CUET Preparation Today
The smaller of the areas enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is |
$2(\pi-2)$ $\pi-2$ $2 \pi-1$ $2 \pi+2$ |
$\pi-2$ |
Circle in Ist quadrant → $y = \sqrt{4 - x^2}$ area of portion = $\frac{\text{area of circle}}{4} - \text{area of triangle}$ $=\frac{\pi \times 4}{4}-\frac{1}{2} \times 2 \times 2$ $=\pi-2$ Option: 2 |
