Step 1: Picture the setup.
A rectangle is drawn inside a circle of radius $r$ so that all four corners touch the circle. We want the largest possible area of such a rectangle.
Step 2: Use an angle variable.
Let one corner make an angle $\theta$ at the centre. Then the half-width is $r\cos\theta$ and the half-height is $r\sin\theta$. The full sides are $2r\cos\theta$ and $2r\sin\theta$.
Step 3: Write the area.
Area is length times breadth: \[ A=(2r\cos\theta)(2r\sin\theta)=4r^2\sin\theta\cos\theta. \]
Step 4: Simplify with a double angle.
Using $2\sin\theta\cos\theta=\sin2\theta$, we get \[ A=2r^2\sin2\theta. \]
Step 5: Maximise the sine.
The largest value of $\sin2\theta$ is $1$, which happens at $2\theta=90^\circ$, that is $\theta=45^\circ$. At that angle the rectangle becomes a square.
Step 6: Get the maximum area.
Putting $\sin2\theta=1$ gives \[ A=2r^2(1)=2r^2. \] So the biggest rectangle has area $2r^2$.
\[ \boxed{2r^2} \]