Theorem: “A perpendicular from the Centre of a circle to a chord, bisects the chord.” (Refer: Proof)
In this case, if a perpendicular is drawn from the center of the circle, it will bisect the base side of the rectangle.
The half-length of this chord would be: $a\times \cos 45^{\circ}=\frac{a}{\sqrt{2}}$.
Hence the length of the rectangle would be $2\times \frac{a}{\sqrt{2}}=a\sqrt{2}$
Area of circle $: \pi a^{2}$ and the area of rectangle $: 2a^{2}$
$\therefore$ Area of remaining portion would be $:\pi a^{2}-2a^{2}$
Option C is correct.