Please help to obtain.. in as elegant a form as possible.. the locus of point $P$ equations if its distances to two circles:
$$ (x-h)^2 + y^2 = a^2;\;(x+h)^2 + y^2 = b^2 ;\;$$
are in a constant ratio $e,$ or
$$\dfrac{ \sqrt{(x + h)^2 + y^2} - a} { \sqrt{(x - h)^2 + y^2} - b}=e.$$
I am looking at a generalization to conic sections with $ (0,\infty)$ domains for circle radii $(a,b)$.
When $ a=0,b=0 $ we have Apollonius Circles as loci. When $ a > 0,b >0 $ the fixed base focal points of Apollonius Circles are expanded to bigger circles and the circles themselves become distorted as shown below.
Required to obtain a parametrization of the Ovals or equations in terms of any known functions.
The set of new Ovals shown have been obtained by contour plots. Distances are positive when normal distance from point P is outside a circle ( and so $e$), and negative when inside.
EDIT1:
Constants used for plot of new Ovals loci when $(2h < a+b)$ for intersecting circles:
$$\;h=1.8,\;a=3,\;b=2,\;e= (-2,-1,0,1,4).$$
In particular I wish to know if the case of equal distances $e=-1$ is an ellipse or if the circle centers are its foci.
Let the center of $C1$ be $(x1,y1)$ and radius $r1$.
The center of $C2$ is $(x2,y2)$ and radius $r2$.
Z is the locus $(x,y)$.
The distance from $Z$ to the center of $C1$ is:
$$D1 = \sqrt{(x - x1)^2 + (y-y1)^2} \tag{1}$$
The closest point from $C1$ to $Z$ is extended from the radius.
The point on the perimeter is $P1$.
The distance from the perimeter of $C1$ to $Z$ is $L1$, from $P1$ to $Z$:
$$L1 = \left\lvert \sqrt{(x - x1)^2 + (y-y1)^2} - r1 \right\rvert \tag{2}$$
Similarly for $C2$:
$$L2 = \left\lvert \sqrt{(x - x2)^2 + (y-y2)^2} - r2 \right\rvert \tag{3}$$
$Z$ is at a distance ratio $e$ : $L1 = e L2$
$$\left\lvert \sqrt{(x - x1)^2 + (y-y1)^2} - r1 \right\rvert = e \left\lvert \sqrt{(x - x2)^2 + (y-y2)^2} - r2 \right\rvert \tag{4}$$
$(x1,y1) = (h,0)$ , $r1 = |a|$.
$(x2,y2) = (-h,0)$, $r2 = |b|$
The sign changes if $Z$ is inside the circle.
Let
$S1 = \sqrt{(x - x1)^2 + (y-y1)^2} \tag{5}$
$S2 = \sqrt{(x - x2)^2 + (y-y2)^2} \tag{6}$
For $Z$ outside both circles the absolute signs are both $+$.
$S1 - r1 = e(S2 - r2) \tag{7}$
$S1-e S2 = r1 - e r2 \tag{8}$
$(S1-e S2)^2 = (r1 - er2)^2 \tag{9}$
$S1^2 + e^2 S2^2 -2 e S1 S2 = (r1 - e r2)^2 \tag{10}$
$S1^2 + e^2 S2^2 - (r1 - e r2)^2 = 2 e S1 S2 \tag{11}$
$(S1^2 + e^2 S2^2 - (r1 - e r2)^2)^2 = 4 e^2 S1^2 S2^2 \tag{12}$
$(S1^2 + e^2 S2^2 - (r1 - e r2)^2)^2 - 4 e^2 S1^2 S2^2 = 0 \tag{13}$
Maxima code:
Finally with all the substitutions:
$$\left(e-1\right)^2\,\left(e+1\right)^2\,y^4+2\,\left(e-1\right)^2\, \left(e+1\right)^2\,x^2\,y^2+4\,\left(e-1\right)\,\left(e+1\right)\, \left(e^2+1\right)\,h\,x\,y^2+2\,\left(e^4\,h^2-2\,e^2\,h^2+h^2-b^2 \,e^4+2\,a\,b\,e^3-b^2\,e^2-a^2\,e^2+2\,a\,b\,e-a^2\right)\,y^2+ \left(e-1\right)^2\,\left(e+1\right)^2\,x^4+4\,\left(e-1\right)\, \left(e+1\right)\,\left(e^2+1\right)\,h\,x^3+2\,\left(3\,e^4\,h^2+2 \,e^2\,h^2+3\,h^2-b^2\,e^4+2\,a\,b\,e^3-b^2\,e^2-a^2\,e^2+2\,a\,b\,e -a^2\right)\,x^2+4\,\left(e-1\right)\,\left(e+1\right)\,h\,\left(e^2 \,h^2+h^2-b^2\,e^2+2\,a\,b\,e-a^2\right)\,x+\left(e\,h-h-b\,e+a \right)\,\left(e\,h-h+b\,e-a\right)\,\left(e\,h+h-b\,e+a\right)\, \left(e\,h+h+b\,e-a\right) = 0$$
