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# Choose the correct answer in the smaller area enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ is $\begin{array}{1} (A)\;2(\pi - 2) \qquad & (B)\;\pi - 2 \qquad &(C)\;2\pi - 1 \qquad & (D)\;2 ( \pi + 2) \end{array}$

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• If we are given two or more curves represented by y=f(x) and y=g(x),where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by x=a and x=b by taking the common values of y from the equation of the two curves.
The smaller area enclosed by the circle $x^2+y^2=4$ and the line x+y=2,is represented by the shaded region in the fig.
Let the area enclosed by the circle and the x-axis be $y_1$.since $x^2+y^2=4 \Rightarrow y=\sqrt{4-x^2}.$
Area of $y_1=\int_0^2\sqrt {4-x^2}dx.$
on integrating we get
Area of $y_1=\begin{bmatrix}\frac{x}{2}\sqrt{4-x^2}+\frac{y}{2}\sin^{-1}\frac{x}{2}\end{bmatrix}_0^2$
On applying limits we get,
Area of $y_1=\begin{bmatrix}\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)\end{bmatrix}$
$\qquad\quad\;\;\;\;=2.\frac{\pi}{2}=\pi$ sq.units.
Let the area enclosed by the straight line and the x-axis be $y_2.$
since x+y=2$\Rightarrow y=2-x,$
Area of $y_2=\int_0^2(2-x)dx.$
on integrating we get,
Area of $y_2=\begin{bmatrix}2x-\frac{x^2}{2}\end{bmatrix}_0^2$
On applying limits we get,
$\qquad\quad4-2=2$ sq.units.
Hence the required area is $y_1-y_2=\pi-2$
Hence the required area is $(\pi-2)$ sq. units.