Question

The area bounded by the circle $x^2+y^2=16$ and the line $y=x$ in the first quadrant is

• A. $4\pi squnits$
• B. $8\pi squnits$
• C. $2\pi squnits$
• D. $\pi squnits$

Answer 1

Answer: (C)

Given, circle is $x^2+y^2=16$ ..(i)
and line is $y=x$ ..(ii)
On solving both equations, we get
$x^2+x^2=16$
$\Rightarrow$ $2x^2=16\Rightarrow x^2=8$
$\Rightarrow$ $x=\pm 2\sqrt{2}$
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