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

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$\pi$ − 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( x/2 \right) + \left( x^$\frac{2}{2}$ - 2x \right) \right]_0^2 = $\frac{1}{2}$ $\times$ 2$\sqrt$(4 - 2^2) + 2 $\times$ \sin^{- 1} \left( $\frac{2}{2}$ \right) + \left( 2^$\frac{2}{2}$ - 2 $\times$ 2 \right) - 0 = 0 + 2 $\times$ $\pi$/2 + \left( 2 - 4 \right) = \left( $\pi$ - 2 \right)\text{ sq units } [Solution diagram: 44145_solution_1.png]

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