Explanation

(b)
Since, $E$ is the mid point of $AC$, therefore, the coordinates of $D$ are $\left( \frac{a}{2},\frac{a}{2} \right)$
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Now, $AC = \sqrt{a^{2} + a^{2}} = \sqrt{2}a$
$\therefore\ \ \ AE = \frac{1}{2}AC = \frac{a}{\sqrt{2}}$
$\therefore$ The equation of circle whose centre is $\left( \frac{a}{2},\frac{a}{2} \right)$ and radius $\frac{a}{\sqrt{2}}$, is
$\left( x - \frac{a}{2} \right)^{2} + \left( y - \frac{a}{2} \right)^{2} = \left( \frac{a}{\sqrt{2}} \right)^{2}$
$\Rightarrow \ \ \ x^{2} + y^{2} = a(x + y)$