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

 A: Indentifying $(x,y)$ with $z=x + y \, \mathrm{i}$ the differential equation reads $$\frac{\partial z}{\partial t}=z^2.$$ Rewriting this we find $$\frac{-\partial z^{-1}}{\partial t} = \frac{\partial z}{\partial t} z^{-2} = 1$$ and so $$-z^{-1}= t+z_0 \textrm{ or } z = -\frac1{t+z_0}$$ for some fixed initial value $z_0 \in \mathbb{C}$. For $t \in \mathbb{R}$ these describe circular trajectories as expected.
Alternative 1: transform the vector field under the inversion
$$\sigma(x, y) = \left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
This map is an involution ($\sigma^2 = \mathrm{id}$) and its Jacobian is $$D\sigma_{(x,y)}=\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix}.$$
Therefore it maps the element $$\begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix}$$ at point $(x, y)$ to the element
$$\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix} \begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$$ at point $\sigma(x,y)$. The trajectories of the transformed vector field are easy to find. Then apply the inversion $\sigma$ to find the trajectories of the original vector field.
Alternative 2: Since you know that the trajectories are circular start with the ansatz $$\begin{cases} x = -R \sin \alpha \\ y = R + R \cos \alpha \end{cases}$$ for some function $\alpha$ of $t$. Then the given differential equations are consistent and both lead to $$\frac{\partial \alpha}{\partial t} = 2R \, (1 + \cos\alpha).$$ One solution to this equation is $\alpha(t) = 2 \arctan(2R \, t)$. Substituting this in the expressions for $x$ and $y$ gives the trajectory
$$\left(\frac{-t}{t^2 + (2R)^{-2}}, \frac{(2R)^{-1}}{t^2+(2R)^{-2}}\right).$$
A: 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.) 
A: $$X(x,y)=(x^2-y^2,2xy)$$
$$
\begin{cases}
\dfrac{dx}{dt}=x^2-y^2 \\
\dfrac{dy}{dt}=2xy
\end{cases}
$$
$$y'=\frac{2xy}{x^2-y^2}.$$
Consider $x'$ instead of $y'$:
$${x^2-y^2}={2x}x'(y)y$$
Substitute $x^2=z$
$$z-y^2=z'y$$
It's a first order linear DE:
$$z'y-z=-y^2$$
$$ \left ( \frac z y\right)'=-1$$
Integrate:
$$ \left ( \frac z y\right)=-y+C$$
Finally:
$$ \boxed {x^2 +y^2=Cy}$$
