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 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.
