Question

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

• A. $9Ï€+2:3π−2$
• B. $9π−2:3Ï€+2$
• C. $7π−2:2π−3$
• D. $7Ï€+2:3Ï€+2$

Answer 1

Answer: (B)

$Areaofcircle=Ï€{\left(\sqrt{2}\right)}^2=2Ï€$
Area of $OCADO=2\left\{AreaofOCAO\right\}$
$=2\left\{areaofOCB+areaofBCA\right\}$
$=2\underset{0}{\overset{1}{∫}}{y}_{p}dx+2$$\underset{1}{\overset{\sqrt{2}}{∫}}{y}_{c}dx$
where $y_p=\sqrt{x}$ and $y_c=\sqrt{2−x^2}$
$∴$ Required Area
$=2\underset{0}{\overset{1}{∫}}$$\sqrt{x}dx+2$$\underset{1}{\overset{\sqrt{2}}{∫}}$$\sqrt{2−x^2}dx$
$=\left[\frac{2}{3}.1−0\right]+2{\left[\frac{x\sqrt{2−x^2}}{2}+si{n}^{−1}\frac{x}{\sqrt{2}}\right]}_{{}_{{}_1}}^{{}^{\sqrt{2}}}$
$=\frac{4}{3}+2\left\{\frac{π}{2}−\frac{π}{4}−\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{π}{4}−\frac{1}{2}\right\}=\frac{3π+2}{6}$
Bigger area $=2π−\frac{3π+2}{6}=\frac{9π−2}{6}$
$∴$ Required Ratio $=\frac{9π−2}{3π+2}i.e.,9π−2:3π+2$