Question

The parabola $y^2=x$ divides the circle $x^2 +y^2 = 2$ into two parts whose areas are in the ratio

• A. $9\pi + 2 : 3\pi -2$
• B. $9\pi - 2 : 3\pi +2$
• C. $7\pi - 2 : 2\pi -3$
• D. $7\pi + 2 : 3\pi +2$

Answer 1

Answer: (B)

$9\pi - 2 : 3\pi +2$

Answer 2

$Area of circle =\pi\left(\sqrt{2}\right)^{2}=2\pi$ Area of OCADO = 2 \left\\{Area\, of\, OCAO\right\\} =2\left\\{area of OCB + area of BCA\right\\} $=2\int\limits^{1}_{{0}}y_pdx+2$ $\int\limits^{{{\sqrt{2}}}}_{{1}}y_cdx$ where $y_{p}=\sqrt{x}$ and $y_{c}=\sqrt{2-x^{2}}$ $\therefore$ Required Area $=2\int\limits^{1}_{{0}}$ $\sqrt{x} dx+2$ $\int\limits^{{{\sqrt{2}}}}_{{1}}$ $\sqrt{2-x^{2}}dx$ $=\left[\frac{2}{3}.1-0\right]+2\left[\frac{x\sqrt{2-x^{2}}}{2}+sin^{-1} \frac{x}{\sqrt{2}}\right]^{^{\sqrt{2}}}_{_{_1}}$ =\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} Bigger area $=2\pi-\frac{3\pi+2}{6}=\frac{9\pi-2}{6}$ $\therefore$ Required Ratio $= \frac{9\pi-2}{3\pi+2} i.e., 9\pi-2 : 3\pi+2$