The circumcenter of the triangle formed by lines $x \mathrm{y}+2 x+2 \mathrm{y}+4=0$ and $x+\mathrm{y}+2=0$ is

The given equation $xy + 2x + 2y + 4 = 0$ factors into $(x + 2)(y + 2) = 0$, yielding the lines $x = -2$ and $y = -2$. A third line is given as $x + y + 2 = 0$.

Find the vertices of the triangle:
Intersection of $x = -2$ and $y = -2$ is $(-2, -2)$.
Intersection of $x = -2$ and $x + y + 2 = 0$ gives $y = 0$, vertex $(-2, 0)$.
Intersection of $y = -2$ and $x + y + 2 = 0$ gives $x = 0$, vertex $(0, -2)$.

With vertices at $(-2, -2)$, $(-2, 0)$, and $(0, -2)$, the triangle is right-angled at $(-2, -2)$. The hypotenuse is between $(-2, 0)$ and $(0, -2)$, so its midpoint is the circumcenter.

Midpoint $= \left( \frac{-2 + 0}{2}, \frac{0 + (-2)}{2} \right) = (-1, -1)$.

Circumcenter: $(-1, -1)$