Suppose mass per unit area of the disc =m
∴ Mass of original disc M=πR2×m
Mass of hole removed from the disc,
M′=π(

R

2

)2×m =

πR2

4

m=

M

4

In figure, mass M is concentrated at O and mass M′ is concentrated at O′, where OO′=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=

M×0−M′× R 2

M−M′

=

− M 4 × R 2

M− M 4

=

−MR

8

×

4

3M

=

−R

6

Negative sign shows that P is to the left of O.