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.
 A: Start with the simpler case of the parabola $y=ax^2$. Let the center of the circle be $C(0,h)$. The line segments $\overline{AC}$ and $\overline{BC}$ are radii of the circle and so are normal to both the circle and parabola. The equation of the normal at a point $A(x_A,ax_A^2)$ on the parabola is $x+2ax_Ay=x_A(1+2a^2x_A^2)$, which has $y$-intercept $\frac1{2a}+ax_A^2$. The square of the distance between $A$ and this point is $x_A^2+\frac1{4a^2}$. Setting this equal to $r^2$ and substituting for $x_A$ in the expression for the $y$-intercept results in $$h=\frac1{4a}+ar^2.$$  
We’re not quite done, though. For values of $r$ that are small relative to $a$, the resulting circle will have only one or even no intersections with the parabola. This can be determined by making the substitution $y=ax^2$ into the equation of the circle $x^2+(y-h)^2=r^2$ and examining its discriminant.  
For the general parabola, you just need to translate this circle by $\left(-\frac b{2a},c-{b^2\over4a}\right)$, which is the vertex of the original parabola.
A: If the parabola is $y = ax^2$ (for the general parabola see below), and the center of your circle is $(0,h)$, then the circle equation is $(y-h)^2 + x^2 = r^2$. So you need equality of the points (on circle and parabola): 
$(y-h)^2 + y/a = r^2$  (first condition)
and equality of the derivatives:
$y' \cdot (y-h) + x = 0$ and $y' = 2 a x$, so $2 a x  \cdot (y-h) + x = 0$ or 
$2 a   \cdot (y-h) + 1 = 0 \qquad (1)$
Squaring you have 
$4 a^2  \cdot (y-h)^2 = 1$ and using the first condition:
$4 a^2 (r^2 - y/a) = 1$ 
So $y = a(r^2 - 1/(4 a^2))$, plugging into  (1) :
$a(r^2 - 1/(4 a^2))-h  = -1/(2a) $
which solves for $h = (4 a^2 r^2 + 1)/(4 a)$.
For the general parabola, we need no new calculations. Transform 
$$
y = ax^2 + b x + c = a (x-b/(2 a))^2 + c - b^2/(4a)
$$
and with this x-shift and y-shift, also the point (0,h) which we just calculated transforms into $(b/(2 a), h + c - b^2/(4a)) $.
A: First of all let's not use "$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 $(\frac b{2a}, d+\frac {b^2}{4a}-c)=(\frac b{2a}, \frac {1+4a^2r^2}{4a}+\frac {b^2}{4a}-c)$.
A: The general equation of a circle is $$(x-x_0)^2+(y-y_0)^2=r^2$$
Due to the symmetry of the parabola around $-\frac{b}{2a}$, you can just use that value as $x_0$. So your unknowns are $x$, $y$, and $y_0$. You have already one equation $$\left(x+\frac{b}{2a}\right)^2+(y-y_0)^2=r^2$$
You also know that your intersection is on the parabola: $$ax^2+bx+c=y$$
The last equation is that the tangents are equal. Calculating the tangent of the parabola is trivial $y'=2ax+b$. For the circle, we would need to rewrite the equation as $$y=y_0\pm\sqrt{r^2-\left(x+\frac{b}{2a}\right)^2}$$
The minus sign corresponds to the bottom half of the circle, the plus sign corresponds to the top. You need to choose the one that has opposite sign to $a$. That means that if the parabola is open on the top, you need to choose the bottom side of the circle. The derivative equation is then:$$2ax+b=\frac{x+\frac{b}{2a}}{\sqrt{r^2-\left(x+\frac{b}{2a}\right)^2}}$$
You should be able to calculate $x$, $y$, and $y_0$ from these equations. For any $y_0$, you will have a $y$ and two $x$ values, symmetric around $-\frac{b}{2a}$
A: If $F$ is the focus and $HK$ the directrix of the parabola, $C$ the center of the circle tangent at $A$, then $ACFH$ in diagram below is a parallelogram. 
Remembering that $FK=1/(2a)$, we have:
$$
CF=AH=a\cdot HK^2+{FK\over2}=a(FH^2-FK^2)+{1\over4a}=ar^2.
$$

