radius of circle tangent to the parabola y=x^2 In the following diagram, what's the amount of r=?

I supposed the center of the small circle to be $(0,y)$. So I have
its distance from $(x, x^2)$ is $(x-0)^2+(y-x^2)^2=1400^2$ this question must have just one answer so the discriminant of the equation must be zero. But then I don't have any idea. how can I relate this to the bigger circle :(

You have the lower circle of radius $1400$. Assuming its center is $(0, a)$ then its equation is
$ x^2 + (y - a)^2 = 1400^2 $
At the tangency points $y = x^2 $, therefore
$ x^2 + (x^2 - a)^2 = 1400^2 $
Expanding, this becomes,
$ x^4 + x^2 (1 - 2 a) + a^2 - 1400^2 = 0 $
Since we want only one intersection point (actually two, but a single value of $x^2$), the discriminant must be zero, hence
$ (1 - 2 a)^2 - 4 (a^2 - 1400^2 ) = 0 $
Simplifying this becomes
$ 1 - 4 a + 4 (1400^2 ) = 0 $
From which,
$ a = \dfrac{ 1 + 4 (1400^2) }{4} = 1400^2 + \dfrac{1}{4} $
Now, we can move to the upper circle, its center is at $( 0, a + r )$ and radius $r$, thus its equation is
$ x^2 + ( y - (a + r) )^2 = r^2$
Again at the points of tangency, $y = x^2 $, so
$ x^2 + (x^2 - (a + r) )^2 = r^2 $
which simplifies to
$ x^4 + x^2 (1 - 2 (a + r) ) + (a + r)^2 - r^2 = 0 $
And we want the value of $x^2$ that solves this quadratic to be a single value, so again, the discriminant must be zero,
$ ( 1 - 2(a + r) )^2 - 4 ( a^2 + 2 a r ) = 0 $
Expanding,
$ 1 - 4 (a + r) + 4 (a^2 + 2 a r + r^2 ) - 4 (a^2 + 2 a r ) = 0 $
and this simplifies to
$ 1 - 4 (a + r) + 4 r^2 = 0 $
Knowing $a$ from the first circle, we can proceed to solve for $r$, there will be two values, and we should take the positive one. The other negative value corresponds to the circle that can drawn tangent to the '$1400$' circle but from below.
Can you take it from here ?
