Ellipse tangent to two circles We're given two circles with radii $p$ and $q$, one centered at the origin, and one centered at the point $(w,0)$. 
I want to construct an ellipse of the form
$$
\frac{(x-c)^2}{a^2} + \frac{y^2}{b^2} = 1
$$
that is tangent to the two circles, as shown here:

If $b$ is known, can we obtain closed form expressions for $a$ and $c$ as functions of $p$, $q$, $w$ and $b$.

For ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Equation of normal at $(x_1,y_1)$:
$$a^2y_1(x-x_1)=b^2x_1(y-y_1)$$
Put $y=0$,
$$x=\frac{a^2-b^2}{a^2}x_1=e^2x_1 \tag{$y_1 \ne 0$}$$
which is the $x$-intercept.
Radius $p$:
\begin{align*}
  p^2 &= (x_1-e^2x_1)^2+y_1^2 \\
  &= \frac{b^4x_1^2}{a^4}+y_1^2 \\
  &= \frac{b^4x_1^2}{a^4}+b^2\left( 1-\frac{x_1^2}{a^2} \right) \\
  &= b^2-\frac{b^2e^2x_1^2}{a^2} \tag{1}
\end{align*}
Similarly,
$$q^2=b^2-\frac{b^2e^2x_2^2}{a^2} \tag{2} $$
$(1)-(2)$,
$$p^2-q^2=\frac{b^2e^2}{a^2}(x_2^2-x_1^2) \tag{3}$$
Also $$w=e^2(x_2-x_1) \tag{4}$$
On solving,
$$x_1=\frac{a^2(p^2-q^2)}{2b^2w}-\frac{a^2w}{2(a^2-b^2)}$$
$$x_2=\frac{a^2(p^2-q^2)}{2b^2w}+\frac{a^2w}{2(a^2-b^2)}$$
and from $(1)$,
$$a=\frac{b\sqrt{(p^2-q^2)^2+w^2
\left( \sqrt{b^2-p^2}+\sqrt{b^2-q^2} \right)^2}}{p^2-q^2}$$
Note that $$c=-e^2x_1$$
The required ellipse is
$$\frac{1}{a^2}
\left[
  x-\frac{w}{2}+\frac{(a^2-b^2)(p^2-q^2)}{2b^2w}
\right]^2+\frac{y^2}{b^2}=1$$
provided $p,q \le b$

A good combination of $p$, $q$, $w$ and $b$:



A bad combination of $p$, $q$, $w$ and $b$:



A special combination of $p$, $q$, $w$ and $b$:


