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

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.

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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.
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.