Box of Circles

Let label the figure as shown. Let the radius of the small circles be 1. Since $AD||BC$ , $\angle DEC = \angle BCE$ and let it be $\theta$ and $t = \tan \frac \theta 2$ .

Then we note that $DE = \dfrac 1t + 1 = \dfrac {1+t}t$ . Since $CD = DE \tan \theta = \dfrac {1+t}t \cdot \dfrac {2t}{1-t^2} = \dfrac 2{1-t}$ . As $CD = 2r$ , where $r$ is the radius of the big circle, implying $r = \dfrac 1{1-t}$ .

Now we have:

$\begin{aligned} \frac {OF}{CO} & = \sin \frac \theta 2 \\ \color{#3D99F6} \frac {\frac 1{1-t}}{\frac {\sqrt{1+t^2}}t + 1 + \frac 1{1-t}} & = \frac t{\sqrt{1+t^2}} & \small \color{#3D99F6} \text{Multiple up and down of LHS by }t(1-t) \\ \color{#3D99F6} \frac t{(1-t)\sqrt{1+t^2} + 2t-t^2} & = \frac t{\sqrt{1+t^2}} & \small \color{#3D99F6} \text{Rearrange} \\ 2 - t & = \sqrt{1+t^2} & \small \color{#3D99F6} \text{Square both sides and rearrange} \\ \implies t & = \frac 34 \end{aligned}$

Then $r = \dfrac 1{1-t} = 4$ , $AB=CD=8$ , $DE = \dfrac {1+t}t = \dfrac 73$ , and $BC = BF+FC = r+\dfrac rt = \dfrac {28}3$ . And:

$\begin{aligned} \frac {A_{\color{#3D99F6}blue}}{A_{\color{#D61F06}red}} & = \frac {[ABCE]-\pi (1^2) - \pi r^2}{[CDE]-\pi(1^2)} \\ & = \frac {[ABCD]-[CDE] - 17\pi}{\frac 12 DE \cdot CD - \pi} \\ & = \frac {\frac {28}3 \cdot 8 - \frac 12 \cdot \frac 73 \cdot 8 - 17 \pi}{\frac 12 \cdot \frac 73 \cdot 8 - \pi} \\ & = \boxed{\dfrac {196-51\pi}{28-3\pi}} \end{aligned}$