From a uniform disk of radius $R$,a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

(N/A) Let the mass per unit area of the original disc be $\sigma$.
The radius of the original disc is $R$. The mass of the original disc is $M = \pi R^2 \sigma$.
The radius of the smaller cut-out disc is $R/2$. The mass of the smaller disc is $M' = \pi (R/2)^2 \sigma = \frac{1}{4} \pi R^2 \sigma = M/4$.
Let $O$ be the centre of the original disc and $O'$ be the centre of the cut-out disc. The distance $OO' = R/2$.
We treat the remaining body as a system of two masses: mass $M$ at $O$ and mass $-M' = -M/4$ at $O'$.
Taking $O$ as the origin $(0,0)$,the centre of mass $x_{cm}$ of the remaining body is given by:
$x_{cm} = \frac{M(0) + (-M')(R/2)}{M - M'} = \frac{0 - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -R/6$.
The negative sign indicates that the centre of mass shifts by $R/6$ from the original centre $O$ in a direction opposite to the centre of the cut-out portion.