Example 11 A set of points $M$ on a plane and 7 different circles $C_{1}, C_{2}, \cdots, C_{k}$ satisfy: circle $C_{k}$ has exactly $k$ points from $M$ $(k=1,2, \cdots, 7)$, then how many points are there in $M$ at least?
Solve: $C_{7}$ has at least $M$ with 7 points, $C_{6}$ has $M$ with 6 points, among which at least $6-2=4$ points are not on $C_{7}$ (since two circles have at most two common points), $C_{5}$ has $M$ with 5 points, among which at least $5-2 \times 2=1$ point is not on $C_{6}$ and $C_{7}$, so the number of points in $M$ $|M|$ satisfies
$$
|M| \geqslant 7+(6-2)+(5-4)=12 \text {. }
$$

On the other hand, as shown in Figure 4-1, 7 circles $C_{1}, C_{2}, \cdots, C_{7}$ satisfy that $C_{k}$ has exactly $M=\left\{p_{1}, p_{2}, \cdots, p_{12}\right\}$ with $k$ points $(k=1,2, \cdots, 7)$.
Therefore, the minimum value of $|M|$ is 12.
$$
\begin{array}{l}
C_{1} \cap M=\left\{P_{3}\right\}, C_{2} \cap M=\left\{P_{9}, P_{10}\right\}, \\
C_{3} \cap M=\left\{P_{3}, P_{4}, P_{12}\right\}, C_{4} \cap M=\left\{P_{2}, P_{4}, P_{9}, P_{10}\right\}, \\
C_{5} \cap M=\left\{P_{1}, P_{5}, P_{8}, P_{11}, P_{12}\right\}, \\
C_{6} \cap M=\left\{P_{6}, P_{7}, P_{8}, P_{9}, P_{10}, P_{11}\right\}, \\
C_{7} \cap M=\left\{P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}\right\} .
\end{array}
$$