$$
\begin{aligned}
& \text { Area of circle }=\pi(\sqrt{2})^2=2 \pi \\
& \text { Area of } O C A D O=2\{\text { Area of } O C A O\} \\
& \quad=2 \text { area of } O C B+\text { area of } B C A\} \\
& =2 \int_0^1 y_p d x+2 \int_1^{\sqrt{2}} y_c d x \\
& \text { where } y_p=\sqrt{x} \text { and } y_c=\sqrt{2-x^2} \\
& \therefore \text { Required Area } \\
& =2 \int \frac{1}{x} d x+2 \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \\
& =2\left[\frac{2}{3} \cdot 1-0\right]+2\left[\frac{x \sqrt{2-x^2}}{2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3 \pi+2}{6} \\
& \text { Bigger area }=2 \pi-\frac{3 \pi+2}{6}=\frac{9 \pi-2}{6} \\
& \therefore \text { Required Ratio }=\frac{9 \pi-2}{3 \pi+2} \text { i.e., } 9 \pi-2: 3 \pi+2
\end{aligned}
$$