The circumcentre of a triangle with vertices $\left(-2,3\right)\left(2,-1\right)$  and $\left(4,0\right)$  is

$D=$  midpoint of $\overline{AB}$  $=\left(\frac{-2+2}{2},\frac{3-1}{2}\right)=\left(0,1\right)$

Slope of $\overline{AB}=\frac{3+1}{-2-2}=\frac{4}{-4}=-1$

Slope of $l_1=1$                   $\left(\because l_1\perp \overline{AB}\right)$

Equation of $l_1:\left(y-1\right)=1\left(x-0\right)$

$y-1=x$

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

$E=$  midpoint of $BC=\left(\frac{2+4}{2},\frac{-1+0}{2}\right)=\left(3,\frac{-1}{2}\right)$

Slope of $\overline{BC}=\frac{1}{2}$

Slope of $l_2=-2$

$\Rightarrow$  Equation of $l_2:\left(y=\frac{1}{2}\right)=-2\left(x-3\right)$

$2y+1=-4\left(x-3\right)$

$2y+1=-4x+12$

$4x+2y=11\to \left(2\right)$

$\begin{array}{c}\left(2\right)\times \left(1\right)\Rightarrow 2x-2y=-2\\ \underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\left(2\right)\Rightarrow 4x+2y=11}\\ 6x=9\end{array}$

$x=\frac{9}{6}=\frac{3}{2}$

$\Rightarrow x=\frac{3}{2}$

$x-y=-1$

$x+1=y$

$\frac{3}{2}+1=y$

$\frac{5}{2}=y$

$\therefore$  circum centre $=\left(\frac{3}{2},\frac{5}{2}\right)$