\(x^2+y^2-2\left(2x-3y\right)+13=0\)
\(\Leftrightarrow x^2+y^2-4x+6y+13=0\)
\(\Leftrightarrow x^2-4x+4+y^2+6y+9=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=2\\y=3\end{matrix}\right..\)
Vậy \(\left(x;y\right)=\left(2;-3\right)\).