To solve the problem, we start by analyzing the given conditions and the circles formed:

Step 1: Number of Circles

There are 5 points, no three collinear and no four concyclic. Each circle is formed by 4 points (missing one of the 5 points). The number of such circles is $\binom{5}{4}=5$.

Step 2: Pairs of Circles

The number of pairs of these circles is $\binom{5}{2}=10$.

Step 3: Intersection of Any Two Circles

Take any two circles. Let the 5 points be $A,B,C,D,E$. For example:

• Circle 1: $ABCD$ (missing $E$)
• Circle 2: $BCDE$ (missing $A$)

These two circles share points $B,C,D$? No—since 3 common points would imply the circles are identical (contradicts "no four concyclic"). Thus, they share exactly 2 distinct points (non-collinear, so distinct).

All pairs of circles are distinct and cannot share 3 points (due to no four concyclic), so each pair intersects at exactly two points.

Step 4: Count of Valid Pairs

All $\binom{5}{2}=10$ pairs intersect at exactly two points.

Answer: $\boxed{10}$

$\boxed{10}$

作者声明:本文包含人工智能生成内容。