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The smaller area of the region cut off by $x + y = 2$ from the circle $x^{2} + y^{2} = 4$ is

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$4 π - 2$

$π - 2$

$2 π - 4$

$π - 4$

answer is A.

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Detailed Solution

$x^{2} + y^{2} = 4$

$x + y = 2$

R.A = Area of sector $O A B - A r e a o f Δ O A B$

$= \frac{π (4)}{4} - \frac{1}{2} (2) (2)$

$= π - 2$

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