= Solution
For $z=x+iy\ne0$, $z^{-1}=(x-iy)/(x^2+y^2)$. The first constraint is $x=x^2+y^2$ and the second is $-y=x^2+y^2$. Completing the squares gives
$$
\boxed{\left(x-\frac12\right)^2+y^2=\frac14,\qquad
x^2+\left(y+\frac12\right)^2=\frac14,\qquad z\ne0.}
$$
These <circles> have centres $(1/2,0)$ and $(0,-1/2)$ and radius $1/2$. Their algebraic intersections satisfy $x=-y$ and $2x^2=x$, giving zero and $(1-i)/2$. Zero is excluded from both original <loci>, so their only actual intersection is
$$
\boxed{z=\frac{1-i}{2}.}
$$
At $(x,y)=(1/2,-1/2)$ the two circle normals are $(0,-1)$ and $(1,0)$, so the tangent directions are perpendicular. The intersection angle is $\boxed{\pi/2}$. This is also an instance of the <orthogonality of reciprocal-coordinate circles> under complex inversion.
The first small circle touches the unit <circle> at $z=1$, and the second at $z=-i$. On the first, $r^2=x\le r$; on the second, $r^2=-y\le r$. Thus each lies entirely within $r\le1$, attaining equality only at its stated point. They are internally tangent to the unit <circle> and do not cross it. The original sketch marks the excluded common origin with an open symbol and the genuine intersection with a filled symbol.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-reciprocal-coordinate-circles.png]
{title=Original <Argand diagram> of reciprocal-coordinate <circles>, their orthogonal intersection, and tangency to the unit <circle>}
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