Find the area of the region lying inside,

$ \begin{aligned} x^{2}+(y-1)^{2} &=1 \text { and outside,} \\\\ c^{2} x^{2}+y^{2} &=c^{2} \text { where } c=(\sqrt{2}-1) \end{aligned} $

Solution:

$ x^{2}+(y-1)^{2}=1....(1) \\ $

$ c=\sqrt{2}-1....(2) \\ $

Solving (1) and (2), we get,

$ x \text {-coordinate of } \mathrm{A}=1 / \sqrt{2} \text {. } \\ $

Due to symmetry about $y$-axis, Area, $ O A B C O \ $ $$ \begin{aligned} &=2 \int_{1}^{1 / \sqrt{2}}\left[c \sqrt{1-x^{2}}-\left(1-\sqrt{1-x^{2}}\right)\right] d x \\ &=2 \int_{1}^{1 / \sqrt{2}}\left[(c+1) \sqrt{1-x^{2}}-1\right] d x \\ &=2 \int_{1}^{1 / \sqrt{2}}\left[\sqrt{2} \sqrt{1-x^{2}}-1\right] d x \end{aligned} \\ $$

$$ \begin{aligned} &=2\left\{\frac{\sqrt{2}}{2}\left(x \cdot \sqrt{1-x^{2}}+\sin ^{-1} x\right)-x\right\}_{0}^{1 / \sqrt{2}} \\\\ &=2\left\{\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2} \cdot \frac{\pi}{4}-\frac{1}{\sqrt{2}}\right\} \\\\ &=2\left\{\frac{\sqrt{2} \pi}{8}-\frac{\sqrt{2}}{4}\right\}=\left(\frac{\sqrt{2} \pi}{4}-\frac{\sqrt{2}}{2}\right) \\\\ &\therefore \quad \text { Area inside the circle and outside of ellipse } \\\\ &=\pi-\left(\frac{\sqrt{2} \pi}{4}-\frac{\sqrt{2}}{2}\right)=\left[\frac{(4-\sqrt{2}) \pi}{4}+\frac{\sqrt{2}}{2}\right] \text { sq.units. } \end{aligned} \\ $$