solution: 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]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 12
extra_info.question: Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?
extra_info.split: test