Explanation

(c)
The coordinates of the centre and radius of given circle are (1, 1) and 2 respectively. Let $AB$ be the chord subtending an angle of $120{^\circ}$ at the centre. Let $M$ be the mid point of $AB$ and let its coordinates be $(h,\ k)$
In $\mathrm{\Delta}OAM$,
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$AM = OA\sin{60{^\circ}}$
$= 2.\frac{\sqrt{3}}{2} = \sqrt{3}$
$\therefore\ \ OM^{2} = OA^{2} - AM^{2}$
$= 4 - \left( \sqrt{3} \right)^{2} = 1$
But $OM^{2} = (h - 1)^{2} + (k - 1)^{2}$
$\therefore\ \ \ (h - 1)^{2} + (k - 1)^{2} = 1$
Hence, locus of $(h,\ k)$ is $(x - 1)^{2} + (y - 1)^{2} = 1$
or $x^{2} + y^{2} - 2x - 2y + 1 = 0$