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 $A_1=2\left[\underset{0}{\overset{2}{\int }}\sqrt{2x}dx+\underset{2}{\overset{2\sqrt{2}}{\int }}\sqrt{8-x^2}dx\right]$

$\Rightarrow A_1=2{\left[\sqrt{2}\cdot \frac{2}{3}{x}^{3/2}\right]}_0^2+2{\left[\frac{x}{2}\sqrt{8-x^2}+\frac{8}{2}si{n}^{-1}\frac{x}{2\sqrt{2}}\right]}_2^{2\sqrt{2}}$

$=\frac{16}{3}+2\left[2\pi -\left(2+4\times \frac{\pi }{4}\right)\right]=\left(\frac{4}{3}+2\pi \right)$ 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
$\frac{A_1}{A_2}=\frac{\frac{4}{3}+2\pi }{6\pi -\frac{4}{3}}$
$=\frac{2+3\pi }{9\pi -2}$