Suppose, the mass of circular disc is \(M\). A square portion of diagonal \(r\) is cut from it. Assuming centre of mass of circle to be origin. Centre of mass of square portion will be at distance \(\frac{r}{2}\) from it.
\(\therefore\) Side of square \(=\frac{r}{\sqrt{2}}\)
Area of square \(=\left(\frac{r}{\sqrt{2}}\right)^2=\frac{r^2}{2}\)
Mass of square, \(m=M \times \frac{r^2 / 2}{\pi r^2}\)
\(m=\frac{M}{2 \pi}\)
Centre of a mass of remaining disc from the centre of disc,
\(\begin{aligned}
x & =M \times 0-\frac{m \times r / 2}{M-m}=\frac{-\frac{M}{2 \pi} \times \frac{r}{2}}{M-\frac{M}{2 \pi}}=\frac{-r}{4 \pi-2} \\
\Rightarrow \quad x & =\frac{r}{2-4 \pi}
\end{aligned}\)