To determine which graph has an Euler circuit, we need to recall that an Euler circuit exists in a graph if and only if:

1. The graph is connected.
2. Every vertex has an even degree.

Let's analyze each option:

Option (A)

• Vertices: The degrees of the vertices in (A) are:
  • Vertex $a$ = 2
  • Vertex $b$ = 2
  • Vertex $d$ = 2
  • Vertex $w$ = 4
  • Vertex $l$ = 2
  • Vertex $q$ = 2
  • Vertex $x$ = 2

All vertices have even degrees, and the graph appears to be connected. Therefore, (A) has an Euler circuit.

Option (B)

• Vertices: The degrees of the vertices in (B) are:
  • Vertex $m$ = 3 (odd)
  • Vertex $n$ = 3 (odd)
  • Vertex $o$ = 3 (odd)
  • Vertex $p$ = 3 (odd)
  • Vertex $r$ = 2 (even)

This graph has several vertices with odd degrees, so it does not have an Euler circuit.

Option (C)

• Vertices: The degrees of the vertices in (C) are:
  • Vertex $f$ = 3 (odd)
  • Vertex $g$ = 2 (even)
  • Vertex $h$ = 2 (even)
  • Vertex $i$ = 2 (even)
  • Vertex $j$ = 3 (odd)
  • Vertex $k$ = 2 (even)

This graph also has vertices with odd degrees, so it does not have an Euler circuit.

Conclusion

The correct answer is (A), as it is the only graph where all vertices have even degrees and the graph is connected.

Would you like further details or have any questions about this solution?

Here are 5 related questions for further exploration:

1. What is the difference between an Euler circuit and an Euler path?
2. How would you determine if a graph has an Euler path but not an Euler circuit?
3. What conditions must be met for a directed graph to have an Euler circuit?
4. Can a graph have multiple Euler circuits? If so, how can we find them?
5. How do Euler circuits relate to Hamiltonian circuits in graph theory?

Tip: Remember that for an Euler circuit, all vertices must have even degrees, but for an Euler path, only two vertices can have odd degrees.