Coupled differential equation arising in flow line. So, I ran (certainly not literally) across these two coupled differential equation given by:

$$x'(t)=\left(x(t)\right)^2-\left(y(t)\right)^2   $$
   $$ y'(t)=2x(t)y(t)$$

These equation occurred when I was trying to figure flow lines, to a vector field given by. 
$\vec F:\mathbb{R}^2\to \mathbb{R}^2\;\; \mid \; \vec{F}(x,y)=(x^2-y^2,2xy) $
And the flow line is given by $\vec{r}'(t)=\vec{F}(\vec{r}(t))$  , where $t$ is any introduced parameter. And obviously $\vec{r}(t)=(x(t),y(t))$ , Right?
Here's further background if it might be useful , http://ocw.mit.edu/courses/mathematics/18-022-calculus-of-several-variables-fall-2010/lecture-notes/MIT18_022F10_l_17.pdf. Example 17.10
And now for what have I tried.
I differentiated equation in terms of $x'(t)$, to get,
$$x''(t)=2x(t)x'(t)-2y(t)y'(t)$$
From original equation we have $y(t)=\pm \sqrt{(x(t))^2-x'(t)}$
Using this and above, to eliminate $y'(t)$ and $y(t)$ from the equation $y'(t)=2x(t)y(t)$, yet not really yielding good expression. 
Any help, to make the solution bit easier?

what happens if we make a change of variables $$ x = r \cos \theta, y = r \sin \theta.  $$  we have $$\begin{align} r'\cos \theta-r \sin \theta  \, \theta' &= r^2 \cos (2\theta)\\
r'\sin \theta+r \cos \theta\, \theta' &= r^2 \sin 2 (\theta)\\\end{align} $$
from these we get $$r' = r^2\cos \theta, \theta' = r\sin \theta $$  if we divide one by the other we have a separable equation $$\frac{dr}r = \frac{\cos \theta \, d\theta}{\sin \theta}$$ the solution is $$r = c\sin \theta \to x^2 + y^2 =cy$$ which represents a circle. 
