The parabola $y^2=x$ divides the circle $x^2+y^2=2$ into two parts whose areas are in the ratio
- $9 \pi+2: 3 \pi-2$
- $9 \pi-2: 3 \pi+2$
- $7 \pi-2: 2 \pi-3$
- $7 \pi+2: 3 \pi+2$
Solution
$ \begin{aligned} & \text { Area of circle }=\pi(\sqrt{2})^2=2 \pi \\ & \text { Area of } O C A D O=2\{\text { Area of } O C A O\} \\ & \quad=2 \text { area of } O C B+\text { area of } B C A\} \\ & =2 \int_0^1 y_p d x+2 \int_1^{\sqrt{2}} y_c d x \\ & \text { where } y_p=\sqrt{x} \text { and } y_c=\sqrt{2-x^2} \\ & \therefore \text { Required Area } \\ & =2 \int \frac{1}{x} d x+2 \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \\ & =2\left[\frac{2}{3} \cdot 1-0\right]+2\left[\frac{x \sqrt{2-x^2}}{2}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}} \end{aligned} $ $ \begin{aligned} & =\frac{4}{3}+2\left\{\frac{\pi}{2}-\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi}{4}-\frac{1}{2}\right\}=\frac{3 \pi+2}{6} \\ & \text { Bigger area }=2 \pi-\frac{3 \pi+2}{6}=\frac{9 \pi-2}{6} \\ & \therefore \text { Required Ratio }=\frac{9 \pi-2}{3 \pi+2} \text { i.e., } 9 \pi-2: 3 \pi+2 \end{aligned} $
Asked in: JEE Main 2012 (07 May Online)