Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance $\frac{\mathrm{R}}{2}$ from the centre of the disc.

For a disc, mass is proportional to surface area. So
Mass of cut portion: $m_1 = k{\mathrm{\pi r}}^2$,
Mass of remaining portion: $m_2 = k\mathrm{\pi }({\mathrm{R}}^2-{\mathrm{r}}^2)$
Apply ${\mathrm{m}}_1{\mathrm{x}}_1 = {\mathrm{m}}_2{\mathrm{x}}_2$ where ${\mathrm{x}}_1 = \frac{\mathrm{R}}{2}$, and solve to get
${\mathrm{x}}_2 = \frac{{\mathrm{Rr}}^2}{2({\mathrm{R}}^2-{\mathrm{r}}^2)}$