A disc of radius r is cut from a larger disc of radius 4r in such a way that the edge of the hole touches the edge of the disc. The centre of mass of the residual disc will be a distance from centre of larger disc :-

Explanation

Method - I : Figure shows the point C as the centre of the whole disc. `C_(1)` is the centre of the hole and `C_(2)` is the centre of mass of the residual disc. If m be the mass of the disc cut from the larger disc, then mass of the whole disc is 16 m, because mass of a disc varies as the square of its radius.
So, the mass of the residual is `m_(2)=16 m-m=15 m`
Clearly, C is the centre of mass of two masses `m_(1)=m` and `m_(2)=15 m` at `C_(1)` and `C_(2)` respectively, where `C C_(1)=4r-r=3r=x_(1)` (say). Let `C C_(2)=x_(2)`. If we take C as the origin, then `x_(cm)=0`.
Now using `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`, we get `0 = (m_(1)x_(1)+m_(2)(-x_(2)))/(m_(1)+m_(2))`
or, `m_(1)x_(1)=m_(2)x_(2)`, Remember as formula
so, `x_(2)=(m_(1)x_(1))/(m_(2))=((m)(3r))/((15 m))=(r )/(5)`
i.e., `C C_(2)=(r )/(5)`

Method - II : We may also use formula `x_(CM)=(x_(1)A_(1)-x_(2)A_(2))/(A_(1)-A_(2))`
Here `A_(1)=16 pi r^(2), A_(2)=pi r^(2)`
`x_(1)=0` and `x_(2)=3r`
`therefore x_(CM)=(0-3r-pi r^(2))/(16pi r^(2)-pi r^(2))=-(r )/(5)`
`therefore C C_(2)=(r )/(5)`