The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations: 
\[x=x^2+y^2 \   \ y=2xy\] 
is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad  \textbf{(E) }4$
If $x=x^2+y^2$ and $y=2xy$, then we can break this into two cases.
Case 1: $y = 0$
If $y = 0$, then $x = x^2$ and $0 = 0$
Therefore, $x = 0$ or $x = 1$
This yields 2 solutions
Case 2: $x = \frac{1}{2}$
If $x = \frac{1}{2}$, this means that $y = y$, and $\frac{1}{2} = \frac{1}{4} + y^2$. 
Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$
This yields another 2 solutions.
$2+2 = \boxed{\textbf{(E) 4}}$