I have the equations of three circles:
$$x^2 + y^2 = r^2$$
$$(x-1.3)^2 + y^2 = (r+m)^2$$
$$x^2 + (y+1.3)^2 = (r+n)^2$$
Where $n$ and $m$ are values that are calculated and will be replaced.
Using these equations, how can I find the point at which the circles are closest to intersecting (they might not have an exact intersection).
Any help or ideas are appreciated!
Subtract the first equation from the second one to get
$$ a^2 - 2 a x = m ^2 +2 m r $$
where $a = 1.3$.
The above gives us the $y$ coordinate of the two possible intersection points between the 1st and 3rd circle.
$$ x_{1,2} = \frac{ a^2 - 2 n r - n^2 }{2 a} $$
The two possible $x$ coordinates of the intersection points are
$$ y_{1,2} = \pm \sqrt{r^2 - x_{1,2}^2} $$
Now subtract the 1st equation from the 3rd equation to get
$$ b^2 - 2 b y = n^2 + 2 n r $$
where $b=-1.3$.
The above gives us the $y$ coordinate of the two possible intersection points between the 1st and 2nd circle.
$$ y_{3,4} = \frac{ b^2 - 2 m r - m^2 }{2 b} $$
The two possible $x$ coordinates of the intersection points are
$$ x_{3,4} = \pm \sqrt{r^2 - y_{3,4}^2} $$
Similarly, subtract the 2nd equation from the 3rd equation to get the equation of the line that connects the two intersection points.
$$2 a x -2 b y -a^2+b^2 = -m^2 -2 m r +n^2 + 2 n r$$
and you can use the 2nd equation and the line above to find the coordinates of the final intersection.
Note that some sets of points might not exist, as those circles might not intersect. Now from the possible 3 sets of 2 intersection points (6 points total), you can take an average, or try to find a circle using least squares that will give you a center and an average radius.
At this point the problem is not well defined, as finding a common point between 6 points is most likely not possible.