The circle $x^2+y^2=1$
The circle is symmetrical with the x-axis and y-axis
Required Area
$\begin{aligned} & =4 \int_0^1\left(\sqrt{1-x^2}\right) d x \\ & {\left[\int \sqrt{a^2-x^2} d x=\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ & =4 \int_0^1\left(\sqrt{1^2-x^2}\right) d x \\ & =4\left[\frac{x \sqrt{1^2-x^2}}{2}+\frac{1^2}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_0^1 \\ & =4\left(0-\frac{\pi}{4}-0-0\right) \\ & =\pi \text { sq.units }\end{aligned}$
Correct Answer is option B.