Up close and personal ....

Clearly, the centers of the two circles will lie on the $y$ -axis. Let the center of a circle with radius $r$ be $(0,k)$ . Then the equation of the circle will be $x^{2} + (y - k)^{2} = r^{2}$ . To find the points of tangency we substitute $y = x^{2}$ into the equation of the circle to find that

$y + (y - k)^{2} = r^{2} \Longrightarrow y^{2} - (2k - 1)y + (k^{2} - r^{2}) = 0$ .

Employing the quadratic formula, we find that

$y = (\frac{1}{2})((2k - 1) \pm \sqrt{(2k - 1)^{2} - 4(k^{2} - r^{2})} \Longrightarrow y = (\frac{1}{2})((2k - 1) \pm \sqrt{-4k + 1 + 4r^{2}})$ .

Now by symmetry the $y$ -coordinates of the two points of tangency must be the same, which implies that the discriminant must equal zero. Thus

$-4k + 1 + 4r^{2} = 0 \Longrightarrow k = (\frac{1}{4})(4r^{2} + 1)$ .

So in general the distance between the centers of circles of radii $a$ and $b$ , where $b \ge a$ , will be

$(\frac{1}{4})((4b^{2} + 1) - (4a^{2} + 1)) = b^{2} - a^{2}$ .

With $a = 2, b = 5$ we end up with a distance between centers of $5^{2} - 2^{2} = \boxed{21}$ .

As a note of interest, the distance between the centers of two circles whose radii differ by 1 will be $(a + 1)^{2} - a^{2} = 2a + 1 = a + (a + 1)$ , implying that two such circles will be (externally) tangent to one another.

By symmetry we observe that the centers of the two circles are on positive y-axis. Let the centers of the two circles be (0, a) and (0, b). Then the equations of the circles are:

$X^{2} + (Y-a)^{2} = 4$

$X^{2} + (Y-b)^{2} = 25$

Now we put the condition of tangency for both these circles with the parabola $Y=X^{2}$ i.e. merging the equations and applying Discriminant of the quadratic, D=0. Hence we get:

$Y^{2} - (2a-1) Y + a^{2}-4=0$ applying D=0, we get a=17/4 and

$Y^{2} - (2b-1) Y + b^{2}-25=0$ applying D=0, we get b=101/4.

Now , distance between the centers of these two circles is b-a = 84/4 = 21. Hence the answer is 21.

As is noted in the solutions below, the y coordinate of the points of tangency must satisfy an equation of the form y + (y-k)^2 = r^2 (where the center of each circle takes the form (0,k)). Differentiating w.r.t. x and yields (dy/dx)*(1+2y - 2k) = 0. As dy/dx is clearly not zero at the points of tangency, y = k - 1/2 at the point of tangency. Since y = x^2 at those points, x = (+/-)(k-1/2)^.5. An application of the Pythagorean Theorem yields k = r^2 - 1/2 + 1/4. The distance between each center is now the difference of the y-coordinates of the centers (since each center is on the y-axis). Therefore, the required distance is 5^2 - 2^2 = 21.