Suppose mass per unit area of the disc $= m$ ∴ Mass of original disc $M = \pi R^2 \times m$ Mass of hole removed from the disc, $M' = \pi \left(\frac{R}{2}\right)^2 \times m$ $= \frac{\pi R^2}{4} m = \frac{M}{4}$ In figure, mass M is concentrated at O and mass $M'$ is concentrated at $O',$ where $OO' = \frac{R}{2}$.After the circular disc of mass $M'$ is removed, the remaining portion can be considered as a system of two masses M at O and $- M'$ at $O'.$If $x$ is the distance of centre of mass (P) of the remaining part, then$x = \frac{M \times 0 - M' \times \frac{R}{2}}{M - M'}$$= \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$ $= \frac{-M R}{8} \times \frac{4}{3 M} = \frac{-R}{6}$Negative sign shows that P is to the left of O.