Problem 784

[Solution] Let $O$ denote the vertex of the right angle, $C$ and $C^{\prime}$ the centers of the circles, and $r$ and $r^{\prime}$ the radii of any two adjacent circles $\left(r>r^{\prime}\right)$. If $T$ is the point of tangency of the two circles, then

and
$$O T=O C^{\prime}+r^{\prime}=(\sqrt{2}+1) r^{\prime},$$

From these two equations, we get the ratio of the radii of adjacent circles
$$\frac{r^{\prime}}{r}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}=(\sqrt{2}-1)^{2} \text {. }$$

If $r$ is the radius of the first circle in the sequence, then its area is $\pi r^{2}$, and the sum of the areas of all other circles, which form a geometric series, is
$$\begin{aligned} & \pi r^{2}\left[(\sqrt{2}-1)^{4}+(\sqrt{2}-1)^{8}+\cdots\right] \\ = & \frac{\pi(\sqrt{2}-1)^{4} r^{2}}{1-(\sqrt{2}-1)^{4}}=\frac{\pi r^{2}}{(\sqrt{2}+1)^{4}-1} . \end{aligned}$$
$\therefore$ The required ratio of the areas is
$$\left[(\sqrt{2}+1)^{4}-1\right]: 1=(16+12 \sqrt{2}): 1 .$$

Therefore, the answer is (C).