Question: If $U$ is the unit disc centered at the origin consider $n$ chords drawn through the interior of $U$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $G$ has a vertex for each intersection point in the interior of $U$ and $2$ vertices for each chord incident to the boundary of $U.$ Naturally $G$ has an edge for each arc directly connecting two intersection points. Is $G$ Hamiltonian ?
Solution: Consider the one-point-extended graph of $G.$ That is the arrangement graph induced by drawing a point $x$ exteriorly to the boundary of $U$ and a line drawn from every point on the boundary of $U$ to $x.$ Denote this graph by $G+x.$ Note that $G+x$ is $4-$connected and planar and by a result of Tutte is Hamiltonian. In proving Plummer's conjecture R. Thomas observes in [1] that the "deletion of any vertex from a 4–connected planar graph results in a Hamiltonian graph." Then delete the point $x$ from $G+x,$ the resulting graph, namely $G,$ is Hamiltonian.
Solution: Consider the one-point-extended graph of $G.$ That is the arrangement graph induced by drawing a point $x$ exteriorly to the boundary of $U$ and a line drawn from every point on the boundary of $U$ to $x.$ Denote this graph by $G+x.$ Note that $G+x$ is $4-$connected and planar and by a result of Tutte is Hamiltonian. In proving Plummer's conjecture R. Thomas observes in [1] that the "deletion of any vertex from a 4–connected planar graph results in a Hamiltonian graph." Then delete the point $x$ from $G+x,$ the resulting graph, namely $G,$ is Hamiltonian.