Find the circumcentre of the triangle formed by the points (1, 3), (0, –2), (–3, 1).

Let the given vertices A(1,3), B(0,–2), C(–3,1) $\text{ and }\mathrm{S}(\mathrm{\alpha },\mathrm{\beta })\text{ be the circumcentre }$
$\begin{array}{l}\therefore \mathrm{SA}=\mathrm{SB}=\mathrm{SC}\\ \Rightarrow {\mathrm{SA}}^2={\mathrm{SB}}^2={\mathrm{SC}}^2\end{array}$

$\mathrm{Now} {\mathrm{SA}}^2={\mathrm{SB}}^2\Rightarrow (\mathrm{\alpha }-1{)}^2+(\mathrm{\beta }-3{)}^2 =(\mathrm{\alpha }-0{)}^2+(\mathrm{\beta }+2{)}^2$
$\Rightarrow \mathrm{\alpha }+5\mathrm{\beta }-3=0 ....\left(1\right)$
$\begin{array}{l}\text{ and }{\mathrm{SB}}^2={\mathrm{SC}}^2\Rightarrow (\mathrm{\alpha }-0{)}^2+(\mathrm{\beta }+2{)}^2=(\mathrm{\alpha }+3{)}^2+(\mathrm{\beta }-1{)}^2\\ \end{array}$
$\Rightarrow \mathrm{\alpha }-\mathrm{\beta }+1=0 ...\left(2\right)$
Solve (1) & (2)
$$\begin{gathered} \therefore \mathrm{\alpha }=-1/3,\mathrm{\beta }=2/3 \\ \therefore \text{ circumcentre }\mathrm{S}=\left(\frac{-1}{3},\frac{2}{3}\right) \end{gathered}$$