Dave Stone's solution...

Let the center of the circle be (a,b) and the point of touching the 
curve f(x)=x^2 be (x,y).  Then we know that y= x^2 and b=1.  We also 
know that the distance between (a,b) and (x,y) is 1, OR square root of 
[(x-a)^2 + (x^2-1)^2] = 1. Squaring both sides and expanding, we get 
x^2-2ax+a^2+x^4-2x^2+1 = 1. Simplifying, we have 

  (1) x^4-x^2-2ax + a^2 = 0. 

Shifting gears, now consider the tangent at the point where circle 
and curve meet.  Elementary calculus tells us that the slope of the 
tangent to the curve is 2x; elementary geometry tells us that the slope 
of tangent to the circle is (a-x)/(y-b), or (a-x)/(x^2-1). Since the 
tangent is the same, 2x = (a-x)/(x^2-1).  Cross-multiplying and 
simplifying, we get 2x^3-2x=a-x, or 2x^3-x=a, Now substitute back into 
equation (1) for a, getting 

  x^4-x^2-2x(2x^3-x)+(2x^3-x)^2=0 
  x^4 - x^2 + 2x^2 - 4x^4 + x^2 - 4x^4 + 4x^6 = 0 
  4 x^6 -7 x^4 + 2 x^2 = 0 
  x^2 (4x^4 -7 x^2 + 2) = 0 

Now consider the zeroes of this equation.  x = 0 is the 
trivial case of the circle centered on the y-axis and tangent to the 
parabola.  We can use the quadratic equation on the term in parentheses 
to find that x^2 = 7 +- sqrt (49-32) / 8, OR x^2= 7 +- sqrt (17) / 8.  
The 7 - sqrt 17 case clearly doesn't work, so we're left with a point of 
intersection of

  x = sqrt [ (7 + sqrt (17)) / 8], 
  y = (7 +sqrt(17) / 8, 

or numerically approx (1.179, 1.390) To find the circle's center, we 
know that b = 1, and a=2x^3-x, the expansion of which is left as an 
exercise for the reader. Numerically, it would be approx. (2.099, 1).