169 Given that the equation $2 x^{2}+2 x y+y^{2}+2 x+2 y-4=0$ represents an ellipse in the Cartesian coordinate system $x O y$, then the coordinates of the center of this ellipse are $\qquad$
$169(0,-1)$. Translate the coordinate system to eliminate the linear terms. Let $x=x^{\prime}+h, y=y^{\prime}+k$, substitute into the equation, and set the coefficients of the linear terms $x^{\prime}$ and $y^{\prime}$ to 0, we get
$$
\left\{\begin{array} { l } 
{ 4 \cdot h + 2 \cdot k + 2 = 0 , } \\
{ 2 \cdot h + 2 \cdot k + 2 = 0 . }
\end{array} \text { that is } \left\{\begin{array}{l}
2 \cdot h+k=-1, \\
h+k=-1 .
\end{array}\right.\right.
$$

So $h=0, k=-1$. Therefore, the center of the ellipse before translation is at $(0,-1)$.