If a straight line $y-x=2$ divides the region $x^2+y^2 \leq 4$ 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

  $3 \pi-4: \pi+4$
• Correct answer
  D

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

Solution

Let I be the smaller portion and II be the greater portion of the given figure then,
Area of $I=\int_{-2}^0\left[\sqrt{4-x^2}-(x+2] d x\right.$ $$ \begin{aligned} & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{-2}^0-\left[\frac{x^2}{2}+2 x\right]_{-2}^0 \\ & =\left[2 \sin ^{-1}(-1)\right]-\left[-\frac{4}{2}+4\right]=2 \times \frac{\pi}{2}-2=\pi-2 \end{aligned} $$ Now, area of $\mathrm{II}=$ Area of circle $-$ area of $\mathrm{I}$. $$ \begin{aligned} & =4 \pi-(\pi-2) \\ & =3 \pi+2 \end{aligned} $$ Hence, required ratio $=\frac{\text { area of I }}{\text { area of II }}=\frac{\pi-2}{3 \pi+2}$

Hence, the correct answer is option: 'D'.