= Solution
For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot y-y\dot x)/r^2$. The <polar coordinates> satisfy
$$
\boxed{\dot r=r(1-r^2),\qquad\dot\theta=-1.}
$$
Hence the rotation is clockwise. The origin is an <unstable equilibrium>, while $r=1$ is a <limit cycle> of period $2\pi$. Writing $u=r^2$ gives the <logistic differential equation> $\dot u=2u(1-u)$. For $r(t_0)=r_0>0$,
$$
\boxed{r(t)^2=\frac1{1+(r_0^{-2}-1)e^{-2(t-t_0)}},\qquad
\theta(t)=\theta_0-(t-t_0).}
$$
All nonzero solutions tend to the <unit-circle attracting limit cycle> as $t\to\infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to-\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes at
$$
\boxed{t_*=t_0+\frac12\log(1-r_0^{-2}),\qquad r(t)\to\infty\text{ as }t\downarrow t_*.}
$$
This <finite-time blow-up of an ordinary differential equation> occurs backward in time. Such exterior solutions have no $t\to-\infty$ behavior: their maximal time interval is $(t_*,\infty)$. The zero solution exists for all time and has no defined polar angle.
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