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

Let I be the smaller portion and II be the greater portion of the given figure then,

Area of $\displaystyle I=\int_{-2}^0\left[\sqrt{4-x^2}-(x+2] d x\right.$I=∫−20​[4−x2​−(x+2]dx $$\begin{aligned} & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{-2}^0-\left[\frac{x^2}{2}+2 x\right]_{-2}^0 \\ & =\left[2 \sin ^{-1}(-1)\right]-\left[-\frac{4}{2}+4\right]=2 \times \frac{\pi}{2}-2=\pi-2 \end{aligned}$$​=[2x​4−x2​+24​sin−1(2x​)]−20​−[2x2​+2x]−20​=[2sin−1(−1)]−[−24​+4]=2×2π​−2=π−2​ Now, area of $\mathrm{II}=$II= Area of circle $-$− area of $\mathrm{I}$I. $$\begin{aligned} & =4 \pi-(\pi-2) \\ & =3 \pi+2 \end{aligned}$$​=4π−(π−2)=3π+2​ Hence, required ratio $=\frac{\text { area of I }}{\text { area of II }}=\frac{\pi-2}{3 \pi+2}$= area of II  area of I ​=3π+2π−2​