<s>[INST] 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?

 [/INST] 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.
</s>