(b) A circle of radius 2 and centre at $\mathrm{O}$ is drawn. The line $\mathrm{AB}: \mathrm{x}+\mathrm{y}=2$ is passed through $(2,0)$ and $(0,2)$. Area of the region $\mathrm{ACB}$
$=$ Area of quadrant $\mathrm{OAB}-$ Area of $\triangle \mathrm{OAB} \ldots$...(i)
Now $x^2+y^2=4$
$\therefore y^2=4-x^2$ or $y=\sqrt{4-x^2}$
$\therefore$ Area of quadrant $\mathrm{OAB}$
$$
=\int_0^2 y d x=\int_0^2 \sqrt{4-x^2} d x
$$
$=\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_0^2=2 \times \frac{\pi}{2}=\pi$ sq. units
Area of the $\triangle \mathrm{ABO}=$ Area of the region bounded by $A B$.
$A B: x+y=2$ or $y=2-x$ and $x=0, y=0$
$$
=\int_0^2(2-x) d x=\left[2 x-\frac{x^2}{2}\right]_0^2=\left(4-\frac{4}{2}\right)-(0)=2
$$
Putting these values in (i),
Area of region $\mathrm{ACB}=\pi-2$