Adjacent circle propagator is bigger in tikz-feynman

Question

I'm trying to reproduce the following Feynman diagram using tikz-feynman.

This is the code that I have tried

\begin{tikzpicture}[baseline=(current bounding box.center)]
\begin{feynman}
\vertex (x);
\vertex [right =of x] (y);
\vertex [right =of y] (z);
\vertex [above left=of x] (a) {$p_1$};
\vertex [above right=of z] (b) {$q_1$};
\vertex [below left=of x] (c) {$p_2$};
\vertex [below right=of z] (d) {$q_2$};
\diagram* {
(a) -- [fermion] x [dot] -- [fermion,half left,looseness=1.5,edge label=$k_1$] y [dot] --[fermion,half left,looseness=1.5] x [dot],
y [dot] -- [fermion,half left,looseness=1.5] (z) [dot] -- [fermion,half left,looseness=1.5] y [dot],
(c) --[fermion] x [dot],
z [dot] --[fermion] {(b),(d)},
};
\end{feynman}
\end{tikzpicture}

This is almost ok, but when compiled I get this diagram

I don't know why the second circle is bigger than the first one. Any help will be appreciated.

Thanks.

Answer 1

I don't know if this helps, but I road that invisible edges may help in Overleaf. In this code, the circles have same shape but arrows doesn't. Sorry if I didn't help a lot, I'm new at this.

\begin{feynman}
\end{tikzpicture}
\begin{tikzpicture}[baseline=(current bounding box.center)]
\begin{feynman}
\vertex (x);
\vertex [right =of x] (y);
\vertex [right =of y] (z);
\vertex [above left=of x] (a) {$p_1$};
\vertex [above right=of z] (b) {$q_1$};
\vertex [below left=of x] (c) {$p_2$};
\vertex [below right=of z] (d) {$q_2$};
\diagram* {
x [dot] --[opacity=0.2] y [dot] --[opacity=0.2] z [dot],
(a) -- [fermion] x [dot] -- [fermion,half left,looseness=1.5,edge label=$k_1$] y [dot] --[fermion,half left,looseness=1.5] x [dot],
y [dot] -- [fermion,half left,looseness=1.5] (z) [dot] -- [fermion,half left,looseness=1.5] y [dot],
(c) --[fermion] x [dot],z [dot] --[fermion] {(b),(d)},
};
\end{feynman}
\end{tikzpicture}