What is the locus of points equidistant from two circles? 
What is the locus of points equidistant from two circles?

$$ x^2+y^2+ 2 h x + 2 g y + c =0 ;\; C =0 ;$$

Construction of circles with $(a,b,2h)= (3,2,3.6)$
Please help finding equation of the locus equidistant from two circles $ C_1=0, C_2=0 ,$ if possible in terms of  $ C_1, C_2. $
.. Like  we have radical axis $ C_1 = C_2$ for equal tangents condition.
EDIT1:
It now appears that $ e=\pm 1$ refer to hyperbola and ellipse cases
respectively , but we should express eccentricities in terms of $ a,b,h$ and parametrizations as the next step.
 A: 
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 an equal distance form both circles : $L1 = L2$
$$\left\lvert  \sqrt{(x - x1)^2 + (y-y1)^2} - r1 \right\rvert = \left\lvert  \sqrt{(x - x2)^2 + (y-y2)^2} - r2 \right\rvert \tag{4}$$
The sign changes if $Z$ is inside the circle.
A: The distance of a point $P$ from the circle centre $A,$ radius $r$ is $\|AP\| - r.$
The distance of the same point $P$ from the circle centre $A',$ radius $r'$ is $\|A'P\| - r'.$
(I am assuming here that "negative distances" are allowed, as is implied by the diagram in the question.)
The two distances are equal if and only if $\|AP\| - \|A'P\| = r - r'.$
If $r = r',$ then the locus of $P$ is the perpendicular bisector of $AA'.$
(That is, unless also $A = A',$ in which case any point $P$ satisfies the condition.)
If $r \ne r'$ and $\|AA'\| = |r - r'|,$ i.e., if the circles touch internally, then the condition on $P$ is $\|AP\| - \|A'P\| = \|AA'\|$ (if $r > r'$) or $\|A'P\| - \|AP\| = \|AA'\|$ (if $r < r'$). In either of these cases, the locus of $P$ is the ray from the centre of the smaller circle that passes through the circles' point of contact.
If $r \ne r'$ and $\|AA'\| < |r - r'|,$ i.e., if one circle is contained in the other, then by the Triangle Inequality no point $P$ satisfies the condtion.
If $r \ne r'$ and $\|AA'\| > |r - r'|,$ then the locus of $P$ is one branch of a hyperbola whose foci are $A$ and $A'$ and whose eccentricity is $\|AA'\|/|r - r'|.$
