3. The equation of the circle passing through the intersection points of the curves $y=\frac{1}{r}$ and $\mathrm{y}=\mathrm{x}^{2}+3 \mathrm{x}-7$ is $\qquad$ ;
Solution: The intersection point $(x, y)$ satisfies $\left\{\begin{array}{l}xy=1 \\ y=x^{2}+3x-7\end{array}\right.$, which leads to $\left\{\begin{array}{l}y^{2}=x-7y+3 \\ x^{2}=y-3x+7\end{array}\right.$, adding these equations gives $x^{2}+y^{2}+2x+6y=10$, i.e., $(x+1)^{2}+(y+3)^{2}=20$ is the equation of the required circle.