Maxima can solve quartics:
$$y^2={{2\,b\,e^2\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^ 2}-2\,a\,e\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^2}-e^4 \,x^2+2\,e^2\,x^2-x^2-2\,e^4\,h\,x+2\,h\,x-e^4\,h^2+2\,e^2\,h^2-h^2+ b^2\,e^4-2\,a\,b\,e^3+b^2\,e^2+a^2\,e^2-2\,a\,b\,e+a^2}\over{\left(e ^2-1\right)^2}}$$
$$y^2={{2\,b\,e^2\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^ 2}-2\,a\,e\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^2}-e^4 \,x^2+2\,e^2\,x^2-x^2-2\,e^4\,h\,x+2\,h\,x-e^4\,h^2+2\,e^2\,h^2-h^2+ b^2\,e^4-2\,a\,b\,e^3+b^2\,e^2+a^2\,e^2-2\,a\,b\,e+a^2}\over{\left(e ^2-1\right)^2}}$$
$$y^2={{-2\,b\,e^2\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a ^2}+2\,a\,e\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^2}-e^ 4\,x^2+2\,e^2\,x^2-x^2-2\,e^4\,h\,x+2\,h\,x-e^4\,h^2+2\,e^2\,h^2-h^2 +b^2\,e^4-2\,a\,b\,e^3+b^2\,e^2+a^2\,e^2-2\,a\,b\,e+a^2}\over{\left( e^2-1\right)^2}}$$
$$y^2={{-2\,b\,e^2\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a ^2}+2\,a\,e\,\sqrt{-4\,e^2\,h\,x+4\,h\,x+b^2\,e^2-2\,a\,b\,e+a^2}-e^ 4\,x^2+2\,e^2\,x^2-x^2-2\,e^4\,h\,x+2\,h\,x-e^4\,h^2+2\,e^2\,h^2-h^2 +b^2\,e^4-2\,a\,b\,e^3+b^2\,e^2+a^2\,e^2-2\,a\,b\,e+a^2}\over{\left( e^2-1\right)^2}}$$
There is an $x$ under the square root?.
Does this mean the general case is not a conic section?
For $a= 0$, $b = 0$
Maxima code:
$$\left(e-1\right)^2\,\left(e+1\right)^2\,y^4+2\,\left(e-1\right)^2\, \left(e+1\right)^2\,x^2\,y^2+4\,\left(e-1\right)\,\left(e+1\right)\, \left(e^2+1\right)\,h\,x\,y^2+2\,\left(e^4\,h^2-2\,e^2\,h^2+h^2 \right)\,y^2+\left(e-1\right)^2\,\left(e+1\right)^2\,x^4+4\,\left(e- 1\right)\,\left(e+1\right)\,\left(e^2+1\right)\,h\,x^3+2\,\left(3\,e ^4\,h^2+2\,e^2\,h^2+3\,h^2\right)\,x^2+4\,\left(e-1\right)\,\left(e+ 1\right)\,h\,\left(e^2\,h^2+h^2\right)\,x+\left(e\,h-h\right)^2\, \left(e\,h+h\right)^2 = 0$$
A circle is expected for the previous equation:
Try to fit the form : ${a}^2((y -y_0)^2 + (x-x_0)^2 - r^2)^2 = 0$ to it:
Expand this expression and equate the coefficients of each $x^ny^m$:
Maxima code (all variables are solved by calculation):
The substitutions were:
$$y0 = 0 \tag{14}$$
$${\it x_0}=-{{\left(e^2+1\right)\,h}\over{e^2-1}} \tag{15}$$
$$r^2={{4\,e^2\,h^2}\over{e^4-2\,e^2+1}} \tag{16}$$
$$a^2=e^4-2\,e^2+1 \tag{17}$$
The result equation is:
$$y^2+\left(x+{{\left(e^2+1\right)\,h}\over{e^2-1}}\right)^2={{4\,e^2 \,h^2}\over{e^4-2\,e^2+1}} \tag{18}$$
Centers $(h,0)$ and $(0,0)$ and inverting $\displaystyle e \rightarrow \frac1{e} \:$ produced the standard form:
$$\boxed{ y^2+\left(x+{{he^2}\over{1-e^2}}\right)^2={{e^2\,h^2}\over{(1 - e^2)^2}}} \tag{19}$$
Maxima centers $(0,0)$ $(h,0)$ , $eL1 = L2$
$$y^2+\left(x-{{e^2\,h}\over{e^2-1}}\right)^2={{e^2\,h^2}\over{e^4-2 \,e^2+1}} \tag{20}$$