Matt Hudelson's algebraic solution
Consider the set of unit circles centered on the y-axis. The center of
the desired circle has the maximum y-coordinate among all such unit
circles which intersect the parabola. So, we want the maximum value of k
such that the equations
x^2 + (y-k)^2 = 1
y = x^2
have a common solution (x,y). In this case, the center of the desired
circle will be at (0,k).
Substituting x^2 for y in the first equation yields:
x^2 + (x^2 - k)^2 = 1,
or x^4 - (2k - 1) x^2 + k^2 - 1 = 0.
By the quadratic formula, we find that
x^2 = ( 2k - 1 +/- Sqrt( (2k-1)^2 - 4k^2 + 4 ))/2
= ( 2k - 1 +/- Sqrt( 5 - 4k ) )/2
If k > 5/4, then we are taking the square root of a negative number, and
so there are no intersection points.
If k = 5/4, then
y = x^2 = ( 10/4 - 1 +/- Sqrt(0) )/2
= 3/4
x = +/- Sqrt(3)/2
which is a valid solution and so the circle must be centered at (0,5/4).
Ken Duisenberg's calculus-based solution
Let the point of intersection on the right be (x1, x1^2). The circle is tangent to the parabola at this point. The slope of the tangent line is 2*x1 at this point, and the slope of the perpendicular is (-1/2*x1) and goes through the center of the circle. I let the equation of the line for the perpendicular be y = (-1/2*x1)*x + b, where (0,b) is the center of the circle. Since we know point (x1, x1^2) is on this line, we can substitute and find (x1)^2 = b - 1/2. The equation of the circle is (y-b)^2 + x^2 = 1. Point (x1, x1^2) is also on the circle, and when we substitute, the equation becomes: ((x1)^2 - b)^2 + (x1)^2 = 1. Substituting in the equation for (x1)^2 above, this simplifies to: (b - 1/2 - b)^2 + b - 1/2 = 1 --> 1/4 + b - 1/2 = 1 --> b = 5/4. So the center of the circle is at (0,5/4).