Converse inversion I was puzzling with circle inversions and was wondering about a converse of this:
Given 3 points A, B and M 
Is there only one point R so that $d(R,A) * d(R,B) = d(R,M)^2$ or are there more of them?
(No there are many points R, figured out that much)
And if there a more of these points what curve do they form?
 A: Assuming that we're talking in the plane with the standard (Euclidean) distance, let's go to coordinates:
\begin{align}
R&=(r_1,r_2)\\
A&=(a_1,a_2)\\
B&=(b_1,b_2)\\
M&=(m_1,m_2)
\end{align}
In this case,
\begin{align}
d(R,A)&=\sqrt{(r_1-a_1)^2+(r_2-a_2)^2}\\
d(R,B)&=\sqrt{(r_1-b_1)^2+(r_2-b_2)^2}\\
d(R,M)&=\sqrt{(r_1-m_1)^2+(r_2-m_2)^2}
\end{align}
Therefore, if we square both sides of your given equation, you're asking for solutions to
$$
\left((r_1-a_1)^2+(r_2-a_2)^2\right)\left((r_1-b_1)^2+(r_2-b_2)^2\right)=\left((r_1-m_1)^2+(r_2-m_2)^2\right)^2.
$$
If you multiply this out, the degree $4$ terms ($r_1^4$, $r_1^2r_2^2$, and $r_2^4$) cancel out, leaving a cubic bivariate polynomial whose solutions are the $R$'s of interest.  There are lots of cubic plane curves, so it is unlikely that this curve has a name.
However, this formulation allows us to see that there are multiple solutions to the given equation (in many cases).  Suppose that you have chosen a value for $r_1$, then you are left with a cubic in $r_2$.  Since cubics always have roots real, you can find a value for $r_2$.  Since you can do this for every $r_1$ (except for, perhaps, finitely many), there are infinitely many $R$'s satisfying the equation (unless there is some cancellation due to a special structure leading to cancellation in the cubic between $A$, $B$, and $M$.
A: There are infinitely many. 
The locus of points such that the product of the distances from those points to two fixed points is constant, is called a Cassini oval. Thus, for two fixed points $q_1$ and $q_2$, a point $p$ on the oval fulfills $dist(q_1,p) \cdot dist(q_2,p) = k$. In the link there is a figure showing an example of the ovals. 
On page $9$ in this paper the equation of the ovals is found in the following form: $$y=\pm \sqrt{-(x^2+a^2) + \sqrt{k^2+4a^2x^2}}$$
where $k$ is the constant and the two fixed points are $(-a,0)$ and $(a,0)$. 
In the figure below I've plotted ovals of the two points $(-2,0)$ and $(2,0)$ for constant values $k=1$ (blue), $k=2$ (red) and $k=3$ (green). I've also introduced a third point at $(1,1)$ and plotted circles centered at that point with radii $\sqrt 1$, $\sqrt 2$, and $\sqrt 3$. 

We see that for $k=1$, there is no intersection of the blue circle with the blue ovals. But for $k=2$ (red) and $k=3$ (green), there are intersections. 
Thus, if a circle centered on the point $q_3$ with radius $\sqrt k$ intersects a Cassini oval with constant $k$, the point of intersection $p$ will fulfill your requirement, i.e. $dist(q_1,p) \cdot dist(q_2,p) = dist(q_3,p)^2$. It should be clear that this will always be possible for large enough $k$.  
EDIT
It seems that the set of points which fulfill your requirement are in fact hyperbolas. At least, I can show this is true when the third point is placed at origo. For the oval we have: $$y=\pm \sqrt{-(x^2+a^2) + \sqrt{k^2+4a^2x^2}}\tag{1}$$
and for the circle we have $$x^2+y^2 = k$$
or $$y=\pm \sqrt{k-x^2}\tag{2}$$
Setting equations 1 and 2 equal we find that $$x^2=\frac{2k+a^2}{4}$$
or $$k=\frac{4x^2-a^2}{2}\tag{3}$$
Inserting equation 3 into equation 2 we find, for $k \ge \frac{a^2}{2}$, that $$x^2-y^2=\frac{a^2}{2}$$
which is the equation of a hyperbola. 
