Four mayors decide to build a circular highway that passes within the boundaries of their cities. Since the four cities are not on the same circle, the mayors hire a company to develop a project for the construction of a circular highway equidistant from the four cities. What is the maximum number of geographically distinct projects that the company can develop?
The number of highways is equal to the number of points that can be the center of a circle (highway) that is equidistant from four given points (cities). Since no circle passes through all four points, if any circle is equidistant from the four points, this circle cannot leave all four points on the inside or all on the outside, so it must divide the set of four points into two, without passing through any of them. Thus, we can only have three types of configurations, depending on the number of points inside and outside the circle. In the first type, the equidistant circle leaves three points inside and one outside; in the second type, two inside and two outside; and in the third type, one inside and two outside.

In the figures below, the continuous circle is the equidistant one.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-068.jpg?height=434&width=914&top_left_y=868&top_left_x=594)

In the first figure, the center of the equidistant circle coincides with the center of the circumcircle of the triangle formed by the three internal points. This same configuration occurs in the third type, where the center of the equidistant circle coincides with the center of the circumcircle of the triangle formed by the three external points. Thus, in these two types, the number of equidistant circles is equal to the number of triangles that can be formed with three of the four points, which is four.

In the second figure, the center of the equidistant circle is on the perpendicular bisector of the two internal points and also on the perpendicular bisector of the two external points. Thus, in this type, the number of equidistant circles is equal to the number of ways to divide the set of four points into two sets of two points each, which is three.

Therefore, the number of possible circular highway projects equidistant from the four cities is \(4 + 3 = 7\).