I have 3 circles in a 2D space with known center coordinates ((xa,ya),(xb,yb),(xc,yc)).
The circles don't intersect at a unique point but they should be close to it. I would like to find the point (x,y) at which they would intersect if we were to change their radiuses the slightest possible to make them intersect in only one point.
eg: circle A has a radius of Ra, B of Rb, C of Rc. I would like to find the intersection of 3 circles with radiuses (1+Da)Ra, (1+Db)Rb, (1+Dc)Rc with the minimum values for {Da, Db, Dc} (according to norm2?).
I can't get my head around it, I have
(1+Da)^2+(1+Db)^2+(1+Dc)^2 = ((y-ya)^2+(x-xa)^2)/Ra^2+((y-yb)^2+(x-xb)^2)/Rb^2+((y-yc)^2+(x-xc)^2)/Rc^2
But if I minimize the right part of the equation I am actually finding the point which would minimize the circles' radiuses, not the deformation of the radiuses, and I don't get a satisfying result.
Is it possible to solve this problem?
A beginning:
Let $(x_i,y_i)$ be the centers of the three given circles, $r_i$ their radii, $s_i$ the envisaged correction of $r_i$, and $(x,y)$ the prospective point of intersection of the corrected circles. Then you want to minimize $$f(s_1,s_2,s_3):=\sum_i s_i^2$$ under the three constraints $$(x-x_i)^2+(y-y_i)^2-(r_i+s_i)^2=0\qquad(1\leq i\leq3)\ .\tag{1}$$You therefore have to set up the Lagrangian $$\Phi(x,y,s_1,s_2,s_3):=f(s_1,s_2,s_3)-\sum_i\lambda_i\bigl((x-x_i)^2+(y-y_i)^2-(r_i+s_i)^2\bigr)$$ and solve the system consisting of $(1)$ and the five equations $${\partial\Phi\over\partial x}=0,\quad {\partial\Phi\over\partial y}=0,\qquad {\partial\Phi\over\partial s_i}=0 \quad(1\leq i\leq3)\ .$$ Good luck!