If a straight line y - x = 2 divides the regi | vedclass

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(D) The given region is a circle $x^2 + y^2 = 4$ with radius $r = 2$ and center at $(0, 0)$. The line $y = x + 2$ passes through $(-2, 0)$ and $(0, 2)

Vedclass · Accepted Answer

$\pi - 2 : 3\pi + 2$

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(D) The given region is a circle $x^2 + y^2 = 4$ with radius $r = 2$ and center at $(0, 0)$. The line $y = x + 2$ passes through $(-2, 0)$ and $(0, 2)$.Let $I$ be the smaller portion and $II$ be the greater portion of the circle.The area of the smaller portion $I$ is given by the integral:Area of $I = \int_{-2}^{0} [\sqrt{4 - x^2} - (x + 2)] dx$$= [\frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2})]_{-2}^{0} - [\frac{x^2}{2} + 2x]_{-2}^{0}$$= [0 + 2 \sin^{-1}(0)] - [(-1) \sqrt{0} + 2 \sin^{-1}(-1)] - [0 - (\frac{4}{2} - 4)]$$= [0 - 2(-\frac{\pi}{2})] - [0 - (-2)]$$= \pi - 2$Now,the area of the greater portion $II$ is:Area of $II = 	ext{Area of circle} - 	ext{Area of } I$$= 4\pi - (\pi - 2) = 3\pi + 2$Therefore,the required ratio is:$\frac{	ext{Area of } I}{	ext{Area of } II} = \frac{\pi - 2}{3\pi + 2}$

If a straight line $y - x = 2$ divides the region $x^2 + y^2 \le 4$ into two parts,then the ratio of the area of the smaller part to the area of the greater part is

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