5. On a plane, there is a set of points $M$ and seven different circles $C_{1}, C_{2}, \cdots, C_{7}$, where circle $C_{7}$ passes through exactly 7 points in $M$, circle $C_{6}$ passes through exactly 6 points in $M$, $\cdots \cdots$, and circle $C_{1}$ passes through exactly 1 point in $M$. The minimum number of points in $M$ is
(A) 11
(B) 12
(C) 21
( D) 28
Answer $(\quad)$
(B)
5.【Analysis and Solution】To minimize the number of points in $M$, the circles should pass through as many common points in $M$ as possible. Since two distinct circles can have at most two common points, and $C_{7}$ passes through 7 points in $M$, $C_{6}$ must pass through at least four other points in $M$. $C_{5}$, in addition to having two common points with $C_{7}$ and $C_{6}$, must also pass through one point in $M$, so $M$ must have at least 12 points.
On the other hand, if there are 12 points, the requirements can be met:
First, construct three circles that intersect pairwise, denoted as $C_{7}$, $C_{6}$, and $C_{5}$. Then construct circle $C_{4}$ to intersect with two circles $C_{6}$ and $C_{7}$ at additional points (but not intersecting with $C_{5}$). Next, construct circle $C_{3}$ passing through one intersection point of $C_{5}$ and $C_{7}$, such that it intersects with $C_{5}$ and $C_{7}$ at new points. Count all the intersection points as points in $M$. At this point, $M$ has 12 points. $C_{7}$ intersects with $C_{6}$, $C_{5}$, and $C_{4}$ at two points each, and intersects with $C_{3}$ at one new point, totaling 7 points; $C_{6}$ intersects with $C_{7}$, $C_{5}$, and $C_{4}$ at two points each, totaling 6 points; $C_{5}$ intersects with $C_{7}$ and $C_{6}$ at two points each, and intersects with $C_{3}$ at one new point, totaling 5 points; $C_{4}$ intersects with $C_{7}$ and $C_{6}$ at two points each, totaling 4 points; $C_{3}$ passes through one intersection point of $C_{7}$ and $C_{5}$, and intersects with each at one new point, totaling 3 points. Finally, selecting two points and one point from the 12 points in $M$ to construct the required circles $C_{2}$ and $C_{1}$ is straightforward. Therefore, the answer is $(\mathrm{B})$.