The area of the smaller region of the circle x^2 + y^2 = 8 cut off by…

The correct answer is Option (1) → $2(\pi-2)$ sq units

Equation of circle: $x^{2}+y^{2}=8$

Center $(0,0)$, radius $r=\sqrt{8}=2\sqrt{2}$

Line: $x=2$

Perpendicular distance of line from center:

$d=|2-0|=2$

Using formula for area of circular segment:

$A = r^{2}\cos^{-1}\!\left(\frac{d}{r}\right) - d\sqrt{r^{2}-d^{2}}$

Substitute $r=2\sqrt{2}$, $d=2$:

$A = (2\sqrt{2})^{2}\cos^{-1}\!\left(\frac{2}{2\sqrt{2}}\right) - 2\sqrt{(2\sqrt{2})^{2}-2^{2}}$

$A = 8\cos^{-1}\!\left(\frac{1}{\sqrt{2}}\right) - 2\sqrt{8-4}$

$A = 8\left(\frac{\pi}{4}\right) - 2\sqrt{4}$

$A = 2\pi - 4$

Area = $2\pi - 4$ sq. units