Circumcentre of the triangle formed by the points  (2, 3), (0, 1), (4, 1) is  

$Let A (4,1) B(0,1) C(2,3)$

                $Midpoint of BC=(\frac{2+0}{2},\frac{1+3}{2})=(1,2)$

                Slope of BC $=\frac{1-3}{0-2}=1$

$Slope of perpendicular bisector of BC is -1$

                Eq of $\perp r$ bisector  of BC is

                $y-2=-1\left(x-1\right)$

                $\Rightarrow x+y-3=0$ -------(1)

                $N=\left(\frac{4+0}{2},\frac{1+1}{2}\right)=\left(2,1\right)$

                Slope of AB $=\frac{1-1}{4-0}=0$

Slope of perpendicular bisector is $\frac{1}{0}$

                Eq. of bisector of AB is    

                x= 2 -------(2)

                Substituting in  (1)

                x = 2, y = 1

                $\therefore$ circumcentre =  (2,1)