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If a straight line $y-x=2$ divides the region $x^2+y^2 \leq 4$ 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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The correct answer is:
$\pi-2: 3 \pi+2$
$\pi-2: 3 \pi+2$
Let I be the smaller portion and II be the greater portion of the given figure then,
Area of $I=\int_{-2}^0\left[\sqrt{4-x^2}-(x+2] d x\right.$
$$
\begin{aligned}
& =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{-2}^0-\left[\frac{x^2}{2}+2 x\right]_{-2}^0 \\
& =\left[2 \sin ^{-1}(-1)\right]-\left[-\frac{4}{2}+4\right]=2 \times \frac{\pi}{2}-2=\pi-2
\end{aligned}
$$
Now, area of $\mathrm{II}=$ Area of circle $-$ area of $\mathrm{I}$.
$$
\begin{aligned}
& =4 \pi-(\pi-2) \\
& =3 \pi+2
\end{aligned}
$$
Hence, required ratio $=\frac{\text { area of I }}{\text { area of II }}=\frac{\pi-2}{3 \pi+2}$
Area of $I=\int_{-2}^0\left[\sqrt{4-x^2}-(x+2] d x\right.$
$$
\begin{aligned}
& =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{-2}^0-\left[\frac{x^2}{2}+2 x\right]_{-2}^0 \\
& =\left[2 \sin ^{-1}(-1)\right]-\left[-\frac{4}{2}+4\right]=2 \times \frac{\pi}{2}-2=\pi-2
\end{aligned}
$$
Now, area of $\mathrm{II}=$ Area of circle $-$ area of $\mathrm{I}$.
$$
\begin{aligned}
& =4 \pi-(\pi-2) \\
& =3 \pi+2
\end{aligned}
$$
Hence, required ratio $=\frac{\text { area of I }}{\text { area of II }}=\frac{\pi-2}{3 \pi+2}$
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