jmc

First, consider the tangent line to the parabola at $(2,4).$ The equation of this tangent is of the form $$y-4=m(x-2).$$Setting $y=x^2,$ we get $x^2-4=m(x-2),$ or $x^2-mx+2m-4=0.$ Since we have a tangent, $x=2$ is a double root of this quadratic. In other words, this quadratic is identical to $(x-2{)}^2=x^2-4x+4.$ Hence, $m=4.$

Let the center of the circle be $(a,b).$ The line joining the center $(a,b)$ and $(2,4)$ is the perpendicular to the tangent line, which means its slope is $-\frac{1}{4}.$ This gives us the equation $$\frac{b-4}{a-2}=-\frac{1}{4}.$$Since the points $(2,4)$ and $(0,1)$ are on the circle, they must be equidistant from its center. The set of all points equidistant from $(2,4)$ and $(0,1)$ is the perpendicular bisector of the line segment joining $(2,4)$ and $(0,1)$. Therefore, the center of the circle must lie on the perpendicular bisector of the line segment joining $(2,4)$ and $(0,1)$. The midpoint of this line segment is $\left(1,\frac{5}{2}\right),$ and its slope is $$\frac{4-1}{2-0}=\frac{3}{2}.$$Hence, $(a,b)$ must satisfy $$\frac{b-5/2}{a-1}=-\frac{2}{3}.$$So, $$\begin{array}{rl}b-4& =-\frac{1}{4}(a-2),\\ b-\frac{5}{2}& =-\frac{2}{3}(a-1).\end{array}$$Solving this system, we find $(a,b)=\boxed{\left(-\frac{16}{5},\frac{53}{10}\right)}.$