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. $2(\pi +2)$

Answer 1

The correct option is B $\pi -2$

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 shown in the diagram.
It can be observed that,
$AreaACBA=AreaOACBO-Area\left(\Delta OAB\right)$
$={\int }_0^2\sqrt{4-x^2}dx-{\int }_0^2(2-x)dx$
$={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^2-{[2x-\frac{x^2}{2}]}_0^2$
$=[2\cdot \frac{\pi }{2}]-[4-2]$
$=(\pi -2)$sq. units
Thus, the correct answer is $B$.