How is this statement not contradictory? Let $$C = v_0, v_1,...,v_k, v_0$$ be a longest circuit in $G$. Suppose $C$ is not a Eulerian circuit.
By longest circuit don't they mean every vertex is visited? How is this different than a Eulerian circuit?
PS I think $G$ is connected, unless I misunderstood the text.
 A: No, longest circuit means exactly what it says: $G$ does not contain any circuit longer than $C$ (though it may contain some other circuit that is equally long. Even if $G$ is connected and has a circuit, it need not have an Euler circuit, but it will definitely have a longest circuit. The graph on five vertices shown below has circuits of lengths $3$ and $4$, so its longest circuit if of length $4$ (around the diamond), but it has no Euler circuit.
             *  
            /|\  
           / | \  
          *  |  *-----*  
           \ | /  
            \|/  
             *

A: You are right, each Eulerian circuit is trivially a longest circuit, as it omits no edges and hence can not be enlarged. 
But note, that, if a graph has a circuit, then there is always a longest circuit (as the size of a circuit is bounded by the number of edges). Hence, each non-Eulerian forest (a graph, that has no Eulerian circuit and is not circle-free) yields a counterexample, e.g:
 A--B--D
 | /
 |/
 C

Certainly, $A-B-C-A$ is a longest circuit, but not Eulerian, since it does not use the edge $BD$.
