Example 2. For $(x, y)$ satisfying the equation $x^{2}+(y-1)^{2}=1$, the inequality $x+y+c \geqslant 0$ always holds, find the range of the real number $c$.


The key to solving the problem is to understand the meaning of each formula in the question using the idea of combining numerical and geometric concepts. For example, "the inequality $x+y+c \geqslant 0$ always holds," means that the point $(x, y)$ is always above or on the line $x+y+c=0$. Here, the point $(x, y)$ refers to the points on the circle $x^{2}+(y-1)^{2}=1$. Therefore, as long as $c$ ensures that all points on the circle are on or above the line, it will suffice.

Solution: Let the line $x+y+c=0$ be tangent to the circle $x^{2}+(y-1)^{2}=1$,

i.e.,
$$
\begin{aligned}
& \left\{\begin{array}{l}
x^{2}+(y-1)^{2}=1, \\
y=-x-c
\end{array}\right. \\
\Rightarrow & c_{1}=-1+\sqrt{2}, \\
& c_{2}=-1-\sqrt{2} .
\end{aligned}
$$

Thus, there are two lines tangent to the circle, with y-intercepts of $-c_{1}=1-\sqrt{2}$ and $-c_{2}=1+\sqrt{2}$, respectively. (See Figure 2) According to the problem, we take $-c_{1}=1-\sqrt{2}$, so $-c \leqslant 1-\sqrt{2}$, which means $c \geqslant \sqrt{2}-1$.