A disc of radius 'r' is removed from the disc of radius 'R' then

a) The minimum shift in centre of mass is zero

b) The maximum shift in centre of mass cannot be greater than $\frac{{\mathrm{r}}^2}{\left(\mathrm{R}+\mathrm{r}\right)}$

c) Centre of mass must lie where mass exists

d) The shift in centre of mass is $\frac{{\mathrm{r}}^2}{\left(\mathrm{R}+\mathrm{r}\right)}$

There won't be a movement in the mass axis if we separate the smaller disc from the larger disc so that they are both concentric. So there is no minimal shift in the centre of mass. The position from where we remove the smaller disc from the bigger disc determines the value of the shift in the centre of mass. The smaller must be eliminated as shown below to provide the greatest change in the centre of mass.

$$\begin{gathered} ∆X_{cm}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2} \\ ∆X_{cm}=\frac{\sigma \left(\pi R^2\right)\left(0\right)-\sigma \pi r^2(R-r)}{\sigma \pi R^2-\sigma \pi r^2} \\ ∆X_{cm}=-\frac{r^2(R-r)}{R^2-r^2} \\ ∆X_{cm}=-\frac{r^2}{(R+r)} \end{gathered}$$

Hence the correct answer is only an and b are correct.