5. Regardless of the real number $m$, the circle $x^{2}+y^{2}+2 m x-m y-25=0$ always passes through two fixed points, their coordinates are $\qquad$ .
5. $(\sqrt{5}, 2 \sqrt{5}) 、(-\sqrt{5},-2 \sqrt{5})$. Hint: The equation can be transformed into: $x^{2}+y^{2}-25+m(2 x-y)=0$. Let $\left\{\begin{array}{l}2 x-y=0 \\ x^{2}+y^{2}=25\end{array}\right.$, and you will get the result.