$5 \mathrm{D}$ Hint: Note that point $P$ satisfies the planar region which is part of an annulus centered at the origin:
$$
1 \leqslant x^{2}+y^{2} \leqslant 9(x \geqslant 0, y \geqslant 0),
$$

Therefore, the minimum value of $|\overrightarrow{P M}|$ is
$$
\sqrt{(1+5)^{2}+(0+5)^{2}}-1=\sqrt{61}-1 \text {. }
$$
(5) Point $P(x, y)$ satisfies the planar region: $\left\{\begin{array}{l}\cos \theta \leqslant x \leqslant 3 \cos \theta, \\ \sin \theta \leqslant y \leqslant 3 \sin \theta,\end{array}(\theta \in \mathbf{R})\right.$ Point $M(x, y)$ satisfies: $(x+5)^{2}+(y+5)^{2}=1$, then the minimum value of $|\overrightarrow{P M}|$ is
(A) $2 \sqrt{2}-1$
(B) $4 \sqrt{2}-1$
(C) $6 \sqrt{2}-1$
(D) $\sqrt{61}-1$