5. The eccentricity of the conic section $\sqrt{x^{2}+y^{2}+6 x-2 y+10}-$ $|x-y+3|=0$ is
5. Solution: The original expression is transformed into $\sqrt{(x+3)^{2}+(y-1)^{2}}=|x-y+3|$, which is $\sqrt{(x+3)^{2}+(y-1)^{2}}=\sqrt{2} \frac{|x-y+3|}{\sqrt{2}}$. Therefore, the ratio of the distance from the moving point $(x, y)$ to the fixed point $(-3,1)$ to its distance to the line $x-y+3=0$ is $\sqrt{2}$. Hence, the locus of this moving point is a hyperbola with an eccentricity of $\sqrt{2}$.