What does $ \lvert z-a \rvert = \mathit Re(z)+a $ look like? What does a loci with the equation look like?

$ \lvert z-a \rvert = \mathit Re(z)+a $

This is for the applying complex numbers topic of an advanced HSC maths course. I was asked to describe the loci.

I know that $ \lvert z-a \rvert $ would get me either a perpendicular bisector or a circle. I also know that $ \mathit Re(z) $ refers to the horizontal values on the complex plane. But I just can't imagine what it looks like.
Given $|z-a|$= $\Re(z)+a$
$$|x+iy-a|=x+a\\|(x-a)+iy|=x+a\\ $$

$$\sqrt{(x-a)^2+y^2}=x+a\\$$

taking square on both side

$$(x-a)^2+y^2=(x+a)^2\\x^2+a^2-2ax+y^2=x^2+a^2+2ax$$

we get $$y^2=4ax$$

This is a right handed parabola with focus
 $(a,0)$
We can treat complex numbers $z = x + iy$ as equations over $(x, y) \in \mathbb R^2$, and use a geometry plotter to plot them. In this case, the equation system is:

$$
\begin{align*}
|x + iy - a| &amp;= Re(x+iy) + a \\
|(x - a) + iy| &amp;= x + a\\
\sqrt{(x-a)^2 + y^2} &amp;= x + a \\
(x-a)^2 + y^2 &amp;= (x + a)^2
\end{align*}
$$

One can use a tool like Desmos for plotting curves like these. In this case, here is a playable version of the graph with $a$ as a parameter.

The image for one choice of $a$ is: