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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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