<s>[INST] For a radius $r$ of a circle, find the center of it in order to it to be touching a parabola exactly twice. So, here goes the problem:
I have a circle $C$ of radius $r$, and a parabola defined by the equation $f(x)=ax^2+bx+c$. I need the center of the circle for it to be touching exactly twice the parabola, this would be the image describing it

And I would need to find the point $(0,a)$ in the image, but note that in an arbitrary parabola the $x$ value of the center can be non-zero. The only thing that's fixed is the radius of the circle.
One thing I noticed is that the derivatives of points $A$ and $B$ must be the same in the parabola and in the circle, but I don't know how to follow.
 [/INST] First of all let's not use "$a$" as the center as you are using that for the parabola. 
Second of all for an arbitrary circle and an arbitray parabola you can't have the symmetric diagram.  But we can "normalize" a parabola.  Let's do that first.
$y = ax^2 + bx + c$
$y-c = a(x^2 + \frac ba x)$
$y-c +\frac {b^2}{4a} = a(x^2 + \frac ba + \frac {b^2}{4a^2})$
$y - c + \frac {b^2}{4a} = a(x+\frac b{2a})^2$.
Now let's "shift" our coordinate system so the $x$ access is raised by $\frac {b^2}{4a} - c$ and the $y$ axis is shifted to the left by $\frac b{2a}$.  In other words $(x_{new}, y_{new}) = (x_{old} - \frac b{2a}, y_{old}+c - \frac{b^2}{4a}$.
In the new system the parabola equation is
$y = ax^2$
And the circle has been shifted to have a center at $(0,d)$.
So you have the equation of the circle is $x^2 + (y-d)^2 = r^2$
So the system of equations $y=ax^2; x^2 + (y-d)^2=r^2$ has exactly two solutions.
So $x^2 + (ax^2 - d)^2 = r^2$ has exactly two solutions.  And by symmetry if $w$ is one of the solutions $-w$ will be the other.
$x^2 + a^2x^4 - 2adx^2 + d^2 =r^2$
$a^2x^4 + (1 - 2ad)x^2 + (d^2 - r^2) = 0$ 
Lew $v = x^2$.  Then the two solutions will be $\pm \sqrt{v}$.
$a^2v^2 + (1-2ad)v + (d^2-r^2) = 0$ will have exactly one solution.
$v = \frac {(2ad-1) \pm \sqrt{(2ad-1)^2 - 4(d^2-r^2)a^2}}{2a^2}$ will have exactly one solution.  That can only happen if
$(2ad-1)^2 - 4(d^2-r^2)a^2 = 0$
So solve for $d$.
$4a^2d^2 - 4ad + 1 - 4a^2d^2 +4a^2r^2 = 0$
$-4ad + 1 + 4a^2r^2 =0$
$d = \frac {1+4a^2r^2}{4a}$
The center in the new coordinate system will be $(0,d)=(0,\frac {1+4a^2r^2}{4a})$.
So the center in the original system will be $(\frac b{2a}, d+\frac {b^2}{4a}-c)=(\frac b{2a}, \frac {1+4a^2r^2}{4a}+\frac {b^2}{4a}-c)$.
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