895...$$
I just find it odd two of the solutions are so simple, and then the third one is so complex. 

It seems that there are three non-negative solutions: the complicated $r\approx0.166$ illustrated above, $r=0$ (which has a simple interpretation, as the partial sums are all 1), and $r=1/\varphi$ the reciprocal of the golden ratio. In this final case, the small appended circles actually cross the larger circle and touch it on the other side.  Here's an illustration where most of the circles are highly translucent while the path that leads to just touching is highlighted: