Question

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is

विकल्प

• 2 (π − 2)

• π − 2

• 2π − 1

• 2 (π + 2)

Answer 1

π − 2

We have, x² + y² = 4 represents a circle with centre at O(0,0) and radius 2
x + y = 2 represents a straight line  cutting the x-axis at A(2, 0) and y axis at B(0, 2) 
Thus , A (2,0) and B(0,2) are also the points of intersection of the straight line and the circle
Smaller area enclosed by the curve and straight line is the shaded area
\[\text{ Shaded area }\left( ABCA \right)\]
\[ =\text{ area }\left( OBCA \right) - \text{ area }\left( OBAO \right)\]
\[ = \int_0^2 \sqrt{4 - x^2} dx\] - \[\int_0^2 \left( 2 - x \right)dx  ..............\left[ \because x^2 + y^2 = 4 \Rightarrow y = \sqrt{4 - x^2} \text{ and }x + y = 2 \Rightarrow y = 2 - x \right] \]
\[ = \int_0^2 \left[ \left( \sqrt{4 - x^2} \right) + x - 2 \right]dx\]
\[ = \left[ \frac{1}{2}x\sqrt{4 - x^2} + \frac{1}{2} \times 4 \times \sin^{- 1} \left( \frac{x}{2} \right) + \left( \frac{x^2}{2} - 2x \right) \right]_0^2 \]
\[ = \frac{1}{2} \times 2\sqrt{4 - 2^2} + 2 \times \sin^{- 1} \left( \frac{2}{2} \right) + \left( \frac{2^2}{2} - 2 \times 2 \right) - 0 \]
\[ = 0 + 2 \times \frac{\pi}{2} + \left( 2 - 4 \right) \]
\[ = \left( \pi - 2 \right)\text{ sq units }\]