Past exam of the mathematical methods I course of the University of Oxford / 2013 / 10 / d / Solution

For $z=x+iy\ne 0$, $z^{-1}=(x-iy)/(x^2+y^2)$. The first constraint is $x=x^2+y^2$ and the second is $-y=x^2+y^2$. Completing the squares givesThese 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 isAt $(x,y)=(1/2,-1/2)$ 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 $r\le 1$, 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.