#### Answer

(a) If we remove the edge AB and the edge FJ, then the resulting graph will have no zero odd vertices, so the modified graph will have at least one Euler circuit.
(b) A,D,I,H,G,F,C,D,E,H,J,G,B,C,A is an Euler circuit.

#### Work Step by Step

(a) We need to verify the number of odd vertices in the graph. Vertex A, vertex B, vertex F, and vertex J are odd vertices and the other vertices are even. For a graph to have at least one Euler circuit, the number of odd vertices must be 0.
If we remove the edge AB and the edge FJ, then vertex A, vertex B, vertex F, and vertex J will be even vertices. The resulting graph will have no zero odd vertices, so the graph will have at least one Euler circuit.
(b) To find an Euler circuit, we can start at any vertex.
Let's start at vertex A. According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges AC and AD are not bridges, we can choose either of these edges as the next step in the path.
From vertex A, the path can travel to vertex D, then to vertex I, then to vertex H, then to vertex G, then to vertex F, and then to vertex C.
At this step, we can see that the edge AC is a bridge, so according to Fluery's Algorithm, we must choose a different edge.
The path can then travel to vertex D. After this step, the path must then travel to vertex E, then to vertex H, then to vertex J, then to vertex G, then to vertex B, then to vertex C, and finally back to vertex A, because these are the only available edges.
This path is A,D,I,H,G,F,C,D,E,H,J,G,B,C,A. This path travels through every edge of the modified graph exactly once, so it is an Euler path. Since it starts and ends at the same vertex, this path is an Euler circuit.
This is one Euler circuit but there are other Euler circuits in this modified graph also.