Smaller area enclosed by the circle x^2+y^2=4 and the line x+y=2 is

The correct answer is Option (1) → \( (\pi-2) \) Sq. units

The line x + y = 2 meets the circle of radius 2 at (2, 0) and (0, 2). The smaller region lies in the first quadrant between the arc and the chord.

    \( \displaystyle \text{Area}=\int_{0}^{2}\sqrt{4-x^2}\,dx-\int_{0}^{2}(2-x)\,dx \)

    \( \displaystyle \int_{0}^{2}\sqrt{4-x^2}\,dx=\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}\right]_{0}^{2} \)

    \( \displaystyle =2\cdot\frac{\pi}{2}=\pi \)

    \( \displaystyle \int_{0}^{2}(2-x)\,dx=\left[2x-\frac{x^2}{2}\right]_{0}^{2} \)

    \( \displaystyle =4-2=2 \)

    \( \displaystyle \text{Area}=(\pi-2)\ \text{sq. units} \)

Shortcut: The region is a quarter circle minus a right triangle: \( \frac{1}{4}\pi(2)^2-\frac{1}{2}(2)(2)=\pi-2 \).