A connected Graph has Euler Circuit $\iff$ `all of its vertices`

have even degree

A connected Graph has Euler Path $\iff$ `exactly 2 of its vertices`

have odd degree

(a) *k*-regular graph where *k* is even number.

a k-regular graph need not be connected always. Example : The given below graph is a $2$ regular graph is not a Euler graph. This is so because there is no single walk which covers all edges.

(b) the complete graph of $90$ vertices

In such a graph `every vertex will have an odd degree`

= 89, Hence it cannot have a Euler path/Circuit.

**(c) **to get degree of all vertices of the complement of cycle on 25 vertices we need to `subtract the degree of a complete graph of 25 vertices with `

`degree`

` of vertices in the `

`original`

` given graph`

i.e. cycle on 25 vertices.

Degree of complement $= 24 - 2 = 22$. Since, every degree is Even, and it is connected also, therefore `Graph Might be a Euler Cycle`

but we need to check graph is connected or not.

Now Question comes it is connected or not :

`It is connected`

because, there is a theorem which says, "$G$ be a graph with $n$ vertices and if every vertex has a degree of at least $\frac{n-1}{2}$ then $G$ is connected." [check this]

Here Degree of each vertex is $22$, which is, of course, greater than $\frac{25-1}{2}=12 $.

Hence Graph must be Euler graph.

`Option C`