Question

Choose the correct answer in the following questions

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 smaller area enclosed by the circle, $x^2+y^2=4$ and the line x+y=2 is represented by the shaded area ACBA.

The intersection points of circle and line are A(2, 0) and B(0, 2).

$\therefore$ Required area (shown is shaded region)

= Area OACBO - Area $\left(\Delta OAB\right)$

$={\int }_0^2\sqrt{4-x^2}dx-{\int }_0^2(2-x)dx$

$[\because x^2+y^2=4\Rightarrow y=\sqrt{4-x^2}andx+y=2\Rightarrow y=2-x]$

$=[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}si{n}^{-1}+\frac{4}{2}{]}_0^2-[2x-\frac{x^2}{2}{]}_0^2=[\frac{2}{2}\sqrt{4-4}+\frac{4}{2}si{n}^{-1}(1)-0-\frac{4}{2}si{n}^{-1}]-[4-2][\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}]$
$=\left[2\frac{\pi }{2}\right]-[4-2]=(\pi -2)$ sq unit. Thus, the correct option is (b)