$4.0<m<\frac{\sqrt{2}}{2}$.
Let the intersection point be $P(x, y)$. Then
$$x^{2}+y^{2}=r^{2}=(m x+m y-m)^{2} \text {. }$$

Therefore, $\frac{\sqrt{x^{2}+y^{2}}}{|x+y-1|}=m$, which means
$$\frac{\frac{\sqrt{x^{2}+y^{2}}}{|x+y-1|}}{\sqrt{2}}=\sqrt{2} m \text {. }$$

Thus, the ratio of the distance from point $P$ to the fixed point $(0,0)$ to the distance from $P$ to the fixed line $x+y=1$ is $\sqrt{2} m$.
According to the problem, $0<\sqrt{2} m<1$, which means $0<m<\frac{\sqrt{2}}{2}$.