Here, $\mathrm{A}(-2,3), \mathrm{B}(6,-1), \mathrm{C}(4,3)$ are the vertices of $\triangle \mathrm{ABC}$.
Let $\mathrm{F}$ be the circumcentre of $\triangle \mathrm{ABC}$.
Let $\mathrm{FD}$ and $\mathrm{FE}$ be the perpendicular bisectors of the sides $\mathrm{BC}$ and $\mathrm{AC}$ respectively.
$\therefore \quad \mathrm{D}$ and $\mathrm{E}$ are the midpoints of side $\mathrm{BC}$ and $\mathrm{AC}$ respectively.
$\begin{array}{ll}\therefore & D \equiv\left(\frac{6+4}{2}, \frac{-1+3}{2}\right) \\ \therefore & D=(5,1) \\ & \text { and } E \equiv\left(\frac{-2+4}{2}, \frac{3+3}{2}\right)\end{array}$
$\therefore \quad \mathrm{E}=(1,3)$


Now, slope of $B C=\frac{3-(-1)}{4-6}=\frac{4}{-2}=-2$
$\therefore \quad \text { Slope of } \mathrm{FD}=\frac{1}{2} \quad \ldots[\because \mathrm{FD} \perp \mathrm{BC}]$
Since FD passes through $(5,1)$ and has slope $\frac{1}{2}$, equation of $\mathrm{FD}$ is
$\begin{aligned}
& y-1=\frac{1}{2}(x-5) \\
\therefore \quad & 2(y-1)=x-5 \\
\therefore \quad & 2 y-2=x-5 \\
\therefore \quad & x-2 y-3=0
\end{aligned}$
Since both the points $\mathrm{A}$ and $\mathrm{C}$ have same $y$ co-ordinates i.e. 3 ,
the given points lie on the line $y=3$.
Since the equation $\mathrm{FE}$ passes through $\mathrm{E}(1,3)$, the equation of $\mathrm{FE}$ is $x=1$.
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of $x$ in (i), we get
$\begin{array}{ll}
& 1-2 y-3=0 \\
\therefore \quad & y=-1 \\
\therefore \quad & \text { Co-ordinates of circumcentre } \mathrm{F} \equiv(1,-1) .
\end{array}$