= Solution
For $z=x+iy$, the equation becomes $\sqrt{(x-2)^2+y^2}=x$. Its right side must be nonnegative. Squaring then gives $y^2=4(x-1)$, whose points already have $x\ge1$, so no extraneous branch has been introduced:
$$
\boxed{x=1+\frac{y^2}{4}.}
$$
This is a <parabola> opening along the positive real axis, with vertex $(1,0)$, focus $(2,0)$ and directrix $x=0$. The original <modulus> equation equates distance from the focus with distance from that directrix.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit-complex-loci.png]
{title=The three complex-plane loci: a single point, four fourth roots, and a right-opening parabola}
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