locus problem in analytical geometry asking about a constant sum of two tangents to two identical circles yielding an ellipse You are given two circles:
Circle G: $(x-3)^2 + y^2 = 9$
Circle H: $(x+3)^2 + y^2 = 9$
Two lines that are tangents to the circles at point $A$ and $B$ respectively intersect at a point $P$ such that $AP + BP = 10$
Find the locus of all points $P$.


This problem is solvable if we set point $P = (x,y)$ and solve the equation $AP + BP = 10$.  After substituting  $GP^2 = AP^2 + 3^2$ and $HP^2 = BP^2 + 3^2$ and getting the following equation for an ellipse 
$16x^2 +25y^2 = 625$
That's a lot of math and algebra to do, so my question is:  What is the geometric reasoning behind why is the locus an ellipse (without using analytical geometry) or is there any other elegant proofs that lack heavy calculations?
 A: Not answering the question but giving further observation,
\begin{align}
  \sqrt{(x-r)^2+y^2-r^2} \pm \sqrt{(x+r)^2+y^2-r^2} &= 2s \\
  \sqrt{x^2-2rx+y^2} \pm \sqrt{x^2+2rx+y^2} &= 2s \\
  2(x^2+y^2) \pm 2\sqrt{(x^2+y^2)^2-4r^2x^2} &= 4s^2 \\
  (x^2+y^2)^2-4r^2x^2 &= 4s^4-4s^2(x^2+y^2)+(x^2+y^2)^2 \\
  (s^2-r^2)x^2+s^2y^2 &= s^4 \\
\end{align}


*

*Positive sign is taken when $s^2>r^2$ giving an ellipse.

*Negative sign is taken for constant difference instead, the locus can be two horizontal lines $(s^2=r^2)$, a hyperbola $(s^2<r^2)$ or a vertical line $(s=0)$.

*The loci always pass through the point $(0,\pm s)$.

*The loci don't have contact with two circles when $s^2>2r^2$.

*The eccentricity is given by $e=\dfrac{r}{|s|}$.
A: Not really an answer to the question, but I wanted to post this gif that shows the ellipse being formed

A: 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
 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 3By 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
For more details, see Apostol, Tom M., and Mamikon A. Mnatsakanian. “New Descriptions of Conics via Twisted Cylinders, Focal Disks, and Directors.”.
