7. In the Cartesian coordinate system $x O y$, point $A(-1, -1)$, let $B$ and $C$ be two different points on the curve $x y=1(x>0)$, and $\triangle A B C$ is an equilateral triangle. Then the area of its circumcircle is $\qquad$
$7.8 \pi$.
Since points $B$ and $C$ are symmetric with respect to the altitude through point $A$ on side $BC$, and point $A$ lies on the axis of symmetry of the curve $xy=1 (x>0)$, the slope of line $BC$ must be -1.
Let the line $l_{BC}: y=-x+b (b>0)$.
Substituting into $xy=1$, we get $x^2 - bx + 1 = 0$.
Thus, $\Delta = b^2 - 4 > 0$.
Solving, we get $b > 2$.
Also, $x_1 + x_2 = b, x_1 x_2 = 1$, where $x_1, x_2$ are the x-coordinates of points $B$ and $C$. Let $D$ be the midpoint of side $BC$. Then $D\left(\frac{b}{2}, \frac{b}{2}\right)$.
Therefore, $|AD| = \frac{\sqrt{3}}{2} |BC|$,
$$
\begin{array}{l}
|AD|^2 = 2\left(\frac{b}{2} + 1\right)^2, \\
|BC|^2 = 2\left(x_1 - x_2\right)^2 = 2\left[\left(x_1 + x_2\right)^2 - 4 x_1 x_2\right] \\
= 2\left(b^2 - 4\right).
\end{array}
$$

Solving, we get $b = 4$ (discard $b = -2$).
Thus, point $D(2,2), |AD| = 3\sqrt{2} = \frac{3}{2} R$.
Therefore, $R = 2\sqrt{2}$.
Hence, $S = 8\pi$.