Circular orbits of ODE System

Question

I want to show that the orbits of $$\dot x=y\exp(1-x^2-y^2) \\ \dot y=-x\exp(1-x^2-y^2)$$ are circles (or subsets of circles).

1. I tried to transform that system into polar coordinates:

$x=r\cos\varphi,y=r\sin\varphi$: $$\dot x=\dot r\cos\varphi-r\dot\varphi\sin\varphi\\ \dot y=\dot r\sin\varphi +r\dot\varphi\cos\varphi$$ by chain rule. So I would get $$\dot r\cos\varphi-r\dot\varphi\sin\varphi=r\sin\varphi\exp(1-r^2)\\\dot r\sin\varphi +r\dot\varphi\cos\varphi=-r\cos\varphi\exp(1-r^2)$$

but I'm not sure if I did that correctly.

2. I wanted to show that $r\dot r=0$. If I isolate $\dot r$ in the first equation and multiply it by $r$ I get $$r\dot r=r(\tan\varphi(re^{1-r^2}+r\dot\varphi))=r^2\tan\varphi (e^{1-r^2}+\dot\varphi)$$

but I don't see how that helps.

Answer 1

From the given ODEs one immediately reads off $$(r^2)^\cdot=(x^2+y^2)^\cdot=2(x\dot x+y\dot y)\equiv0\ .$$

Answer 2

$$\begin{cases}\frac{dx}{dt}=y\exp(1-x^2-y^2) \\ \frac{dy}{dt}=-x\exp(1-x^2-y^2)\end{cases}$$ $$\frac{dy}{dx}=\frac{-x\exp(1-x^2-y^2)}{y\exp(1-x^2-y^2)}=-\frac{x}{y}$$ $$ydy+xdx=0$$ $$y^2+x^2=\text{constant}>0$$ $$y^2+x^2=R^2$$