The area of the region lying above $X$ -axis, and included between the circle $x^{2}+y^{2}=2 a x$ and the parabola $y^{2}=a x, a>0$ is

• A

  $8 \pi a^{2}$
• Correct answer
  B

  $a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$
• C

  $\frac{16 \pi a^{2}}{9}$
• D

  $\pi\left(\frac{27}{8}+3 a^{2}\right)$

Solution

Given, equation of circle $x^{2}+y^{2}=2 a x$ $\Rightarrow$ $(x-a)^{2}+y^{2}=a^{2}$ and equation of parabola is $y^{2}=a x, a>0$ Intersection points of circle and parabola $\begin{array}{lr}\Rightarrow & x^{2}+a x=2 a x \\ \Rightarrow & x^{2}=a x\end{array}$ $\Rightarrow \quad x^{2}-a x=0$ $\Rightarrow \quad x(x-a)=0$ $\Rightarrow \quad x=0, a$ Intersecting points are $(0,0)$ and $(a, a)$ $\therefore$ Required area $=\frac{\pi a^{2}}{4}-\int_{0}^{a} \sqrt{a x} d x$ $=\frac{\pi a^{2}}{4}-\sqrt{a}\left(\frac{x^{3 / 2}}{3 / 2}\right)_{0}^{a}$ $=\frac{\pi a^{2}}{4}-\frac{2 a^{2}}{3}=a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

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