Prove/disprove: In a graph $G$ with at least one component that does not contain a Hamiltonian circuit, we can add a vertex $x$ and certain edges that connect it with certain vertices in the graph, such that we get a graph where every component of the graph has a Hamiltonian circuit.
My answer was:
Disprove. Take for example the claw graph with 3 vertices. Any addition of $x$ and certain edges will not make a Hamiltonian circuit. (it does make a Hamiltonian path, but not a circuit.)
Is that correct or am I missing something?