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,

$\text{Area}\, \text{of} \,I = \int\limits^{0}_{-2} \left[\sqrt{4-x^{2}} - \left(x+2\right)\right]dx$
$ = \left[\frac{x}{2} \sqrt{4-x^{2} } + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}\right)\right]^{0}_{-2} - \left[\frac{x^{2}}{2} + 2x\right]^{0}_{-2} $
$= \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} $