Place three circles such that they uniquely intersect at each point in the plane

Is it possible to place three circle centers in a plane such that there is a single unique three-way intersection between the three circles for any given set of circle radii? For example, see the image below:

I've chosen the three locations shown by cyan dots. In this case, the arrangement appears to have two viable sets of three-way intersections at two separate scalings (as shown by the blue circle and smaller black circle). One set of is shown by the pink circles and another scale by the light blue circles.

In this case, I'm using fixed-point math to determine that an intersection is "close enough" such that it is considered a three-way intersection, but I thought perhaps it was possible to show analytically that no matter what positions you place the three circle centers you'd never be able to have only one three-way intersection given different scalings. If somebody could show that then I can stop trying different arrangements of the three centers!

Thanks!

EDIT: There seems to be some confusion about what I'm asking. Here's another way to put it: can we place three circles in a plane such that every point in the plane has a unique combination circle radii such that when they intersect at that point they intersect nowhere else. Hopefully that clears things up a bit?

EDIT 2: Let me try describing it another way. Imagine I have some hidden point. In this case, say it's the black point. I tell you the distance up to a scale factor from each of your three circle centers. Is it possible for you to place those circle centers to guarantee that you can always uniquely identify where my hidden point is? In this case, I'm showing a counterexample. It's not a good placement because there are two possible places that hidden point could be. I'm asking whether there's a "good" arrangement such that this case never happens.

EDIT 3: Ugh, I'm doing a terrible job here. Let's try a fuller story. Imagine I have three signal receivers placed on the ground in some configuration. These are the three circle centers. I'm also holding some beacon that emits a signal to the receivers. This is the black point. This beacon has an unknown signal quality from, say, [0.5 1.5]. If the signal quality is 1.0, the number the three receivers show me is 1.0 * distance to the beacon. If the signal quality is 0.5, then the receivers show me 0.5 * distance to the beacon. Imagine we do not know the signal quality -- it is hidden from us. Is it possible to uniquely determine the actual position of the beacon given only the three (scaled by quality) receiver distances? In the image I've shown, this placement of the circle centers appears to be bad, because the beacon could be at the blue location with a lower signal quality OR the black location with a higher signal quality. Is it possible to place the receivers in some other configuration (all in a line? along a circle?) such that we're guaranteed to have a unique solution up to a scale?

• I think it is better to reformulate your question using the term "radius" instead of "scale". – TZakrevskiy Nov 7 '14 at 21:29
• OK, that sounds like a good idea. I'll make the changes. – aardvarkk Nov 7 '14 at 21:29
• I have absolutely no idea what the question is even after the 2 edits. – user2345215 Nov 7 '14 at 21:43
• Damn! I'm really sorry, I'm trying my best to be clear. My edits are probably confusing the matter more. I'll try one more edit and describe the underlying problem so hopefully you can see where I'm trying to go with it. – aardvarkk Nov 7 '14 at 21:47

After the comment below, I interprete the question differently, again. The equstion then seems to be whether there are points $A,B,C$ such that the map $$P\mapsto |P-A|:|P-B|:|P-C|$$ is injective. The answer is: No. If $A,B,C$ are collinear, we have reflection symmetry, hence always at least two solutions to a (valid) given proportion $|P-A|:|P-B|:|P-C|$. Therefore assume $A,B,C$ are not collinear. Assume for the moment that $\Delta ABC$ is not isosceles. Relabel the points so that $\angle CBA$ is one of the acute angles of the triangle. Then $\angle$ the prolonged ray $BC$ intersects the perpendicular bisector $\ell$ of $AB$ (i.e. the locus of points $P$ with $|P-A|:|P-B|=1$) in some point $D$ that is $\ne B$. The locus of points $P$ with $|P-B|:|P-C|=|D-B|:|D-C|$ is a circle $\odot$ with center on the line $BC$, hence intersects the line $\ell$ in a second point $E$ (apart from $D$). We conclude that $$|D-A|:|D-B|:|D-C| = |E-A|:|E-B|:|E-C|$$ (There are many more such pairs, this was just especially easy to construct).
The above construction fails only if the circle $\odot$ degenerates to a point, i.e. if $\Delta ABC$ is isosceles. In that case we can make a minor adjustment to move $D$ a little bit away from $B$ by replacing $\ell$ with the locus of points $P$ with $|P-A|:|P-B|=q$ where $q$ is only approximately $1$; this is a (large) circle that intersects $BC$ not orthogonally, hence this leads to a point $D\ne B$ and then $E\ne D$.