Each pair of circles has at most two intersection points. There are $\binom{4}{2} = 6$ pairs of circles, so there are at most $6\times 2 = 12$ points of intersection. The following configuration shows that $\boxed{12}$ points of intersection are indeed possible:

[asy]
draw(Circle((0,0),2));
draw(Circle((1,1.3),2));
draw(Circle((-1,1.3),2));
draw(Circle((0,0.7),2));
[/asy]
Question:
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
Answer:
Let's think step by step.

user
12
Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
test