Change of variables and circular solutions Consider the equations:
$ \frac{dx}{dt} = y$  and  $ \frac{dy}{dt} = -x $
By transforming variables, obtain $ \frac{dr^2}{dt}=0 $ and $ \frac{d \theta}{dt} = -1$
I know that if I say let $r^2 =x^2 +y^2 \cdot \frac{dr^2}{dt} = tx \frac{2x}{dt} + 2y \frac{dy}{dt} =2xy =2xy-2xy = 0$, which would mean $r^2$ is neutrally stable, but I don't understand where this gets me, much less what a change of variables is. 
Moreover, what can one conclude about whether these are circular solutions, their direction, and their stability. I have not seen this mentioned before.

The best I can interpret the situation is as follows:
$dx/dt = y$ and $dy/dt = -x$ when solved (accompanied with initial and/or boundary conditions, which are absent from your problem statement), will give a solution of the form $x = x(t)$ and $y = y(t)$. When you plot this on x-y plane, you will get a curve, with $t$ being a parameter.
Every curve on x-y plane can be expressed by a polar equation as well, with $x = r \cos \theta$ and $y = r \sin \theta$. Now, in your case, $r$ and $\theta$ will also be functions of $t$. Use these transformations along with chain and product rules for differentiation to get $dr^2/dt = 0$ and $d \theta / dt = -1$.
The parameter $t$ in fact controls how fast a particle moves on that curve. You can have the exact same curve traced by two different particles at different speeds. A simple ODE in $x$ and $y$ will give you only the curve without any information about the speed of the particle.
