5. Given that $P$ is a point on the line $y=x+1$, and $M, N$ are points on the circles $C_{1}:(x-4)^{2}+(y-1)^{2}=4$ and $C_{2}: x^{2}+(y-2)^{2}=1$ respectively. Then the maximum value of $|P M| - |P N|$ is $(\quad)$.
(A) 4
(B) 5
(C) 6
(D) 7
5.C.

As shown in Figure 2, it is easy to see that the circle $C_{3}: x^{2}+(y-5)^{2}=4$ is symmetric to the circle $C_{1}$:
$$
\begin{array}{l}
(x-4)^{2}+ \\
(y-1)^{2}=4 \text { with respect to the line } y=x+1.
\end{array}
$$

Therefore, for any point $M$ on $C_{1}$, there exists a point $M^{\prime}$ on circle $C_{3}$ such that $|P M| = |P M^{\prime}|$. Thus, we only need to find the maximum value of $|P M^{\prime}| - |P N|$. 
Notice that $|P M^{\prime}| - |P N| \leq |M^{\prime} N| \leq 4 + 2 = 6$. When the points are $P(0,1)$, $M^{\prime}(0,7)$, and $N(0,1)$, the equality holds.