In this answer, I want to present another proof for Option C being Euler Graph.

If a vertex $u$ has degree $d$ in graph $G$ then in $\overline{G},$ $degree(u) = (n-1) – d.$

Since, every vertex in $C_n$ has degree $2$, so, degree of every vertex in $\overline{C_n}$ in $n-3.$

So, degree of every vertex in $\overline{C_{25}}$ in $22.$

**Theorem :**

If $C_n$ is cycle graph on $n$ vertices then $\overline{C_n}$ is connected if $n \geq 5.$

Proof :

Take any two vertices $a,b$ in $\overline{C_n}.$ We show that they will be connected by a path.

Case 1 : $a,b$ are Not adjacent in $C_n :$

In this case, $a,b$ will be adjacent in $\overline{C_n}.$ So, they are connected by a path (of length one)

Case 2 : $a,b$ are adjacent in $C_n :$

Since $n \geq 5$ and every vertex in $C_n$ has degree 2, also $a,b$ are adjacent, so, there is at least one vertex, say $x$ which is Not adjacent to $a,b$, So, in $\overline{C_n}$, $ax,bx$ will be edges, so, we have a path between $a,b.$

Hence proved.

Since, we can manually check that $\overline{C_3}$, $\overline{C_4}$ are disconencted, so, we can say the following theorem :

**Theorem :**

**If $C_n$ is cycle graph on $n$ vertices then $\overline{C_n}$ is connected if and only if $n \geq 5.$**

So, $\overline{C_{25}}$ is connected and has even degree for every vertex, so, it is Euler graph.

We can generalize it as follows :

**Complement of a cycle on $n$ vertices has Euler Circuit if and only if $n$ is odd and $n\geq 5.$**