Question

平面上画出四个不同的圆。至少两个圆相交的点最多有多少个？

Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$8$

$9$

$10$

$12$

$16$

两个不同的圆最多相交于 $2$ 个点。

Two distinct circles meet in at most $2$ points

四个圆有 $\binom{4}{2}$ 对；再乘以 $2$ 。

There are $\binom{4}{2}$ pairs of circles; multiply by $2$

每一对圆最多有 $2$ 个交点，而四个圆共有 $\binom{4}{2}=6$ 对，因此最多有 $6\cdot2=12$ 个交点。

取四个位置一般的圆，使每一对圆都相交两次，且没有三个圆经过同一点，就能达到这个上界。

所以正确答案是 D。

Each pair of circles meets in at most $2$ points, and there are $\binom{4}{2}=6$ pairs, giving at most $6\cdot2=12$ intersection points.

The bound is attainable by taking four circles in general position so that every pair crosses twice and no three pass through the same point.

Thus, the correct answer is D.