Is it possible disconnected graph has euler circuit? I have doubt !

Wikipedia says :


An Eulerian graph is one in which all vertices have even degree;
  Eulerian graphs may be disconnected.


What I know :

Defitition of an euler graph 
"An Euler circuit is a circuit that uses every edge of a graph exactly once. ▶ An Euler path starts and ends at different vertices. ▶ An Euler circuit starts and ends at the same vertex."
According to my little knowledge "An eluler graph should be degree of all vertices is even, and should be connected graph".
I am asking :Is it possible disconnected graph has euler circuit ? If it is possible show an example .
EDITED : Here is my suplimentary problem , I voted for the anwser .

Which of the following graphs has an Eulerian circuit?

 A: Here we go:
$$\huge\cdot\qquad\cdot$$
remember that $0$ is even. The circuit is the "empty circuit"
Since the graph has no edges, we've already passed every edge if we don't even move :D
A: The other answers answer your (misleading) title and miss the real point of your question.
Yes, a disconnected graph can have an Euler circuit. That's because an Euler circuit is only required to traverse every edge of the graph, it's not required to visit every vertex; so isolated vertices are not a problem. A graph is connected enough for an Euler circuit if all the edges belong to one and the same component.
But that's not what was confusing you five years ago when you asked the question. The confusion arises from the fact that some people use the term "Eulerian graph" to mean a graph that has an Euler circuit, while other people (deplorably) use the term "Eulerian graph" for any graph in which all vertices have even degree, regardless of whether or not it has a Eulerian circuit.
From the Wikipedia article Eulerian path:

The term Eulerian graph has two common meanings in graph theory. One meaning is a graph with an Eulerian circuit, and the other is a graph with every vertex of even degree. These definitions coincide for connected graphs.

People using the second definition would consider the graph $K_3\cup K_3$ to be "Eulerian", although of course it has no Euler circuit.
A: It's only possible for a disconnected graph to have an Eulerian path in the rather trivial case of a connected graph with zero or two odd-degree vertices plus vertices without any edges. An Eulerian path for the connected graph is also an Eulerian path for the graph with the added edge-free vertices (which clearly add no edges that need to be traversed). Whoop-te-doo! The whole issue seems pretty nit picky and pointless to me, though it appears to fascinate certain Wikipedia commenters.
