Given three ellipses in the plane specified as follows
$(r - C_1)^T Q_1 (r - C_1) = 1$,
$(r - C_2)^T Q_2 (r - C_2) = 1$
$(r - C_3)^T Q_3 (r - C_3) = 1$
I want to find the circle that is externally tangent to all three ellipses.
My attempt: is detailed in my solution below.
Your comments on both the question and the answer are valuable to me, and highly appreciated, so are your answers.
First of all, the equations of the three ellipses can be written as follows:
$ r^T Q_1 r + b_1^T r + c_1 = 0 $
$ r^T Q_2 r + b_2^T r + c_2 = 0 $
$ r^T Q_3 r + b_3^T r + c_3 = 0 $
where $ b_i = - 2 Q_i C_i $, and $c_i = C_i^T Q_i C_i - 1 $.
Let $r_1$ , $r_2$ , and $r_3$ be the tangency points between the circle to be found and the three ellipses respectively, and let $C_4$ be the center of the circle. Then the following $8$ equations can be written. The first three state that these points belong to the respective ellipses:
$ r_i^T Q_i r_i + b_i^T r_1 + c_i = 1, i = 1,2,3 $
The next three state that
$ (r_i - C_4) = \alpha_i ( 2 Q_i r_i + b_i ) , i=1,2,3 $
Therefore, if anti-symmetric matrix $E$ is defined as
$ E = e_1 e_2^T - e_2 e_1^T
Then, we will have
$ (r_i - C_4)^T E (2 Q_i r_i + b_i) = 0 , i = 1, 2, 3$
And finally
$(r_1 - C_4)^T (r_1 - C_4) = (r_2 - C_4)(r_2 - C_4) $
and
$(r_1 - C_4)^T (r_1 - C_4) = (r_3 - C_4) (r_3 - C_4)$
These are $8$ quadratic equations in $8$ unkowns what are the coordinates of $r_1, r_2, r_3, C_4$.
I solved them using the Newton-Raphson multivariate method, with initial guess of $C_4$ being the centroid of $C_1, C_2, C_3$, and initial guess of $r_1, r_2, r_3$ being $\dfrac{2}{3}$ of the way between the initial guess of $C_4$ and $C_1, C_2, C_3$ respectively.
The method converged very quickly, in just $5$ iterations. The three ellipses I created and the found circle are shown below.