Let us take the circle to be a circle with center (0,0) and radius r .

Let P(x,y) be the vertex of the rectangle that lies on the first quadrant.

Let $\theta$ be the angle made y OP with the x-axis.

Then $x=r \cos \theta, y= r \sin \theta$

The dimensions of the rectangle are $2x=2r \cos \theta,2y =2r \sin \theta$

Area of the rectangle are $A=4r^2 \sin \theta \cos \theta$

$A(\theta)= 2r^2 \sin 2 \theta$

$A'(\theta)= 4r^2 \cos 2 \theta=0$

$A''(\theta)= -8r^2 \sin 2 \theta=0$

For max/ min $A'(\theta)=0 => 4r^2 \cos 2 \theta$

=> $\cos 2 \theta=0 => 2 \theta = \large\frac{\pi}{2}$

=> $\theta =\large\frac{\pi}{4}$

when $\theta =\large\frac{\pi}{4}$

$A''(\theta)= -8r^2 \sin 2 \bigg (\large\frac{\pi}{4} \bigg)$

$\qquad= -8r^2 \sin 2 \bigg (\large\frac{\pi}{2} \bigg)$

$\qquad = -8r^2 < 0 $

$\therefore$ A is larger

when $\theta =\large\frac{\pi}{4}$

$2x= 2r \times \large\frac{1}{\sqrt 2}$$=\sqrt 2 r$

$2y=2r \times \large\frac{1}{\sqrt 2}$$=\sqrt 2 r$

The dimensions of the rectangle are $\sqrt 2r$ and $ \sqrt 2r$

The rectangle is square also.