Exam

JEE Main

Year

2012

Date

12 May

Mode

Online

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2012 · 12 May · Online · Mathematics · Area Under Curves

If a straight line y-x=2 divides the region x^2+y^2 ≤ 4 into two...

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

Step-by-step 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}