Решите уравнение x^2+y^2-12x+4y+40=0.

\[x^{2} + y^{2} - 12x + 4y + 40 = 0\]

\[\left( x^{2} - 12x + 36 \right) + \left( y^{2} + 4y + 4 \right) = 0\]

\[(x - 6)^{2} + (y + 2)^{2} = 0\]

\[\left\{ \begin{matrix}
x - 6 = 0 \\
y + 2 = 0 \\
\end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix}
x = 6\ \ \ \\
y = - 2 \\
\end{matrix} \right.\ \]

\[Ответ:(6;\ - 2).\]