Find the circumcentre of the triangle with the vertices (–2, 3), (2, –1) and (4, 0).

Let A(–2, 3), B(2, – 1), C(4, 0) and S($\mathrm{\alpha },\mathrm{\beta }$) be the circumcentre
$\therefore \mathrm{SA}=\mathrm{SB}=\mathrm{SC}\Rightarrow {\mathrm{SA}}^2={\mathrm{SB}}^2={\mathrm{SC}}^2$
$\text{ Now }{\mathrm{SA}}^2={\mathrm{SB}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{\alpha }+2{)}^2+(\mathrm{\beta }-3{)}^2=(\mathrm{\alpha }-2{)}^2+(\mathrm{\beta }+1{)}^2\\ \Rightarrow \mathrm{\alpha }-\mathrm{\beta }+1=0 ....\left(1\right)\end{array}$
$\text{ and }{\mathrm{SB}}^2={\mathrm{SC}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{\alpha }-2{)}^2+(\mathrm{\beta }+1{)}^2=(\mathrm{\alpha }-4{)}^2+(\mathrm{\beta }-0{)}^2\\ \Rightarrow 4\mathrm{\alpha }+2\mathrm{\beta }-11=0\dots \dots \left(2\right)\end{array}$
By Solving (1) & (2) $\mathrm{\alpha }=\frac{3}{2};\mathrm{\beta }=\frac{5}{2}$
$\therefore \text{ circumcentre =}\left(\frac{3}{2},\frac{5}{2}\right)$