The smaller area enclosed by the circle x^{2} | vedclass

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(A) The circle is $x^{2}+y^{2}=2^{2}$,which has a radius of $2$ and center at $(0,0)$. The line $x+y=2$ passes through $(2,0)$ and $(0,2)$.The smaller

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$\pi-2$

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(A) The circle is $x^{2}+y^{2}=2^{2}$,which has a radius of $2$ and center at $(0,0)$. The line $x+y=2$ passes through $(2,0)$ and $(0,2)$.The smaller area enclosed by the circle and the line is the area of the circular segment in the first quadrant.Area $= \int_{0}^{2} (y_{	ext{circle}} - y_{	ext{line}}) dx$$= \int_{0}^{2} (\sqrt{4-x^{2}} - (2-x)) dx$$= \int_{0}^{2} \sqrt{4-x^{2}} dx - \int_{0}^{2} (2-x) dx$Using the formula $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:$= [\frac{x}{2}\sqrt{4-x^{2}} + \frac{4}{2}\sin^{-1}(\frac{x}{2})]_{0}^{2} - [2x - \frac{x^{2}}{2}]_{0}^{2}$$= [0 + 2\sin^{-1}(1)] - [0 + 2\sin^{-1}(0)] - [(4 - 2) - (0 - 0)]$$= 2(\frac{\pi}{2}) - 2 = \pi - 2$ sq. units.Thus,the correct answer is $A$.

The smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is

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