I want to find the circle which exclusively touches 3 other circles.

This is essentially the classic problem of Apollonius. I use the following equation to find the center of that circle and its radius:

$\left( \begin{matrix} 2(x_2 -x_1) & 2(y_2 - y_1) & 2(r_2 -r_1) \\ 2(x_3 -x_2) & 2(y_3 - y_2) & 2(r_3 -r_2) \\ 2(x_1 -x_3) & 2(y_1 - y_3) & 2(r_1 -r_3) \end{matrix} \right) \cdot \left( \begin{matrix} x_s \\ y_s \\ r_s \end{matrix} \right) = \left( \begin{matrix} x_2^2 -x_1^2 +y_2^2-y_1^2+r_1^2-r_2^2 \\ x_3^2 -x_2^2+y_3^2-y_2^2+r_2^2-r_3^2 \\ x_1^2 - x_3^2 +y_1^2-y_3^2 +r_3^2-r_1^2 \end{matrix} \right) $

where $x_s , y_s$ and $r_s$ are the coordinates of the center point and the radius of the solution circle. The other points are the coordinates and the radii of the other three circles respectively.

So basically this is a system of equations of the form:

$ A \cdot x = B$

with the solution

$x = A^{-1} \cdot B$

If I use three circles with equal radius this system is suddenly not solvable anymore, as the last column in A becomes 0 and $det A = 0$.

But there is clearly a unique solution or am I wrong here? Keep in mind, it is only about exclusively touching circles and only about the problem with 3 circles.

So what am I doing wrong here?

  • $\begingroup$ In general, Apollonius' problem has eight distinct solutions. In the particular case of congruent circles, two of those solutions are identical, so there are only seven distinct solutions. The midpoint of the double solution is in this case exactly the circumcenter of the three circles' midpoints. Can you figure out why this is? $\endgroup$ – Anon Jul 18 '16 at 13:36
  • $\begingroup$ I guess its because in that case they form a triangle, in which the circle centers being the triangle points. The point with equal distance from each of these points is the centroid of the triangle. $\endgroup$ – FreddyKay Jul 18 '16 at 13:40
  • $\begingroup$ And it's clear why the center of the double solution has equal distance from each of the circle centers? $\endgroup$ – Anon Jul 18 '16 at 13:43
  • $\begingroup$ I understand, that the center of the solution circle has to be at equal distance from the other circles as they are congruent. In which case the exclusion circle and the inclusion circle share the same center. Which is why I do not get a unique solution for the above equation? $\endgroup$ – FreddyKay Jul 18 '16 at 13:49

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