A circular disc of diameter d and a square plate of edge d are...

Let the centre of the circular disc be the origin $(0, 0)$.

Let $\sigma$ be the mass per unit area of both plates.

The mass of the circular disc is $m_1 = \sigma \pi \left(\dfrac{d}{2}\right)^2 = \dfrac{\sigma \pi d^2}{4}$.
The centre of mass of the disc is at $x_1 = 0$.

The mass of the square plate is $m_2 = \sigma d^2$.
Since the square plate touches the disc at the midpoint of its edge, the distance from the centre of the disc to the centre of the square plate is the sum of the radius of the disc and half the edge length of the square. Thus, the centre of mass of the square plate is at $x_2 = \dfrac{d}{2} + \dfrac{d}{2} = d$.

The position of the centre of mass of the combination is given by:

$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Substituting the values, we get:

$x_{cm} = \dfrac{\left(\dfrac{\sigma \pi d^2}{4}\right)(0) + (\sigma d^2)(d)}{\dfrac{\sigma \pi d^2}{4} + \sigma d^2}$

$x_{cm} = \dfrac{\sigma d^3}{\sigma d^2 \left(\dfrac{\pi}{4} + 1\right)}$

$x_{cm} = \dfrac{d}{\dfrac{\pi + 4}{4}} = \dfrac{4d}{4 + \pi}$

Therefore, the centre of mass of the combination is at a distance of $\dfrac{4d}{4 + \pi}$ to the right of the centre of the disc.