Find the circumcenter of the triangle whose sides are $\mathrm{x}+\mathrm{y}=0,2\mathrm{x}+\mathrm{y}+5=0,\mathrm{x}-\mathrm{y}=2$

$$\begin{gathered} \text{ Given lines are} \\ \mathrm{x}+\mathrm{y}=0 ....\left(1\right) \end{gathered}$$
$\begin{array}{l}2\mathrm{x}+\mathrm{y}+50=0 .....\left(2\right)\\ \mathrm{x}-\mathrm{y}-2=0 .....\left(3\right)\end{array}$

solving (1) & (2)
$\begin{array}{ccccc} & \mathrm{x} \mathrm{y}& 1& & \\ 1& 0 & 1& 1& \\ 1& 5& 2& 1& \end{array}$
$\begin{array}{l}\frac{\mathrm{x}}{5-0}=\frac{\mathrm{y}}{0-5}=\frac{1}{1-2}\Rightarrow \frac{\mathrm{x}}{5}=\frac{\mathrm{y}}{-5}=\frac{1}{-1}\\ \frac{\mathrm{x}}{5}=\frac{1}{-1}\Rightarrow \mathrm{x}=-5;\frac{\mathrm{y}}{-5}=\frac{1}{-1}\Rightarrow \mathrm{y}=5\\ \mathrm{A}=(-5,5)\end{array}$
Solving (2) & (3)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ 1 5 2 1 \\ -1 -2 1 -1 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{-2+5}=\frac{\mathrm{y}}{5+4}=\frac{1}{-2-1}\\ \frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{9}=\frac{1}{-3}\Rightarrow \frac{\mathrm{x}}{3}=\frac{1}{-3}\Rightarrow \mathrm{x}=-1\\ \frac{\mathrm{y}}{9}=\frac{1}{-3}\Rightarrow \mathrm{y}=-3;\mathrm{B}=(-1,-3)\end{array}$
Solving (3) & (1)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ -1 -2 1 -1 \\ 1 0 1 1 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{0+2}=\frac{\mathrm{y}}{-2-0}=\frac{1}{1+1}\\ \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-2}=\frac{1}{2}\Rightarrow \frac{\mathrm{x}}{2}=\frac{1}{2}\Rightarrow \mathrm{x}=1\\ \frac{\mathrm{y}}{-2}=\frac{1}{2}\Rightarrow \mathrm{y}=-1;\mathrm{C}=(1,-1)\end{array}$
$\mathrm{A}=(-5,5),\mathrm{B}(-1,3),\mathrm{C}=(1,-1)$
$\text{ Let }\mathrm{S}=(\mathrm{x},\mathrm{y})\text{ be a circumcenter }$
$\begin{array}{l}\mathrm{SA}=\mathrm{SB}=\mathrm{SC}\text{ S.O.B.S }\\ {\mathrm{SA}}^2={\mathrm{SB}}^2={\mathrm{SC}}^2\end{array}$
$\text{ Let }{\mathrm{SA}}^2+{\mathrm{SB}}^2$
$\begin{array}{l}(\mathrm{x}+5{)}^2+(\mathrm{y}-5{)}^2=(\mathrm{x}+1{)}^2+(\mathrm{y}-3{)}^2\\ {\mathrm{x}}^2+25+10\mathrm{x}+{\mathrm{y}}^2+25-10\mathrm{y}={\mathrm{x}}^2+1+2\mathrm{x}+{\mathrm{y}}^2+9-6\mathrm{y}\\ \\ \Rightarrow 10\mathrm{x}-10\mathrm{y}+50-2\mathrm{x}+6\mathrm{y}-10=0\\ \Rightarrow 8\mathrm{x}-4\mathrm{y}+40=0\\ \Rightarrow 2\mathrm{x}-\mathrm{y}+10=0\dots \text{ (4) }\end{array}$
$\text{ Let }{\mathrm{SB}}^2={\mathrm{SC}}^2$
$\begin{array}{l}(\mathrm{x}+1{)}^2+(\mathrm{y}-3{)}^2=(\mathrm{x}-1{)}^2+(\mathrm{y}+1{)}^2\\ {\mathrm{x}}^{\text{⧸}2}+1+2\mathrm{x}+{\mathrm{y}}^2+9-6\mathrm{y}={\mathrm{x}}^2+1-2\mathrm{x}+{\mathrm{y}}^2+1+2\mathrm{y}\\ \\ \Rightarrow 2\mathrm{x}+9-6\mathrm{y}+2\mathrm{x}-1-2\mathrm{y}=0\\ \Rightarrow 4\mathrm{x}-8\mathrm{y}+8=0\\ \Rightarrow \mathrm{x}-2\mathrm{y}+2=0 .....\left(5\right)\end{array}$
Solving (4) & (5)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ -1 10 2 -1 \\ -2 2 1 -2 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{-2+10}=\frac{\mathrm{y}}{10-4}=\frac{1}{-4+1}\\ \frac{\mathrm{x}}{18}=\frac{\mathrm{y}}{6}=\frac{1}{-3}\end{array}$
$\mathrm{x}=\frac{18}{-3}=-6;\mathrm{y}=\frac{6}{-3}=-2$
Circum center S = (–6, –2)