= Solution
For $r>0$, differentiate $r^2=x^2+y^2$ and use the two differential equations:
$$
\dot r=\frac{x\dot x+y\dot y}{r}=r(a-f(x,y)),\qquad
\dot\theta=\frac{x\dot y-y\dot x}{r^2}=1.
$$
Substituting $x=r\cos\theta$, $y=r\sin\theta$ gives the required equations in <polar coordinates>. The origin is handled in <Cartesian coordinates>, where it is an <equilibrium point>; an angular coordinate is not defined there.
A <Lyapunov function> near an <equilibrium point> is a continuously differentiable <positive-definite scalar function> $V$ whose <derivative> along solutions is nonpositive. This implies local <Lyapunov stability> under the usual comparison bounds; if its <derivative> is negative definite, the <Second Lyapunov theorem> gives local <asymptotic stability>. Here take
$$
V(x,y)=\frac12(x^2+y^2),\qquad \dot V=r^2(a-f(x,y)).
$$
If $a<f(0,0)$, continuity gives a small ball and $\delta>0$ such that $f(x,y)-a\ge\delta$ inside it. Consequently $\dot V\le-2\delta V$. A sufficiently small initial sublevel set stays inside the ball, and $V(t)\le V(0)e^{-2\delta t}$. \b[The origin is <asymptotically stable>], in fact locally <exponentially stable>. If $a>f(0,0)$, continuity instead gives $\dot V\ge2\delta V$ in a small ball. Every nonzero initial point there exits that ball in forward time, proving \b[<instability> of the origin]. These inequalities hold along all local solutions; continuity of $f$ alone need not supply uniqueness for a general <vector field>.
If $f=g(r)$, a nonzero constant radius $r_0$ satisfying $a=g(r_0)$ gives a <periodic orbit> of period $2\pi$, since $\dot\theta=1$. Linearizing the scalar radial equation there gives
$$
\dot\rho=-r_0g'(r_0)\rho+O(\rho^2).
$$
Thus the <periodic orbit> is <orbitally asymptotically stable> for $g'(r_0)>0$ and <unstable> for $g'(r_0)<0$. The direction along the <periodic orbit> is neutral, as expected for an autonomous <periodic orbit>; <orbital asymptotic stability> means <attraction in a dynamical system> to the orbit with a possible phase shift.
For the specified radial response,
$$
g(0)=\frac32,\qquad g'(r)=\frac{-2(r-1)}{[1+(r-1)^2]^2}.
$$
The origin is therefore <stable> for $0<a<3/2$ and <unstable> for $a>3/2$. The nonzero-radius equation has <roots of an equation>
$$
\boxed{r_\pm=1\pm\sqrt{\frac{2-a}{a-1}},\qquad1<a\le2,}
$$
retaining only positive radii. For $1<a<3/2$, the <square root> exceeds one, so only $r_+>1$ exists and $g'(r_+)<0$: the <periodic orbit> is <unstable>. For $3/2<a<2$, both radii are positive; $0<r_-<1$ gives $g'(r_-)>0$ and a <stable> <periodic orbit>, while $r_+>1$ remains <unstable>.
As $a$ increases through $3/2$, the <stable> small-radius <periodic orbit> emerges from the origin while the origin loses <Lyapunov stability>. This is the supercritical <Hopf-type transition with a nonsmooth radial response> requested. It follows directly from the explicit radial dynamics. There is a regularity qualification: $g(r)=3/2+r/2+O(r^2)$ near zero, so the <vector field> in <Cartesian coordinates> contains $x\sqrt{x^2+y^2}$ and $y\sqrt{x^2+y^2}$ terms. It is continuously differentiable but not generally twice differentiable at the origin, and the <periodic orbit> radius grows as $r_-\sim2(a-3/2)$ rather than the square-root law of a generic smooth <Hopf normal form>. One should therefore use the explicit computation here rather than invoke a smooth <Hopf bifurcation> theorem with unmet hypotheses. At $a=3/2$ itself the origin is still weakly <attracting>, because $a-g(r)<0$ for small positive $r$.
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