jmc

Let one of the points of tangency be $(a,a^2).$ By symmetry, other point of tangency is $(-a,a^2).$ Also by symmetry, the center of the circle lies on the $y$-axis. Let the center be $(0,b),$ and let the radius be $r.$



The equation of the parabola is $y=x^2.$ The equation of the circle is $x^2+(y-b{)}^2=r^2.$ Substituting $y=x^2,$ we get $$x^2+(x^2-b{)}^2=r^2.$$This expands as $$x^4+(1-2b)x^2+b^2-r^2=0.$$Since $(a,a^2)$ and $(-a,a^2)$ are points of tangency, $x=a$ and $x=-a$ are double roots of this quartic. In other words, it is the same as $$(x-a{)}^2(x+a{)}^2=(x^2-a^2{)}^2=x^4-2a^2{x}^2+a^4=0.$$Equating the coefficients, we get $$\begin{array}{rl}1-2b& =-2a^2,\\ b^2-r^2& =a^4.\end{array}$$Then $2b-2a^2=1.$ Therefore, the difference between the $y$-coordinates of the center of the circle $(0,b)$ and the point of tangency $(a,a^2)$ is $$b-a^2=\boxed{\frac{1}{2}}.$$