Why are the integral curves of this vector field easy to find? I'm reading a book of differential geometry that states that it's easy to get the integral curves of this vector field:

$$X(x,y)=(x^2-y^2,2xy)$$

But proceeding the way the book says, first I have to take that equation as 
$$\frac{dx}{dt}=x^2-y^2$$
$$\frac{dy}{dt}=2xy$$
But from that I don't know what more to do to proceed. I know that the integral curves must be circles of center $(0,t)$, and the vector field seems as this


Note that the system implies 
$$y'(x)=\frac{2xy(x)}{x^2-y(x)^2}.$$
(This is the approach mentioned by @Sameh Shenawy in the comments.) 
This can be solved by standard methods. 
For example, setting $y(x)=xf(x)$ we arrive at the separable differential equation 
$$f'(x) = \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2}.$$
The relevant integral may be done by a partial fraction expansion. 
In my opinion the method put forward by @WimC is much preferable. 
Addendum
The gory detail ... 
\begin{align*}
f'(x) &= \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2} \\
\frac{1-f^2}{f(1+f^2)}df &= \frac{dx}{x} \\
\frac{(1+f^2)-2f^2}{f(1+f^2)} df 
 &= \frac{dx}{x} \\
\int\left(\frac{1}{f} - \frac{2f}{1+f^2}\right)df
 &= \int\frac{dx}{x} \\
\ln f - \ln(1+f^2) &= \ln\frac{x}{2r} \\
\ln\frac{f}{1+f^2} &= \ln\frac{x}{2r} \\
\frac{f}{1+f^2} &= \frac{x}{2r} \\
\frac{y/x}{1+y^2/x^2} &= \frac{x}{2r} \\
\frac{y}{x^2+y^2} &= \frac{1}{2r} \\
x^2+y^2-2ry &= 0 \\
x^2 + y^2 - 2ry + r^2 &= r^2 \\
x^2 + (y-r)^2 &= r^2.
\end{align*}
(The integration constant was chosen to have the form $-\ln 2r$ for convenience.) 
