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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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