<s>[INST] 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?
 [/INST] 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 $e=\dfrac{r}{|s|}$.
</s>