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Q : 6 Smaller area enclosed by the circle $\small x^2+y^2=4$ and the line $\small x+y=2$ is

(A) $\small 2(\pi -2)$ (B) $\small \pi -2$ (C) $\small 2\pi -1$ (D) $\small 2(\pi +2)$

Answers (1)

best_answer

So, 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

Thus it can be observed that,

Area of ACBA = Area OACBO - Area of $(\triangle OAB)$

$=\int_0^2 \sqrt{4-x^2} dx -\int_0^2 (2-x)dx$

$= \left ( \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}{\frac{x}{2}} \right )_0^2 - \left ( 2x -\frac{x^2}{2} \right )_0^2$

$= \left [ 2.\frac{\pi}{2} \right ] -[4-2]$

$= (\pi -2) units.$

Thus, the correct answer is B.

Posted by

Divya Prakash Singh

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