4. The set of points $P(x, y)$ that satisfy the equation $\arctan x+\operatorname{arccot} y=\pi$ is part of a quadratic curve $C$. Then the coordinates of the focus of $C$ are $\qquad$
4. $(-\sqrt{2}, \sqrt{2}),(\sqrt{2},-\sqrt{2})$.

From the condition $\arctan x+\operatorname{arccot} y=\pi$, it is easy to see that $x>0, y<0$. Therefore, from $\operatorname{arccot} y=\pi-\arctan x$ we get $y=\cot (\pi-\arctan x)=$ $-\cot (\arctan x)=-\frac{1}{x}$. Hence, the curve $C$ is the hyperbola $y=$ $-\frac{1}{x}$, whose asymptotes are the coordinate axes, and the eccentricity is $\sqrt{2}$. Therefore, the line on which the vertices and foci lie is one of the angle bisectors formed by the asymptotes, $y=-x$.
Thus, the two vertices are $(-1,1)$ and $(1,-1)$, so the two foci are $(-\sqrt{2}, \sqrt{2})$ and $(\sqrt{2},-\sqrt{2})$.