Explanation

(b)
Let mid point of the chord $AB$ is $C(x_{1},y_{1})$
In $\mathrm{\Delta}COB,\ \ \sin\frac{\pi}{4} = \frac{BC}{OB}$
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$\Rightarrow \ \ \frac{1}{\sqrt{2}} = \frac{BC}{2}$
$\Rightarrow \ \ \ BC = \sqrt{2}$
Using Pythagoras theorem,
$OB^{2} = OC^{2} + CB^{2}$
$\Rightarrow (2)^{2} = x_{1}^{2} + y_{1}^{2} + \left( \sqrt{2} \right)^{2}$
$\Rightarrow x_{1}^{2} + y_{1}^{2} = 2$
Hence, locus of mid point of chord is
$x^{2} + y^{2} = 2$