Not Nine Circles Theorem

Let the radius of the four ${\color{#E81990}\text{pink}}$ circles be $1$ and the radius of the four quarter circles be $r$ . So, the radius ratio of one quarter circle to one pink circle is $\dfrac{r}{1}=r$ .

In $\triangle DEF$ , $\begin{aligned} DE^2&=EF^2+FD^2 \\ (1+r)^2&=1^2+FD^2 \\ \implies FD&=\sqrt{r^2+2r} \end{aligned}$ The length of each side of the square is, $AD=AF+FD=1+\sqrt{r^2+2r}$ In $\triangle DCB$ , $DB=r+r-2,\;\; BC=1+\sqrt{r^2+2r},\;\; CD=1+\sqrt{r^2+2r}$ $\begin{aligned} DB^2&=BC^2+CD^2\\ (2r-2)^2&=(1+\sqrt{r^2+2r})^2+(1+\sqrt{r^2+2r})^2 \\ \implies 2r-2&=\sqrt{2}\left(1+\sqrt{r^2+2r}\right) \end{aligned}$ Solving the above equation, we have $r=3+\sqrt{2}+2\sqrt{2+\sqrt{2}}$ Therefore, $A=3,\;B=1,\;C=2,\;D=2,\;E=2,\;F=1,\;G=2\implies A+B+C+D+E+F+G=\boxed{13}$

Suppose the square has side 1, the large circles have radius R, the small circles have radius r. Half a diagonal is $\frac{1}{2}\sqrt{2}=R-r$ We can also identify a right-angled tiangle with hypothenuse R+r and right sides 1-r and r so that $(R+r)^2=(1-r)^2+r^2$ Combining these we get $2r^2+(2\sqrt{2}+2)r-\frac{1}{4}=0$ Taking the positive root: $r=\frac{1}{2}\sqrt{\sqrt{8}+4}-\frac{1}{2}\sqrt{2}-\frac{1}{2},R=\frac{1}{2}\sqrt{\sqrt{8}+4}-\frac{1}{2}$

The ratio can, after some work, be reduced to $\frac{R}{r}=3+\sqrt{2}+2\sqrt{2+\sqrt{2}}$ Here we have $A=3,B=1,C=2,D=2,E=2,F=1,G=2$ These numbers have sum of $\boxed{13}$