A uniform disc has a radius of gyration equal to its radius r about...

Let the mass of the uniform disc be $M$ and its radius be $r$.

The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is $I_{cm} = \dfrac{1}{2} M r^2$.

Let the required axis be at a distance $d$ from the centre. By the theorem of parallel axes, the moment of inertia about this axis is:

$I = I_{cm} + M d^2 = \dfrac{1}{2} M r^2 + M d^2$

The radius of gyration $k$ about this axis is given as $r$. Therefore, the moment of inertia can also be written as:

$I = M k^2 = M r^2$

Equating the two expressions for $I$:

$\dfrac{1}{2} M r^2 + M d^2 = M r^2$

$M d^2 = M r^2 - \dfrac{1}{2} M r^2 = \dfrac{1}{2} M r^2$

$d^2 = \dfrac{r^2}{2}$

$d = \dfrac{r}{\sqrt{2}}$