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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     JEE Main Online Paper (Held On 12 May 2012)

A) \[3\pi -8:\pi +8\]                              

B)                        \[\pi -3:3\pi +3\]

C)                        \[3\pi -4:\pi +4\]                              

D)                        \[\pi -2:3\pi +2\]

Correct Answer: D

Solution :

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