Question

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

• A. $3\pi -8:\pi +8$
• B. $\pi -3:3\pi +3$
• C. $3\pi -4:\pi +4$
• D. $\pi -2:3\pi +2$

Answer 1

Answer: (D)

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

$\text{Area}\text{of}I=\underset{-2}{\overset{0}{\int }}\left[\sqrt{4-x^2}-\left(x+2\right)\right]dx$
$={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_{-2}^0-{\left[\frac{x^2}{2}+2x\right]}_{-2}^0$
$=\left[-2{\sin }^{-1}\left(-1\right)\right]-\left(-\frac{4}{2}+4\right)=2\times \frac{\pi }{2}-2=\pi -2$
Now, area of II = Area of circle - area of I.
$=4\pi -(\pi -2)$
$=3\pi +2$
Hence, required ratio $=\frac{\text{ares}\text{of}I}{\text{area}\text{of}II}=\frac{\pi -2}{3\pi +2}$