question_answer

The area bounded by the circle \[{{x}^{2}}+{{y}^{2}}=16\] and the line \[y=x\]in the first quadrant is

A)  \[4\pi \,\,\,sq\,units\]         

B)  \[8\pi \,\,\,sq\,units\]

C)  \[2\pi \,\,\,sq\,units\]         

D)  \[\pi \,\,\,sq\,units\]

Correct Answer: C

Solution :

Given, circle is \[{{x}^{2}}+{{y}^{2}}=16\] ?.(i) and line is \[y=x\] ?..(ii) On solving both equations, we get \[{{x}^{2}}+{{x}^{2}}=16\] \[\Rightarrow \] \[2{{x}^{2}}=16\,\,\,\,\Rightarrow \,\,\,\,{{x}^{2}}=8\] \[\Rightarrow \] \[x=\pm 2\sqrt{2}\] and \[y=\pm 2\sqrt{2}\] \[\therefore \] Intersection points are \[P(2\sqrt{2},2\sqrt{2})\] and \[Q(-2\sqrt{2},\,-2\sqrt{2})\] \[\therefore \]  Required area = Area of \[\Delta OAP+\] Area of \[\Delta APB\] \[=\frac{1}{2}\times OA+AP+\int_{2\sqrt{2}}^{4}{\sqrt{16-{{x}^{2}}}}\,\,dx\] \[=\frac{1}{2}\times 2\sqrt{2}\times 2\sqrt{2}\] \[+\left[ \frac{x}{2}\,\sqrt{16-{{x}^{2}}}+\frac{16}{2}{{\sin }^{-1}}\left( \frac{x}{4} \right) \right]_{2\sqrt{2}}^{4}\] \[=4+\left[ 0+8\times \frac{\pi }{2}-\sqrt{2}\times 2\sqrt{2}-8{{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right) \right]\] \[=4+\left( 4\pi -4-8\times \frac{\pi }{4} \right)\] \[=4+(4\pi -4-2\pi )\] \[=4+2\pi -4=2\pi \] sq units