To draw a graph with only two eccentricties. I was trying to convert graphs $C_8$ and $C_9$ into graphs $A$ and $B$, by adding two extra vertices, such that both $A$ and $B$  contains exactly two vertices with eccentricity three and rest with eccentricity four, and $C_8$ is induced in $A$ and $C_9$ is induced in $B$. Kindly help me to get the graph (may be by adding 3 vertices). Any hint or suggestion will be helpful. Following is my (failed) attempt.
Graph $A$ 

Graph $B$

 A: I was able to mess around with this and get an answer, but I wasn't able to find a great procedure. I really just arrived at the conclusion that the vertices with eccentricity $3$ probably should be the newly introduced vertices. In order for this to work out any "bridge" that they create in the original $C_8$ (or $C_9$) couldn't be a "shortcut" for any vertices in the cycle. Note how both graphs are structurally similar. I bet this concept could generalize nicely for higher order cycles with a little bit of work.
Here is your graph $A$ starting from a $C_8$. The two bottom vertices colored in green are the ones with eccentricity $3$. The red ones have eccentricity $4$. 

Here is your graph $B$ starting from a $C_9$. Colored similarly as above.

A: Looking at the symmetry of C8->A I found an expansion for C9->B.
 : C8 -> A using symmetry
 : C9 -> B using symmetry and asymmetric ghost point
 : C9 -> B (choose A1 xor A2)
If You are not able to see the images clearly please forgive me. I did not have enough time to make a graphic. I can explain more later if necessary.
