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Yes.

An example:

_____A---------B________ A connected directly to B and D by one path.

_____|_______/|\________ B connected directly to A and E by one path, and to C by two paths.

_____|______/_|_\_______

_____|_____/___\_|______

_____|__E/_____\|______ E connected directly to B and D by one path.

_____|____\_____C______ C connected directly to B and D by two paths.

_____|_____\____|\_____

_____|______\___|__\___

_____|_______\__|__/___

_____|________\_|_/____

_____|_________\|/_____

_____-------------D_____ D connected directly to A and E by one path, and to C by two paths.

There is an Euler circuit: ABCDEBCDA

But a Hamiltonian circuit is impossible: as part of a circuit A can only be reached by the path BAD, but once BAD has been traversed it is impossible to get to both C and E without returning to B or D first. However there is a Hamiltonian Path: ABCDE.

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Q: Can a graph have an Euler circuit but not a Hamiltonian circuit?
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