7. Given the curve $C: y=\sqrt{-x^{2}-2 x}$ and the line $l: x+y-m=0$ intersect at two points. Then the range of $m$ is ( ).
(A) $(-\sqrt{2}-1, \sqrt{2})$
(B) $(-2, \sqrt{2}-1)$
(C) $[0, \sqrt{2}-1)$
(D) $(0, \sqrt{2}-1)$
7.C.

The curve $(x+1)^{2}+y^{2}=1(y \geqslant 0)$ represents the upper half (including endpoints) of a circle with center at $(-1,0)$ and radius 1. The line $l$ has a slope of -1 and an intercept of $m$. Given that the line $l$ intersects the curve $C$ at two points, we have 
$\frac{|-1-m|}{\sqrt{2}}<1$, and $m \geqslant 0$. Solving this, we get $0 \leqslant m<\sqrt{2}-1$.