= Solution
Write $z=x+iy\ne0$. Since $z^{-1}=(x-iy)/(x^2+y^2)$, the two <loci> are
$$
\boxed{\left(x-\frac12\right)^2+y^2=\frac14,\qquad
x^2+\left(y+\frac12\right)^2=\frac14,\qquad z\ne0.}
$$
Thus they are radius-$1/2$ circles centred at $(1/2,0)$ and $(0,-1/2)$, each missing the origin. Their circle equations intersect at $(0,0)$ and $(1/2,-1/2)$, but only the latter belongs to the original loci. At that point the radius vectors are $(0,-1/2)$ and $(1/2,0)$, so the tangents meet at a right angle.
Every point on either circle has distance at most $1/2+1/2=1$ from the origin. Equality occurs only at $(1,0)$ on the first or $(0,-1)$ on the second. The <unit circle> is therefore internally tangent to both and crosses neither. The sketch marks the excluded origin explicitly, rather than treating it as a second intersection.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017-resit-reciprocal-loci-and-hyperbola.png]
{title=Reciprocal complex loci and orthogonal principal axes of a hyperbola}
{description=Original exact geometric plots. The empty point at the origin is excluded from both reciprocal loci; the hyperbola uses the original level six, with its principal axes and asymptotes.}
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