Aspire Faculty ID #16438 · Topic: NIMCET 2009 · Just now
NIMCET 2009

The smaller area bounded by $y = 2 - x$ and $x^2 + y^2 = 4$ is:

Solution

Line intersects circle at two symmetric points. 
Required smaller area = circular segment area. 
 Solve intersection: 
$y = 2 - x$ 
$x^2 + (2 - x)^2 = 4$ 
$x^2 + x^2 - 4x + 4 = 4$ 
$2x^2 - 4x = 0$ 
$2x(x - 2) = 0$ 
 So intersection points at $(0, 2)$ and $(2, 0)$. 
 Using known segment area formula → the smaller area = $\pi - 2$.

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