Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?
$\textbf{(A)}\ \frac{a+b}{2} \qquad \textbf{(B)}\ a+b \qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \sqrt{\frac{a^2+b^2}{2}}\qquad \textbf{(E)}\ \frac{a+b}{2ab}$
Expanding the quadratic and collecting terms gives $y = 2x^2 - (2a + 2b)x + (a^2+b^2)$.  For a quadratic of the form $y = Ax^2 + Bx + C$ with $A>0$, $y$ is minimized when $x = -\frac{B}{2A}$, which is the average of the roots.
Thus, the quadratic is minimized when $x = \frac{2a+2b}{2\cdot 2} = \frac{a+b}{2}$, which is answer $\boxed{\textbf{(A)}}$.