Question

The parabola $y^{2} = 2x$ divides the circle $x^{2 }+ y^{2} = 8$ in two parts. Then, the ratio of the areas of these parts is

• A. $\left(3\pi-2\right) : \left(10\pi+2\right)$
• B. $\left(3\pi+2\right) : \left(9\pi-2\right)$
• C. $\left(6\pi-3\right): \left(11\pi-5\right)$
• D. $\left(2\pi-9\right): \left(9\pi+2\right)$

Answer 1

Answer: (B)

We have $y^{2} = 2x ...\left(i\right)$, and $x^{2} + y^{2} = 8\, ...\left(ii\right)$, a
circle with centre $\left(0,0\right)$ and radius $2\sqrt{2}$.
Let the area of the smaller part of the circle be $A_{1}$ and that of the bigger part be $A_{2}$. We have to find $\frac{A_{1}}{A_{2}}$.

On solving $\left(i\right)$ and $\left(ii\right)$, we get $x = 2, -4$
$x = -4$ is not possible as both the points of intersection have the same positive $x$-coordinate.
Thus, $C\equiv\left(2,0\right)$
Now, $A_{1}=2\left[\text{Area}\left(OBCO\right)+\,\text{Area}\left(CBAC\right)\right]$
or $\quad$ $A_{1}=2\left[\int\limits_{0}^{2}\sqrt{2x}\,dx+\int\limits_{2}^{2\sqrt{2}}\sqrt{8-x^{2}} \,dx\right]$

$\Rightarrow A_{1}=2\left[\sqrt{2}\cdot\frac{2}{3}x^{3 /2}\right]_{0}^{2}+2\left[\frac{x}{2}\sqrt{8-x^{2}}+\frac{8}{2}sin^{-1}\frac{x}{2\sqrt{2}}\right]_{2}^{2\sqrt{2}}$

$=\frac{16}{3}+2\left[2\pi-\left(2+4\times\frac{\pi}{4}\right)\right]=\left(\frac{4}{3}+2\pi\right)$ sq. units

Area of circle $=\pi\left(2\sqrt{2}\right)^{2}=8\pi$ sq. units
Hence, $A_{2}=8\pi-A_{1}=\left(6\pi-\frac{4}{3}\right)$ sq. units
Then, the required ratio is
$\frac{A_{1}}{A_{2}}=\frac{\frac{4}{3}+2\pi}{6\pi-\frac{4}{3}}$
$=\frac{2+3\pi}{9\pi-2}$