Is $\{x2y : |x| = |y|, x\in A, y\in\{0,1\}^*, d(x,y) = k\}$ context-free for some infinite regular language $A$?

Question

For two equal-length binary strings $x$ and $y$, let $d(x,y)$ denote the Hamming distance. Prove or disprove: there exists a positive integer $k$ such that the language $\{x2y : |x| = |y|, x\in A, y\in\{0,1\}^*,d(x,y) = k\}$ is context-free for some infinite regular language $A$ over the alphabet $\{0,1\}$.  

I think the statement is false. Intuitively, context-free grammars cannot keep track of the number of positions where strings differ, but I'm not sure how to use the pumping lemma for this case. If $k=1$, then I'd need show that the given language isn't context free for any infinite regular language $A$. If we consider the regular language generated by $0^*$ for instance, then the language would consist of all strings $0^n 2 y$ where $y$ has length $n$ and exactly one $1$. If $A$ were finite, obviously the given language would be context-free for all $k$ because it would be finite.  

This question was based off of this other post.

Answer 1

Let $F=\{x2y : |x| = |y|,\ x\in \{0\}^*,\ y\in\{0,1\}^*,\ d(x,y) = 1\}$, the language of all strings $0^n2y$ where $y$ consists of $n-1$ $0$s and one $1$.

Note that $F=\{0^p00^q20^q10^p: p\ge0, q\ge0\}$. Here is a context-free grammar for it.
    $S\to 0S0\mid 0T1$
    $T\to 0T0\mid 2$

Since $F$ is context-free, the proposition in the question is true.

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On the other hand, the language $H_{A,k}=\{x2y^R : |x| = |y|,\ x\in A,\ y\in\{0,1\}^*,\ d(x,y) = k\}$ is always context-free when $A$ is regular. Note that $y^R$ instead of $y$ appears at the end of each word, so that we can construct a context-free grammar by synchronizing the generation of $x$ by a right linear grammar and the generation of $y^R$ by a corresponding left linear grammar, as pointed out by Hendrik Jan.