Determine whether a vertex lies on a cycle of given length

You can also use method all_simple_paths, and there si no need for removing v from the graph

def FindCycle(G, v, cycle_len):
    for cycle in G.all_simple_paths(starting_vertices=[v], ending_vertices=[v], max_length=cycle_len):
        if len(cycle) == cycle_len + 1:
            return cycle
    return None

For instance

sage: FindCycle(graphs.CompleteGraph(4), 0, 3)
[0, 1, 2, 0]

It remains to deal with the special cases cycle_len == 1 which can happen if the graph has loops and the case cycle_len == 2 for which the proposed method returns a non simple cycle (e.g., [0, 1, 0]),

Removing a given vertex v and searching for a simple path connecting two vertices from the neighborhood of v should do the job:

def FindCycle(G, v, cycle_len):
    H = G.copy()
    H.delete_vertex(v)
    try:
        return [v] + next(p for p in H.all_paths_iterator(starting_vertices=G.neighbors(v), ending_vertices=G.neighbors(v), simple=True, max_length=cycle_len-2) if len(p)==cycle_len-1)
    except StopIteration:
        return None

For example, FindCycle(graphs.CompleteGraph(4), 0, 3) produces the cycle (as a list of vertices) [0, 3, 1].