Euler Graph-For a graph to be Euler it must have** Euler Line**.

**Euler Line**-It is a closed walk(yes closed walk and not closed path because as per definition of a path you are not allowed to re-visit a vertex) which covers all the edges of graph G.

Now since a Euler graph always has a Euler line, we have a closed walk(walk in which initial and terminal vertices are same), and all the vertices in this closed walk are of degree 2. So, needless to say, every vertex in a Euler graph must have **Even degree**. For one edge enters and one edge leaves this walk.

So, all in all for a graph to be Euler it must certainly have 2 conditions

(1)It must be connected. (Otherwise, you cannot form **Euler Line**)

(2)All vertices must have even degree.

Okay now let's check options one by one

**(A)Any k-regular graph where k is an even-number:**

Okay, we have one condition satisfied for Graph to be Euler but here it is not told whether the graph is connected or not?. So with surety, we cannot comment about this graph being Euler.So, this option is rejected.

**(B)A complete graph on 90 vertices:** This graph would have the degree of every vertex = 90-1=89 which is odd and yes the graph is connected but we don't have even degree for all vertices. So this option rejected.

**(C)The complement of a cycle on 25 vertices.**

The Complement of a graph G is defined as a graph G` with** |V(G)|=|v(G`)|**

and **|E(G`)| **= $\frac{n(n-1)}{2}$ - |E(G)|.

So this graph G` here would have

$\frac{25(24)}{2}-25$ edges

which evaluates to 275 edges.

Now, the degree of all vertices of this graph would be 24-2=22.

We need to check whether this graph is connected or not.

For a complete graph on n vertices, we have $\frac{(n-1)!}{2}$ distinct cycles covering all vertices.

(Ref: https://math.stackexchange.com/questions/249817/how-many-hamiltonian-cycles-are-there-in-a-complete-graph-k-n-n-geq-3-why )

So for K_{25} we must be $\frac{24!}{2}$having cycles from which we are removing only 1 so complement of a cycle on 25 vertices need to be connected.

Hence, the complement of a cycle on 25 vertices must be **Eulerian.**

**Answer-(C)**