Question:
Smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is
A. $2(\pi-2)$
B. $\pi-2$
C. $2 \pi-1$
D. $2(\pi+2)$
Solution:
The smaller area enclosed by the circle, $x^{2}+y^{2}=4$, and the line, $x+y=2$, is represented by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area (ΔOAB)
$=\int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(2-x) d x$
$=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}-\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}$
$=\left[2 \cdot \frac{\pi}{2}\right]-[4-2]$
$=(\pi-2)$ units
Thus, the correct answer is B.