Given two non-intersecting circles (i.e. their centers are separated by a distance larger than the sum of their radii), it is always possible to place a third circle on top of (or below) these circles such that the third circle is tangent to the first two circles. In fact there are infinitely many such third circles.
My question is: what is the curve described by the centers of such third circles?
If the parameters of the first two circles are $x_1, y_1, r_1$ and $x_2, y_2, r_2$ it seems to me that the center of the third circle $(x,y)$ should satisfy the following $$ \sqrt {(x-x_1)^2+(y-y_1)^2} - r_1 = \sqrt {(x-x_2)^2+(y-y_2)^2} - r_2$$ But the solution to this equation eludes me.
HINT
Wlog we can assume
and consider
$$\sqrt {x^2+y^2} - 1 = \sqrt {(x-a)^2+(y)^2} - r$$
$$\sqrt {x^2+y^2} = \sqrt {(x-a)^2+y^2} +1- r$$
$${x^2+y^2} = (x-a)^2+y^2 +(1- r)^2+2(1-r) \sqrt {(x-a)^2+y^2}$$
$$x^2+y^2-x^2+2xa-a^2-y^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$
$$2xa-a^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$
$$[2xa-(a^2 +(1- r)^2)]^2= 4(1-r)^2 [(x-a)^2+y^2]$$
$$4x^2a^2-4xa(a^2 +(1- r)^2)+(a^2 +(1- r)^2)^2=4(1-r)^2x^2-8a(1-r)^2x+4a^2(1-r)^2+4(1-r)^2y^2$$
$$[4a^2-4(1-r)^2]x^2+[8a(1-r)^2-4a(a^2 +(1- r)^2]x-4(1-r)^2y^2=4a^2(1-r)^2-(a^2 +(1- r)^2)^2$$
Note that for $r=1$ we obtain
$$4a^2x^2-4a^3x=-a^4\implies 4x^2-4ax+a^2=0 \implies (2x-a)^2=0 \implies x=\frac a 2$$
For $r=2$ and $a =4$ we have
$$60x^2-240x-4y^2-225=0$$