Given two points, how to find a circle through them that's also tangent to the $x$-axis? A seemingly simple geometry problem that is surprisingly difficult. I want to find the radius of a circle that is tangent to the $x$-axis, but also must contain two given points.  I understand there are many situations that have no solution, but for those that do have a solution, how can I find it? 
For the image below, the circle origin is at $(d_s, R_0)$, the first point is at $(0,h_s)$, and the second point is at $(d,h_r)$.  For valid solutions, $h_s$, $h_r$, and $d$ must all be less than $R_0$.  The known quantities are $h_s$, $h_r$, and $d$, and the unknowns are $R_0$ and $d_s$.

 A: An analytical solution: The midpoint of the two points is $(\frac{d}{2}, \frac{h_s+h_r}{2})$, so the equation of their perpendicular bisector is
$$
y-\frac{h_s+h_r}{2} = \frac{d}{h_s-h_r} \left(x-\frac{d}{2}\right)
$$
The center must lie on that bisector.  It must also be equidistant from the $x$-axis and the given point on the $y$-axis:
$$
y^2 = x^2 + (y-h_s)^2
$$
which yields
$$
y = \frac{x^2+h_s^2}{2h_s}
$$
Substitution into the bisector equation gives us
$$
\frac{x^2+h_s^2}{2h_s}-\frac{h_s+h_r}{2}
    = \frac{d}{h_s-h_r} \left(x-\frac{d}{2}\right)
$$
Multiply both sides by $2h_s(h_s-h_r)$ to get
$$
(x^2+h_s^2)(h_s-h_r)-h_s(h_s^2-h_r^2) = dh_s(2x-d)
$$
or
$$
(h_s-h_r)x^2 - 2dh_sx + d^2h_s-h_rh_s(h_s-h_r) = 0
$$
which yields
$$
d_s = x = \frac{dh_s \pm \sqrt{d^2h_s^2-(h_s-h_r)[d^2h_s-h_rh_s(h_s-h_r)]}}{h_s-h_r}
$$
When there are two solutions, one is "between" the two points, and the other is "on the outside", on the side of the "lower" point.  And then the radius $R_0$ is given by
$$
R_0 = y = \frac{d}{h_s-h_r} \left(x-\frac{d}{2}\right) + \frac{h_s+h_r}{2}
$$
ETA: It seems much less messy in any individual scenario with specific values for $d, h_s, h_r$.
A: Let the points be $(a,b)$ and $(0,c)$. The construction I used was constructing the perpendicular bisector of the two points, and a parabola with directrix as a line $y=-c$ and focus at $(0,c)$. The $x$-axis is parallel to the directrix and the $y$-axis passes through the parabola's focus. Note that I am trying to make a circle passing through the two points tangent to the $-c$ line, not the $x$-axis as in the original question (it makes calculations easier).
The perpendicular bisector contains all points equidistant from the two points, and the circle's center must lie on it. The parabola contains all points equidistant from the $(0,c)$ point and the $-c$ line. The circle's center must also lie on this parabola. Thus the center must be on the intersection of the parabola and the perpendicular bisector.
I got equations for the bisector and the parabola, and then gave them to Wolfram Alpha to solve them. Here's what I got: assuming $b,c\neq 0$ and $b-c\neq 0$ (reasonable assumptions), the two points which can be centers of eligible circles are
$$
x = \frac{\mp \sqrt{2cd}-2ac}{b-c} \\
y = \frac{\pm 2a\sqrt{2cd}+d}{2(c-b)^2} \\
\text{where } d = a^2b + a^2c + b^3 - 3b^2c + 3bc^2 - c^3
$$
with radii equal to their respective $y$ parts, plus $c$ because the original line is at $-c$. The distance from the center of the circle to the $y$-axis is the $x$ part of the center. Whew.
A: Hint:
A circle that has center $(\alpha, \beta)$ and is tangent to the $x$ axis has equation:
$$
(x-\alpha)^2+(y-\beta)^2=\beta^2
$$
Substiute the coordinate of the two points and find $(\alpha,\beta)$.
