Is it possible to draw $6$ circles of equal radius each passing through the centres of exactly three others?

Question

Is it possible to draw $6$ circles of equal radius each passing through the centres of exactly three others?

The problem seems to be a direct application of pigeonhole principle but how to use it? May I get a hint?

Here's my take on the Problem : Consider the nodes of the graph as the centres of the circle and the edges as the connection between two circles. Then there will be 9 such edges. Assume that the radius of each circle is $1$ unit. So, how do we ensure that each edge is indeed $\leq 1$ unit?

Answer 1

Yes! Draw an equilateral triangle with side length $r$. Then translate the triangle by $r$.

Note: This answer uses the conventional definition of a "circle" to mean the set of points of a fixed distance away from a given point.

Answer 2

A cubic (degree 3 for every vertex) graph with 6 vertices and unit distance edges will probably suffice.

In general, a unit distance graph with $n$ vertices and edges with degree $k$ will probably satisfy your problem.

B. Mehta gives one such graph, the prism graph $Y_3$. Though $K_{3,3}$ has 6 vertices and is cubic, it is not a unit distance graph in the plane or 3 dimensions.

Here is the Petersen graph with 10 vertices as a unit distance graph:

Here is the cubic matchstick graph with 8 vertices, the smallest such graph that is also planar:

There are many other examples.