The sum of the vector lengths of the tangents of two circles in a conic section

Question

A few days ago, I came to an amazing intuition while using GeoGebra, and I could not prove whether it was previously discovered or is it new, please mention a source if it already exists...

The sum of the lengths of the tangents of two circles touching a conic section, each of which touches it at two points, and starting from a point on the circumference of the conic section is equal to the distance between the centers of the two circles multiplied by the reciprocal of the coefficient of central difference, where the lengths of the two tangents are two vectors and the shortest of them will take a negative value if the point taken from the segment is not located between the two circles.

These are illustrations that include different relative states:

direct result: The length of a tangent to a circle tangent to a conic at two points that radiates from the point of tangency to another circle tangent to it at two points is equal to the distance between the centers of the two circles.

Answer 1

It turns out to be quite simple to prove, but one should be familiar with the theorem shown in the picture

Reference: https://www.mccme.ru/~akopyan/papers/EnGeoFigures.pdf

which generalizes to all conic sections by multiplying the tangent length by the reciprocal of the center divergence

Also, use the theorem in the following question:

The distance between the center of a circle touching the a conic cut at two points, and between the midpoint of the two tangent points

Whoever wants to add details of the proof, these are the ideas that should be used

Answer 2

Here's a coordinate approach.

We start with this parameterization of a focus-at-origin conic of eccentricity $e$, latus rectum $r$, and major/transverse axis of symmetry along the $x$-axis:

$$C(\theta) = \frac{r}{1-e\cos\theta}\; (\cos\theta,\sin\theta) $$

By symmetry, we need only consider $0\leq \theta\leq \pi$. Moreover, we'll assume throughout that $1-e\cos\theta>0$ (which only matters for hyperbolas) so that $C$ lies on the branch associated with the given focus.

Calculus tells us that a normal vector at $C$ is given by $$n(\theta) = (e-\cos\theta,-\sin\theta)$$

Let $P:=C(p)$ and $Q:=C(q)$ be non-vertex points (ie, not on the $x$-axis) at which two circles are tangent to the conic; without loss of generality, we may take $0<p\leq q<\pi$, so that $\cos p\geq \cos q$. The normal lines at $P$ and $Q$ meet the $x$-axis at the centers —say, $P_\circ$ and $Q_\circ$ of these circles, and we have $$P_\circ = \left(\frac{er}{1-e\cos p},0\right) \quad Q_\circ = \left(\frac{er}{1-e\cos q},0\right) \quad\to\quad |P_\circ Q_\circ| = \frac{e^2 r \,(\cos p - \cos q)}{(1 - e \cos p) (1 - e \cos q)} \tag1$$

Now, let $S:=C(s)$ be some other point on the conic, and let $\overline{SP'}$ and $\overline{SQ'}$ be tangent to $\bigcirc P_\circ$ and $\bigcirc Q_\circ$ at $P'$ and $Q'$. Pythagoras (and some ugly symbol manipulation) tells us $$ |SP'|^2 = |SP_\circ|^2 - |PP_\circ|^2 = \frac{e^2 r^2 (\cos s - \cos p)^2}{(1 - e \cos s)^2 (1 - e \cos p)^2} $$ Thus, for appropriate choices of sign $\pm_1$ and $\pm_2$, we have $$|SP'| = \pm_1\frac{er (\cos p-\cos s)}{(1-e\cos p)(1-e\cos s)} \qquad |SQ'| = \pm_2\frac{er (\cos q-\cos s)}{(1-e\cos q)(1-e \cos s)}$$ Those signs depend upon the position of $S$ relative to $P$ and $Q$. Specifically, we have these cases

• $S$ between $P$ and $Q$ ($p\leq s\leq q$): $\quad\pm_1= +, \quad \pm_2=-$ $$|SP'|+|SQ'|=|P_\circ Q_\circ|$$

• Otherwise ($s \leq p \leq q$ or $p\leq q\leq s$): $\quad\pm_1=\pm_2$ $$\left|\;|SP'|-|SQ'|\;\right|=|P_\circ Q_\circ|$$ This verifies OP's claims ... mostly. Recall that we ignored $P$ and/or $Q$ being a vertex of the conic; this is because the point and its normal do not determine the tangent circle. Consideration of this case is left as an exercise to the reader. $\square$