Unit Circle Internally Tangent to Ellipse

Question

Suppose I have an ellipse (horizontally oriented) with semi major axis $l$ and semi minor axis $s$.

Let me define vertex tangency as the possibility of having a unit circle be internally tangent at one point at the ellipse vertex.

Let this condition also include the case where the circle is an osculating circle of the ellipse at the vertex.

This condition translates to the following inequality, when it holds true - an ellipse allows for vertex tangency.

$$ \frac{l}{s^2}\leq1 \, \text{ (vertex tangency allowed)} $$

Vertex tangent example.

Vertex tangent - osculating example.

Conversely, if we take the negation of the inequality then vertex tangency is not possible. Also, the closest a unit circle can get to the vertex is a point where it lies on the semi major axis and is tangential at two points. $$ \frac{l}{s^2}>1 \, \text{ (vertex tangency forbidden)} $$

Internal tangent at two points example.

Suppose that the ellipse is "large enough" (without being too explicit, let's take it to mean that it can contain the unit circle and have it a some distance away from the center without any intersection). What is the distance from the center on the semi major axis such that the unit circle is at its closest to the vertex? Equivalently, what is the distance on the semi major axis such that the unit circle is internally tangential to the ellipse at two points?

Specifically, looking at the sketch below - can one find any one of $\phi$, $\theta$ or the distance between the ellipse center and unit circle center (well finding one would be the same as finding all of them) given only $l$ and $s$?

Of course assuming they are "large enough" to pose this problem with the unit circle.

I am looking for an analytical solution (perhaps even just using basic geometry). I found an interesting but convoluted way to pose the problem as constrained optimization problem that would allow me (with some tedious work) to find a numerical solution , I stated below just for completeness.

Use a Weierstrass substitution to transform the Cartesian descriptions of ellipse and the unit circle (at arbitrary center within ellipse). 

Then, equate the resulting polynomials to obtain a fourth order polynomial. 

Set y coordinate of circle center to zero.

The discriminant of a quartic polynomial has a specific case where two of its roots are real and distinct and with two other being complex. 

Find the x coordinate of the center coordinate that remains within the ellipse and obtains the aforementioned case of the discriminant

Any help would be appreciated!

Answer 1

Let $C$ be the center of the circle, $P$ a tangency point, $A$ and $B$ the foci of the ellipse. As $PC$ is the bisector of $\angle APB$, from the angle bisector length formula we get:

$$ PC={s\over l}\sqrt{AP\cdot BP}. $$

But $AP/AC=BP/BC=l/f$, where $f=AB/2=\sqrt{l^2-s^2}$, hence:

$$ PC={s\over f}\sqrt{AC\cdot BC}= {s\over f}\sqrt{(f+x)(f-x)}, $$

where $x$ is the distance from $C$ to the centre of the ellipse.

We can then solve for $x$:

$$ x=f\sqrt{1-{r^2\over s^2}}, $$
where $r=PC$ is the radius of the circle (the OP asked for $r=1$). This is the requested formula. Of course, as explained in the question, it works only if $r \le s \le\sqrt{rl}$.

Answer 2

The point of tangency $P(x, y)$ has to satisfy three equations:

$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \hspace{40pt} (1) $ Ellipse equation, $ a \gt b \gt 1 $

$ (x - x_0)^2 + y^2 = r^2 \hspace{40pt} (2)$ Circle equation

In this problem, $r = 1 $ (Unit circle)

$ - \dfrac{(x - x_0)}{y} = - \dfrac{ x }{y} \left(\dfrac{b^2}{a^2} \right) \hspace{40pt} (3) $ Tangency equation

Substituting for $(x - x_0)$ from the third equation into the second equation,

$ x^2 \left( \dfrac{b^4}{a^4} \right) + y^2 = r^2 \hspace{40pt} (4) $

Now it is very easy to solve equation $(4)$ with equation $(1)$ (a linear system in $x^2 $ and $y^2$). The solution is

$ x^2 = \dfrac{ \left(1 - \dfrac{r^2}{b^2} \right) }{ \left(\dfrac{1}{a^2} - \dfrac{b^2}{a^4} \right) }$

$ y^2 = \dfrac{ \left( \dfrac{r^2}{a^2} - \dfrac{b^4}{a^4} \right) } { \left( \dfrac{1}{a^2} - \dfrac{b^2}{a^4} \right) } $

Once $x, y$ are determined, one can determine $x_0$ from equation $(3)$, and then $ \phi$ and $\theta$ can be computed.

Note that the solution for $x^2, y^2$ is valid only if $ r \lt b $ and $r \gt \dfrac{b^2}{a} $

Answer 3

Let us denote $F_1,F_2$ the foci, $P$ the point of tangency and $S$ the center of the circle.

If $l,s$ are the semi-major and semi-minor axes of the ellipse, respectively, then the focal distance $2f=|F_1F_2|$ satisfies $$f^2=l^2-s^2. \quad \tag 1$$ Tangent to the ellipse at $P$ is orthogonal to the angle bisector $\angle F_1PF_2.$
Angle bisector theorem writes $$\frac{|F_1S|}{|F_2S|}=\frac{|PF_1|}{|PF_2|}$$

With the use of labeling as in the picture, we deduce from the theorem
$$ly=fx \quad \tag 2$$

Cosine theorem in triangles $\triangle F_1SP$ and $\triangle F2SP$ gives $$(2l-x)^2=(2f-y)^2+r^2-2(2f-y)r\cos (\pi-\phi) \quad \tag 3$$ $$x^2=y^2+r^2-2yr\cos \phi \quad \tag 4$$ Expanding $(3)$ and subtracting $(4)$ from $(3)$ gives $$4l^2-4lx=4f^2-4fy+4fr\cos \phi.$$

EDIT

Due to $(1),(2)$ and $r=1$ we get $$\cos \phi=\frac{s^2-lx+fy}{f}$$ and further $$\cos \phi=\frac{s^2l-l^2x+f^2x}{fl}=\frac{s^2(l-x)}{l\sqrt{l^2-s^2}}.$$

For the present configuration, the formula contains also the shorter distance from $P$ to a focus.