Explanation

(d)
Given equation of circle and line are
$x^{2} + y^{2} = 1$ …(i)
and $x + y = 1$ …(ii)
From Eqs. (i) and (ii),
$x^{2} + (1 - x)^{2} = 1$
$\Rightarrow \ x^{2} + 1 + x^{2} - 2x = 1$
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$\Rightarrow \ 2x^{2} - 2x = 0\ \Rightarrow \ 2x(x - 1) = 0$
$\Rightarrow \ x = 0,\ x = 1\ \Rightarrow \ y = 1,\ y = 0$
$\therefore$ Point of intersection of circle and line are $A(1,\ 0)$ and $B(0,\ 1)$
$\therefore$ Required area $= \int_{0}^{1}{\left\lbrack \sqrt{1 - x^{2}} - (1 - x) \right\rbrack dx}$
$= \left\lbrack \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\sin^{- 1}{x - x + \frac{x^{2}}{2}} \right\rbrack_{0}^{1}$
$= \frac{1}{2} \bullet \frac{\pi}{2} - 1 + \frac{1}{2}$
$= \left( \frac{\pi}{4} - \frac{1}{2} \right)\ sq\ unit$