Решите уравнение х^2+у^2-8х+12у+52=0.

\[x^{2} + y^{2} - 8x + 12y + 52 = 0\]

\[x^{2} - 8x + 16 + y^{2} + 12y + 36 = 0\]

\[(x - 4)^{2} + (y + 6)^{2} = 0\]

\[x - 4 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ y + 6 = 0\]

\[x = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = - 6.\]

\[Ответ:(4;\ - 6)\text{.\ }\]