Exercise 6. Given a point $P$ and a circle $\mathcal{C}$ in the plane, the distance from $P$ to $\mathcal{C}$ is defined as the minimum length $PM$ between $P$ and a point $M$ on the circle $\mathcal{C}$. For example, if $P$ lies on the circle, then the distance from $P$ to $\mathcal{C}$ is zero, and if $P$ is the center of the circle, then the distance from $P$ to $\mathcal{C}$ is equal to the radius of $\mathcal{C}$.

Given four non-cyclic points $A, B, C, D$, what is the maximum number of circles that pass at equal distance from these four points?
Solution to Exercise 6: Let $\mathcal{C}$ be a circle passing at equal distance from $A, B, C, D$. The four points cannot all be inside the circle, otherwise they would be equidistant from the center, and thus would be concyclic. Similarly, they cannot all be outside.

Determine the number of circles $\mathcal{C}$ passing at equal distance from $A, B, C, D$ such that $A, B, C$ are on the same side of the circle. Suppose $A, B, C$ are not collinear. Let $\Omega$ be the center of the circumcircle of $A, B, C$ and $R$ its radius. Let $r$ be the radius of $\mathcal{C}$ and $d = \Omega D$. The fact that $A, B, C, D$ are equidistant from $\mathcal{C}$ can be written as $|r - R| = |r - d|$. Moreover, $r - R$ and $r - d$ cannot be equal, otherwise we would have $d = R$ and $A, B, C, D$ would be concyclic. The equality $|r - R| = |r - d|$ is therefore equivalent to $r - R = d - r$, or $r = \frac{R + d}{2}$, which proves the existence and uniqueness of the circle $\mathcal{C}_{D}$ such that $A, B, C$ are on one side and $D$ on the other side of the circle.

(If $A, B, C$ are collinear, no point is equidistant from $A, B, C$, so there does not exist a circle passing at equal distance from $A, B, C$ such that $A, B, C$ are all on the same side of the circle.)

We similarly define the circles $\mathcal{C}_{A}, \mathcal{C}_{B}$, and $\mathcal{C}_{C}$ if they exist.

![](https://cdn.mathpix.com/cropped/2024_06_04_6f47bf1fc05b53b522d5g-4.jpg?height=591&width=656&top_left_y=1683&top_left_x=732)

Determine the number of circles $\mathcal{C}$ passing at equal distance from $A, B, C, D$ such that $A, B$ are on one side and $C, D$ on the other. The center of $\mathcal{C}$ must be equidistant from $A$ and $B$, so it lies on the perpendicular bisector of $[A B]$. Similarly, it lies on the perpendicular bisector of $[C D]$.

Suppose that $(A B)$ and $(C D)$ are not parallel. Then the perpendicular bisectors of $[A B]$ and $[C D]$ intersect at a point $\Omega$. Let $R_{1} = \Omega A$ and $R_{2} = \Omega C$. Then $R_{1} \neq R_{2}$ because $A, B, C, D$ are not concyclic. As in the previous case, we see that the circle with center $\Omega$ and radius $\frac{R_{1} + R_{2}}{2}$ is the unique circle equidistant from $A$ and $B$ such that $A$ and $B$ are on one side and $C, D$ on the other. Denote it by $\mathcal{C}_{A B}$.

![](https://cdn.mathpix.com/cropped/2024_06_04_6f47bf1fc05b53b522d5g-5.jpg?height=493&width=526&top_left_y=279&top_left_x=797)

Suppose that the perpendicular bisectors of $[A B]$ and $[C D]$ are parallel but not coincident. Then there does not exist a circle equidistant from $A, B, C, D$ such that $A, B$ are on one side and $C, D$ on the other.

Suppose that the perpendicular bisectors of $[A B]$ and $[C D]$ are coincident. If $A, B, C, D$ are not collinear, then the perpendicular bisectors of $[A B]$ and $[A C]$ intersect at a point $\Omega$ which is equidistant from $A, B, C, D$, which contradicts the fact that $A, B, C, D$ are not concyclic.

Suppose that $A, B, C, D$ are collinear such that the midpoints of $[A B]$ and $[C D]$ are coincident. Then there are infinitely many circles passing at equal distance from $A, B, C, D$, with centers on the common perpendicular bisector.

![](https://cdn.mathpix.com/cropped/2024_06_04_6f47bf1fc05b53b522d5g-5.jpg?height=559&width=609&top_left_y=1165&top_left_x=758)

The discussion above shows that, except in the special case where the four points are collinear such that the midpoint of two points is equal to the midpoint of the other two, there are at most 7 circles passing at equal distance from $A, B, C, D$: these are $\mathcal{C}_{A}, \mathcal{C}_{B}, \mathcal{C}_{C}, \mathcal{C}_{D}, \mathcal{C}_{A B}, \mathcal{C}_{A C}, \mathcal{C}_{A D}$. Equality holds when any two of the six lines determined by $A, B, C, D$ are not parallel.