Let $r$ be the radius of the given fixed circle. Let a rectangle be inscribed in this circle with sides $x$ and $y$.

Since the rectangle is inscribed in the circle, the diagonal of the rectangle equals the diameter of the circle.

Therefore,

$$x^2 + y^2 = (2r)^2 = 4r^2 \quad \text{... (1)}$$

The area of the rectangle is $A = xy$.

From equation (1), we get:

$$y = \sqrt{4r^2 - x^2}$$

Substituting this in the area formula:

$$A = x\sqrt{4r^2 - x^2}$$

Differentiating with respect to $x$:

$$\frac{dA}{dx} = \sqrt{4r^2 - x^2} + x \cdot \frac{(-x)}{\sqrt{4r^2 - x^2}}$$ $$\frac{dA}{dx} = \frac{4r^2 - x^2 - x^2}{\sqrt{4r^2 - x^2}}$$ $$\frac{dA}{dx} = \frac{4r^2 - 2x^2}{\sqrt{4r^2 - x^2}}$$

For maximum area, $\frac{dA}{dx} = 0$:

$$4r^2 - 2x^2 = 0$$ $$x^2 = 2r^2$$ $$x = r\sqrt{2}$$

From equation (1):

$$y^2 = 4r^2 - x^2 = 4r^2 - 2r^2 = 2r^2$$ $$y = r\sqrt{2}$$

Therefore, $x = y = r\sqrt{2}$, which means the rectangle is a square.

To verify this gives maximum area, we find the second derivative:

$$\frac{d^2A}{dx^2} = \frac{d}{dx}\left[\frac{4r^2 - 2x^2}{\sqrt{4r^2 - x^2}}\right]$$

At $x = r\sqrt{2}$:

$$\frac{d^2A}{dx^2} = \frac{2r\sqrt{2}(2r^2 - 6r^2)}{(4r^2 - 2r^2)^{3/2}}$$ $$\frac{d^2A}{dx^2} = \frac{-8r^3\sqrt{2}}{2\sqrt{2}\,r^3}$$ $$\frac{d^2A}{dx^2} = -4 < 0$$

Since the second derivative is negative at $x = r\sqrt{2}$, the area is maximum at this point.

Since $x = y$, the rectangle is a square.

∴ Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.