The coordinates of the center of the circle are (0, 593/16) = (0, 37.0625).
Solution Without Calculus:
The equation of the circle can be expressed as $x^2+(y-a)^2=9$. We must determine the value of $a$ for which the circle and the parabola with equation $y=1+4x^2$ intersect in exactly two locations. In other words, we want the system of equations for the two curves to have exactly two solutions. It is clear that these two solutions will have the same $y$-coordinate, so there must only be one value of $y$ satisfying both equations (along with the associated values of $x$).
Solving the circle equation for $x^2$ and substituting into the parabola equation gives us
$$y=1+4(9-(y-a)^2)\hskip.2in \text{or equivalently} \hskip.2in 4y^2+(1-8a)y+(4a^2-37)=0$$
Because there must only be a single value of $y$ satisfying this equation, we must have the discriminant of this equation (the expression under the radical in the quadratic formula) equal to 0. This gives us:
$$(1-8a)^2-4(4)(4a^2-37)=0 \hskip.2in \text{or equivalently}\hskip.2in -16a+593=0,$$
which gives us $a={593\over 16}$ as claimed.
Solution With Calculus:
For the circle to come to rest inside the parabola, the curves given by $y=1+4x^2$ and $x^2+(y-a)^2=9$ must have the same slope at their points of intersection. Differentiating each equation with respect to $x$ gives us $y’=8x$ for the parabola and $y’={x\over a-y}$ for the circle. Setting these derivatives equal to one another gives
$$8x={x\over a-y}\hskip.2in \text{or equivalently}\hskip.2in a=y+{1\over 8}$$
From the second equation we have $y-a=-1/8$. Substituting this into the equation of the circle gives us
$$x^2+(-1/8)^2=9\hskip.2in \text{or equivalently}\hskip.2in x^2={575\over 64}$$
Substituting into the equation of the parabola then gives us $y=1+4(575/64)={591/16}$ as the $y$-coordinate where the curves intersect. Finally, using the result that $a=y+{1\over 8}$ obtained above, we find that $a={591\over 16}+{1\over 8}={593\over 16}$.