The fact that the axiom of foundation doesn't imply the axiom of choice is pretty standard (the model Cohen created to prove the consistency of $\neg AC$ models the axiom of foundation as well), and it's also known that you can have both of them (in $\mathbb{L}$ for instance), or neither of them (Fraenkel's model where the set of atoms is countable). But can you have $AC$ without the axiom of foundation ?
What might work would be to do just like Fraenkel did and build a model with only one atom, so that you wouldn't have the issue with the set of atoms, but I'm not sure one can prove that choice doesn't fail elsewhere in such a model. 
So the question is : does $AC$ imply $AF$ ?



Relation between AC and the axiom of foundation
Of course the axiom of choice is consistent with the failure of the axiom of foundation.


To get Fraenkel's model with the atoms, you usually start with a model of $\sf ZFA+AC$. You can find the relevant proofs in Jech "The Axiom of Choice" in Chapter 4.


(Note that atoms are usually obtained by weakening extensionality, not foundation. To get to foundation one has some way to go. You can find the details in Felgner's "Models of ZF set theory", and more about independence of various statements.) 


