We have, $x=2+3 \,\cos\, \theta$
and $y=1-3 \,\sin\, \theta$
$\Rightarrow (x-2)=3 \,\cos \,\theta\dots$(i)
and $(y-1)=-3\, \sin\, \theta \dots$(ii)
On squaring and adding Eqs. (i) and (ii), we get
$(x-2)^{2}+(y-1)^{2}=9$
$\therefore $ Centre $=(2,1)$ and radius $=3$