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

The circle has centre O(0, 0) and radius \(2\sqrt{2}\). The line x = 2 meets it where \(y^2 = 8 - 4 = 4\), that is, at A(2, 2) and B(2, −2).

The smaller region lies between x = 2 and \(x = 2\sqrt{2}\), and it is symmetric about the x-axis, so

    \(\displaystyle \text{Area} = 2\int_{2}^{2\sqrt{2}} \sqrt{8 - x^2}\,dx\)

Using the standard integral

    \(\displaystyle \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\)

with \(a = 2\sqrt{2}\):

    \(\displaystyle \text{Area} = 2\left[\frac{x}{2}\sqrt{8 - x^2} + 4\sin^{-1}\frac{x}{2\sqrt{2}}\right]_{2}^{2\sqrt{2}}\)

At \(x = 2\sqrt{2}\): \(0 + 4\sin^{-1}1 = 4 \times \frac{\pi}{2} = 2\pi\).

At \(x = 2\): \(1 \times \sqrt{4} + 4\sin^{-1}\frac{1}{\sqrt{2}} = 2 + 4 \times \frac{\pi}{4} = 2 + \pi\).

    \(\text{Area} = 2[2\pi - (2 + \pi)]\)
    \(= 2(\pi - 2)\) sq units

Shortcut: OA and OB have slopes 1 and −1, so the angle AOB is a right angle. The region is a quarter-circle sector minus triangle OAB:

    \(\displaystyle \text{Sector} = \frac{1}{4}\pi(2\sqrt{2})^2 = 2\pi\)

    \(\displaystyle \text{Triangle} = \frac{1}{2} \times AB \times 2 = \frac{1}{2} \times 4 \times 2 = 4\)

    \(\text{Area} = 2\pi - 4 = 2(\pi - 2)\)

Why the other options are incorrect

• Option (2) \(8\pi\) sq units — Incorrect. This is the area of the whole circle, \(\pi(2\sqrt{2})^2 = 8\pi\).

• Option (3) \((3\pi+2)\) sq units — Incorrect. This is half of the larger region, \(\frac{1}{2}(6\pi + 4)\), that is, its part above the x-axis.

• Option (4) \(2(3\pi+2)\) sq units — Incorrect. This is the larger region: \(8\pi - (2\pi - 4) = 6\pi + 4\).