I know it doesn't have a Hamiltonian circuit because vertices c and f will be traversed twice in order to return to a. Just confirming this.
I mainly want to know whether I have the definition of distinct Euler circuits in a graph right, and whether the graph below is an example of this, i.e. {a,b,c} and {f,g,h}, being the 2 distinct Euler circuits.
Best Answer
An Euler circuit by definition traverses every edge of the graph exactly once, so neither $\langle a,b,c\rangle$ nor $\langle d,e,f\rangle$ is an Euler circuit. Starting at $a$ and counting two circuits as the same if one is simply the reversal of the other, I get the following four distinct Euler circuits:
$$\begin{align*} &a,c,d,f,g,h,f,e,c,b\\ &a,c,d,f,h,g,f,e,c,b\\ &a,c,e,f,g,h,f,d,c,b\\ &a,c,e,f,h,g,f,d,c,b \end{align*}$$
You are correct about the lack of a Hamilton circuit in this graph and the reason for it.