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] …