A circle inside an ellipse Consider an ellipse with semi-axes $a$ and $b$, taller than it is wide with a small circle of radius $r$ inside.  Assume the circle falls to the lowest point possible while staying inside the ellipse.
If $2r\le a-c$ then the circle and ellipse will meet at a single point at the bottom. If $2r>a-c$ the circle and ellipse will intersect at two points on the opposite side, leaving a space between the bottom of the circle and the bottom of the ellipse. For this case, given the radius of the circle and the dimensions of the ellipse how do I calculate the distance $d$ between the bottom of the circle and the bottom of the ellipse?


Never underestimate the power of cartesian coordinates
Consider the problem of finding the intersection points between ellipse and circle of radius $r$ centered in $(0,y_c)$, which leads to the system of equations
$$\begin{cases}\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\\
x^2 + (y-y_c)^2 = r^2.\end{cases} $$
By substitution you obtain the following quadratic equation.
$$ y^2\left(\frac{1}{b^2} -\frac{1}{a^2} \right) + 2y\left(\frac{y_c}{a^2}\right) + \frac{r^2-y_c^2-a^2}{a^2}=0. $$
Since you want the two curves to intersect at only two points with the same ordinate, than this equation must have discriminant equal to $0$, which gives you
\begin{eqnarray} 
\left(\frac{y_c}{a^2}\right)^2-\left(\frac{1}{b^2} -\frac{1}{a^2} \right)\left(\frac{r^2-y_c^2-a^2}{a^2}\right) = 0. \tag{1}\label{discr}
\end{eqnarray}
Now you just need to solve the quadratic equation \eqref{discr} with respect to the unknown $y_c$. This rapidly gives
$$y_c^2 = \frac{(b^2-a^2)(a^2-r^2)}{a^2}.$$
The desired distance is
$$w = b-|y_c| - r,$$
that is

$$ w = b - \frac{\sqrt{(b^2 - a^2)(a^2-r^2)}}{a} -r,$$

as, of course, in the other proposed solutions.
