Minimum vertex cover on k-regular graphs, for fixed k>2 NP-hard proof? Good Evening!

I am aware that the minimum (cardinality) vertex cover problem on cubic graphs ($3$-regular) graphs is $NP$-hard.  Say positive integer $k&gt;2$ is fixed. Has there been any computational complexity proof (aside from the $3$-regular result, note this would be $k=3$,) that shows the minimum (cardinality) vertex cover problem on $k$-regular graphs is $NP$-hard (e.g., $4$-regular)?  Since $k$ is fixed, you aren't guaranteed the cubic graph instances needed to show the classic result I mentioned above.  

Note that this problem would be straightforward to see is $NP$-hard from the result I mentioned at the start if we were to state that this were for any regular graph (since $3$-regular is a special case), we don't get that when $k$ is fixed. 

Does anybody know of any papers that address the computational complexity of minimum (cardinality) vertex cover on a $k$-regular graph, when $k$ is fixed/constant?  I have been having difficulties trying to find papers that address this (in the slew of documents that cover the classic result of minimum (cardinality) vertex cover on cubic graphs being $NP$-hard.) 

Thank you so much!

EDIT (again): @R B gave a construction below!
By adding a small ($O(k)$) sized graph to each vertex you could reduce $VC$ of size $m$ in a 3-regular graph $G$ to $VC$ of size $m+2k|V|$ in a $k-regular$ graph.



Here is a reduction for 5-regular graphs, and I believe you can easily generalize it for every $(2k+1)$-regular graphs, for every integer $k$, and perhaps some trickery would allow similar construction for $2k$-regular graphs;

Consider the graph $H$ in the photo:



(all vertices has degree 5, except for the blue vertex).

For every vertex $v$ in the 3-regular graph $G$, create two copies of $H$ and connect $v$ to their blue vertices.

It is easy to see that regardless of whether $v$ was selected for the cover, you have to include at least $5$ vertices from $H_v$ in the cover (you can take the blue and green ones).

Hence every vertex cover of size $m$ in $G$ corresponds to a $m+10|V|$ sized cover in the new 5-regular graph, and vice versa.



EDIT:

Similar trick could be applied to get $2k$-regular graphs as well. Consider for example the following reduction from 3-regular to 4-regular graphs;

Partition the graph vertices into pairs (duplicate the graph if the number of vertices is odd), and for every pair $u,v$ connect each to a blue vertex in the following structure:





For example, if your initial cubic graph is:



Then the resulting $4$-regular graph is:



And the 5-regular graph would be: