= Solution
Use <completing the square> to obtain
$$
(x-1)^2+y^2=4.
$$
The <locus> is therefore a \b[<circle> with <centre> $(1,0)$ and <radius> $2$]. It meets the horizontal axis at $(-1,0)$ and $(3,0)$ and the vertical axis at $(0,\pm\sqrt3)$; its highest and lowest points are $(1,\pm2)$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-circle-locus.png]
{title=The circle locus, its centre, and its coordinate-axis intersections}
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