I'm trying to solve the following problem algorithmically. It seems it's a bit too math-y for StackOverflow.
Suppose we're given the coordinates of the centers of three circles ($A,B$ and $C$), as well as the radius of each circle. Now, suppose we iteratively add some small amount to the radius of each circle, until all three circles intersect at some point, $[x_i,y_i]$. How would I go about determining their point of intersection? Now, of course, in practice I can only increase the radius by some discrete amount (maybe $0.01$ units), so I'll have to take into account that they may never actually intersect, rather I'll probably want to determine when they're within some threshold (unless there's a better way to accomplish this entirely). With each iteration, I'll also have to be able to determine whether such an intersection even exists.
The best idea I can think of is to somehow compute every possible point along the circumference of each circle (in practice, a large discrete number, of course), and then check to see if there are three points within some threshold (i.e. there exists 3 points, one contributed by each circle, within some small threshold, maybe $0.01$ units).
Most important here is efficiency and ease of implementation.
NOTE: I should've added– I'm aware of the solution involving finding the intersection of two hyperbolas. However, for this exercise, I'd like to avoid that particular solution.
Best Answer
Let the centres be $(x_A,y_A)$, $(x_B,y_B)$, $(x_C,y_C)$, and the initial radii $r_A$, $r_B$, $r_C$. You want to find $(x,y)$ and $r$ such that $$\tag1(x-x_A)^2+(y-y_A)^2=(r_A+r)^2$$ $$\tag2(x-x_B)^2+(y-y_B)^2=(r_B+r)^2$$ $$\tag3(x-x_C)^2+(y-y_C)^2=(r_C+r)^2$$ Expanding and subtracting $(1)-(2)$ and $(1)-(3)$, we have $$\tag4-2x(x_A-x_B)+(x_A^2-x_B^2)-2y(y_A-y_B)+(y_A^2-y_B^2)\\=r_A^2-r_B^2+3(r_A-r_B)r $$ $$\tag5-2x(x_A-x_C)+(x_A^2-x_C^2)-2y(y_A-y_C)+(y_A^2-y_C^2)\\=r_A^2-r_C^2+3(r_A-r_C)r $$ so two linear equations in the unknowns $x,y,r$. Unless we are unlucky, these determine a line, i.e., something of the form $$\tag6 x=x_0+ar, y=y_0+br.$$ Substituting this into $(1)$, say, you obtain (in general) a quadratic equation in $r$ with hopefully two solutions, and use $(6)$ accordingly to find $(x,y)$.