7. Given a line passing through the origin (excluding the $x$-axis) intersects the circle $x^{2}+y^{2}=1$ at points $A$ and $B$, and a fixed point $C(2,0)$, then the minimum value of the area of the circumcircle of $\triangle A B C$ is $\qquad$ .
7. $\frac{25 \pi}{16}$.

Let $A(\cos \theta, \sin \theta)$. Then
$$
B(-\cos \theta,-\sin \theta)(\theta \neq k \pi, k \in \mathbf{Z}) \text {. }
$$

Let the circumcenter of $\triangle A B C$ be $(x, y)$. Then
$$
\left\{\begin{array}{l}
y=-x \cot \theta, \\
y-\frac{\sin \theta}{2}=\frac{2-\cos \theta}{\sin \theta}\left(x-\frac{2+\cos \theta}{2}\right) .
\end{array}\right.
$$

Eliminating $y$ gives $x=\frac{3}{4}$, which means the circumcenter lies on the line $x=\frac{3}{4}$.
Thus, the minimum value of the circumradius is $2-\frac{3}{4}=\frac{5}{4}$. Therefore, the minimum value of the circumcircle area is $\frac{25 \pi}{16}$.