7. The line $x+y-1=0$ passing through the point $M(1,0)$ intersects the circle $x^{2}+y^{2}=r^{2}$ at points $A$ and $B$, and $A M=2 M B$. Then the radius of the circle $r=$ $\qquad$
7. $\sqrt{5}$.

Let the intersection points of the circle with the $x$-axis be
$$
P(-r, 0), Q(r, 0) \text {. }
$$

Then $M A \cdot M B = M P \cdot M Q$.
Since $A M = 2 M B$, we have,
$$
\begin{array}{l}
2 M B^{2} = M P \cdot M Q \\
= (r+1)(r-1) = r^{2} - 1, \\
A B = 3 M B = 3 \sqrt{\frac{r^{2} - 1}{2}} .
\end{array}
$$

It is easy to calculate the distance $d$ from the center of the circle $O(0,0)$ to the line $x + y - 1 = 0$ as $d = \frac{\sqrt{2}}{2}$.
Also, $r^{2} = d^{2} + \left(\frac{A B}{2}\right)^{2}$, so
$$
r^{2} = \frac{1}{2} + \frac{9}{8}\left(r^{2} - 1\right) \Rightarrow r = \sqrt{5} \text {. }
$$