Let $X$ and $Y$ be iid random variables distributed geometrically with probability of success $p$ and support $\mathbb{N}=\{0,1,2,\cdots\}.$ So in particular, letting $q=1-p$, we have that $$\mathbf{P}(X=k)=\mathbf{P}(Y=k)=pq^k\quad\forall k \in \mathbb{N}.$$
I'm trying to find $$\mathbf{P}(X=x\mid X+Y=k)$$ for $x,k \in \mathbb{N}.$ As an intermediate goal, I'm trying to find $$\mathbf{P}(X=x,\, X+Y=k)$$
But the result I'm getting is independent of $x$ for some reason. Here's the logic:
$\mathbf{P}(X=x,\, X+Y=k)=$
$\mathbf{P}(X=x,\, Y=k-x)=$
$\mathbf{P}(X=x)\mathbf{P}(Y=k-x)=$
$pq^x pq^{k-x}=$
$p^2 q^k$
Surely the expression of interest isn't independent of $x$, so why is it vanishing?
The result is perhaps at first a little surprising. But your calculation is perfectly correct. Continue on the path you are on, dividing by the easily computed $\Pr(X+Y=k)$ to find the conditional probability. You will end up with a familiar distribution.