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?
 A: 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. 
A: Alternatively, observe that
$$x'(t) + i y'(t) = (x(t)+iy(t))^2$$
$$x'(t)-iy'(t)=(x(t)-iy(t))^2$$
So,
$$u'(t) = [u(t)]^2$$
$$v'(t)=[v(t)]^2$$
Where $x(t)=\dfrac{1}{2}(u(t)+v(t)), y(t)=\dfrac{1}{2i}(u(t)-v(t))$ are easily found.
A: A variation on abel's answer:
Set $z(t)=x(t)+iy(t)$. Then your pair of real ODEs is equivalent to single complex ODE 
$$
z'(t) = z(t)^2
,
$$
which can be solved by separation of variables:
$$
\int \frac{dz}{z^2} = \int dt
\iff
- \frac{1}{z(t)} = t + C
\iff
z(t) = -\frac{1}{t+C}
,
$$
where $C=a+ib$ is some arbitrary complex constant.
So the solution $z(t)=x(t)+iy(t)$ is the image of a horizontal line in the complex $w$ plane, $w(t)=t+C=(t+a)+ib$, under the Möbius transformation $z=-1/w$. Hence it's a circle through the origin with center on the $y$ axis.
