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

Options

A. `(3pi-2)/(10pi+2)`
B. `(3pi+2)/(9pi-2)`
C. `(6pi-3)/(11pi-5)`
D. `(2pi-9)/(9pi+2)`

Answer

B

Explanation

Let the area of the smaller part of circle be `A_(1)` and that of the bigger part be `A_(2)`. We have to find `(A_(1))/(A_(2))`.

The point B is a point of intersection (lying in the first quadrant) of the given parabola and the circle, whose coordinates can be obtained by solving the two equations `y^(2)=2xandx^(2)+y^(2)=8`.
`rArrx^(2)+2x=8`
`rArr(x-2)(x+4)=0`
`rArr x=2,4`
x = - 4 is not possible as both the points of intersection have the same positive x-coordinate.
Thus, `C-=(2,0)`
Now, `A_(1)=2["Area (OBCO) + Area (CBAC)"]`
`=2[int_(0)^(2)y_(1)dx+int_(2)^(2sqrt(2))y_(2)dx]`
where, `y_(1)` and `y_(2)` are respectively the values of y from the equations of the parabola and that of the circle.
or `A_(1)=2[int_(0)^(2)sqrt(2x)dx+int_(2)^(2sqrt(2))sqrt(8-x^(2))dx]`
`rArrA_(1)=2[sqrt(2).(2)/(3).x^(3//2)]_(0)^(2)+2[(x)/(2)sqrt(8-x^(2))+(8)/(2)sin^(-1).(x)/(2sqrt(2))]_(2)^(2sqrt(2))`
`=(16)/(3)+2[2pi+(2+4xx(pi)/(4))]`
`=((4)/(3)+2pi)` sq units
Area of the circle `=pi(2sqrt(2))^(2)=8pi` sq units
Hence, `A_(2)=8pi-A_(1)=6pi-(4)/(3)`
Then, the required ratio `(A_(1))/(A_(2))=((4)/(3)+2pi)/(6pi-(4)/(3))`
`=(2+3pi)/(9pi-2)`