Golden Circles

Let the radius of circle $C_n$ be $r_n$ . Then we note that the horizontal distance between centers of $C_2$ and $C_3$ is equal to the sum of distance between centers of $C_2$ and $C_4$ and distance between centers of $C_3$ and $C_4$ and by Pythagorean theorem:

$\begin{aligned} \sqrt{(r_2+r_3)^2-(r_2-r_3)^2}&= \sqrt{(r_2+r_4)^2-(r_2-r_4)^2}+ \sqrt{(r_3+r_4)^2-(r_3-r_4)^2} \\ 2 \sqrt{r_2r_3}&= 2 \sqrt{r_2r_4}+ 2 \sqrt{r_3r_4} \quad \quad \small \color{#3D99F6} \text{Divide both sides by }2\sqrt{r_2r_3r_4} \\ \implies \frac 1{\sqrt{r_4}} &= \frac 1{\sqrt{r_3}} + \frac 1{\sqrt{r_2}} \end{aligned}$

Then we have:

$\begin{aligned} \frac 1{\sqrt{r_1}} & = 1 = F_1 & \small \color{#3D99F6} \text{where }F_n \text{ is the }n\text{th Fibonacci number.} \\ \frac 1{\sqrt{r_2}} & = 1 = F_2 \\ \frac 1{\sqrt{r_3}} & = \frac 1{\sqrt{r_2}} + \frac 1{\sqrt{r_1}} = 2 = F_3 \\ \frac 1{\sqrt{r_4}} & = \frac 1{\sqrt{r_3}} + \frac 1{\sqrt{r_2}} = 3 = F_4 \\ \implies \frac 1{\sqrt{r_n}} & = F_n \end{aligned}$

Then we have:

$\begin{aligned} \lim_{n\to \infty} \frac {A_n}{A_{n+1}} & =\lim_{n\to \infty} \frac {πr_n^2}{πr_{n+1}^2} \\ & = \lim_{n\to \infty} \left(\sqrt{\frac {r_n}{r_{n+1}}}\right)^4 \\ & = \lim_{n\to \infty} \left(\frac {F_{n+1}}{F_n}\right)^4 \\ & = \varphi^4 & \small \color{#3D99F6} \text{where }\varphi \text{ is the golden ratio.} \end{aligned}$

Therefore, $m=\boxed 4$ .