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I have three touching circles, and I was trying to find the point inbetween them that is equally far from all three circles. So I created a system of equations:

$$ \begin{array}{lcl} (x-x1)^2 + (y-y1)^2=(r+r1)^2 \\ (x-x2)^2 + (y-y2)^2=(r+r2)^2 \\ (x-x3)^2 + (y-y3)^2=(r+r3)^2 \end{array} $$

$x1, x2, x3, y1, y2, y3, r1, r2, r3$ are the coordinates of the 3 circles and their radii. $x, y$ are the x-and y-coordinate of the point I'm looking for. $r$ is the distance of that point to each circle.

It seems like a simple system, but apparently it's not. Could anyone help me with solving the problem?

Also, a more general solution could be useful, where the initial three circles don't have to touch (but don't overlap). Then I would need to find the biggest circle that can fit in between them, without overlapping any.

Thanks!

The Oddler
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    I think you've underestimated the complexity of the problem. Mathematica compute the result in just under a second but the result is nearly half a megabyte of textual output and is certainly unfit for human consumption. (This is why it WA choked on the return.) You should probably look into the Descartes Circle Theorem and it's generalizations for a more elegant approach to your problem. – Mark McClure Nov 23 '15 at 17:24
  • "Why didn't WA work" is an inappropriate question for this site, although you could ask a question directly about the problem and be fine... – rschwieb Nov 23 '15 at 17:29
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    Mathematica is not very good at simplifying things. The actual answer is quite comprehensible. – Ivan Neretin Nov 23 '15 at 17:43
  • @rschwieb Ok thanks, I'll change my question to that then. – The Oddler Nov 23 '15 at 21:06
  • @IvanNeretin That indeed looks like what I'm looking for. I'll have to check it out in more detail later this week. If you formulate it as an answer, now that I have changed my question to better for math.stackexchange, I will accept it. Thanks a lot for the help! – The Oddler Nov 23 '15 at 21:19

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