This is not a complete answer to your original question, but there is a nice geometrical explanation on why any point $P$ on the ellipse $16x^2+25y^2=625$ satisfies $|\overline{AP}|+|\overline{BP}|=10$. This configuration can be obtained by considering a hyperboloid of revolution, two spheres tangent to it, and an intersecting plane.
![Figure 1](https://cdn.statically.io/img/i.sstatic.net/inaW1.png)
Figure 1
![Figure 2](https://cdn.statically.io/img/i.sstatic.net/05GPB.png)
Figure 2
Suppose we have a cylindrical surface $S:x^2+y^2=25$, two spheres $S_1:x^2+y^2+(z-5)^2=25$ and $S_2:x^2+y^2+(z+5)^2=25$, and a plane $\Pi:3x-4z=0$ (See Figure 1). Then your configuration appears on $\Pi$ as the intersections of $S$,$S_1$,$S_2$ and $\Pi$ (see Figure 2). Let $G=S_1 \cap \Pi$, $H=S_2 \cap \Pi$, $E=S \cap \Pi$. $C_1=S \cap S_1$, and $C_2=S \cap S_2$. Take any point $P$ on $E$. Let $A$ ($B$ resp.) be the point of tangency of a tangent line from $P$ to $G$ ($H$ resp.). Draw a generator line $l$ of $S$ passing through $P$. Denote the intersection of $l$ and $C_1$ ($C_2$ resp.) as $A'$ ($B'$, resp.).
Because Both $\overline{AP}$ and $\overline{BP}$ are tangent line segments from $P$ to $S_1$, they have the same length (see Figure 3; note that two triangles $PAO_{S1}$ and $PA'O_{S1}$ are congruent, where $O_{S1}$ is the center of $S_1$). Thus $|\overline{AP}|=|\overline{A'P}|$. Similarly, $|\overline{BP}|=|\overline{B'P}|$. Trivially $|\overline{A'B'}|=|\overline{A'P}|+|\overline{B'P}|$ is constant. Therefore, we can conclude that $|\overline{AP}|+|\overline{BP}|$ is also constant. So while this is not a rigorous proof of your original question because we need to consider the converse to this proposition, you can intuitively see why conic sections and two double contact circles have such a property.
![Figure 3](https://cdn.statically.io/img/i.sstatic.net/1I7Fy.png)
Figure 3
By the way, let $\Pi_1$ ($\Pi_2$ resp.) be the plane containing $C_1$ ($C_2$ resp.), and let $d_1=\Pi \cap \Pi_1$ and $d_2=\Pi \cap \Pi_2$. Denote the foot of the perpendicular line from $P$ to $d_1$ ($d_2$ resp.) as $C$ ($D$ resp.) Then,
$$\frac{|\overline{AP}|}{|\overline{CP}|}=\frac{|\overline{BP}|}{|\overline{DP}|}=e$$ ,where $e$ is the eccentricity of $E$. This indicates that $d_1$ and $d_2$ have a property analogous to the directrix of a conic. Compare Figure 1 with Figure 4, where two spheres $S_1':x^2+y^2+(z-6.25)^2=25$ and $S_2':x^2+y^2+(z+6.25)^2=25$ are tangent to $\Pi$ (see also Dandelin Spheres).
![Figure 4:Dandelin spheres](https://cdn.statically.io/img/i.sstatic.net/1zhI2.png)
Figure 4
For more details, see Apostol, Tom M., and Mamikon A. Mnatsakanian. “New Descriptions of Conics via Twisted Cylinders, Focal Disks, and Directors.”.