Context
The other day I wanted to answer this question.
Which is now closed so doesn't accept answers (but this isn't the important part).
Since I didn't know the topic I went to look it up but I saw that there isn't much about it online, so out of curiosity I wanted to see what the formula for the center and the radical radius was.
I consider 3 generic circles in the Cartesian plane: $$c_i: (x-x_i)^2+(y-y_i)^2=r_i^2\qquad\text{for }i=1,2,3$$
By doing a few steps I arrived at these formulas: $$\mathbf{C}=(x_C,y_C)=\left(\frac{\begin{vmatrix} y_2y_3+x_1^2-r_1^2&y_1&1\\ y_3y_1+x_2^2-r_2^2&y_2&1\\ y_1y_2+x_3^2-r_3^2&y_3&1 \end{vmatrix}}{2\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}}, \frac{\begin{vmatrix} x_1&x_2x_3+y_1^2-r_1^2&1\\ x_2&x_3x_1+y_2^2-r_2^2&1\\ x_3&x_1x_2+y_3^2-r_3^2&1 \end{vmatrix}}{2\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}}\right)$$ and $$r=\begin{cases}\sqrt{\left(x_C-x_{1}\right)^{2}+\left(y_C-y_{1}\right)^{2}-r_{1}^{2}}\\ \sqrt{\left(x_C-x_{2}\right)^{2}+\left(y_C-y_{2}\right)^{2}-r_{2}^{2}}\\ \sqrt{\left(x_C-x_{3}\right)^{2}+\left(y_C-y_{3}\right)^{2}-r_{3}^{2}}\end{cases}$$
Question
The 3 formulas for calculating the radius are all equivalent and correct, but I would like them not to depend so directly on a specific center.
In simple words: I would like a formula that is "regular/symmetrical" like the one for the center and that it does not directly depend on the fact of necessarily having to calculate $x_C$ and $y_C$ first, because I'm pretty sure it can be written as the ratio of two determinants of some matrix (the denominator matrix is that of the determinant of the coordinates of the centers).
Obviously I tried to develop the formulas but the formulas are excessively long (only the term $(x_C-x_1)^2$ alone contains 224 monomials, so in total there are 449 terms), could anyone help me in some way to obtain the formula efficiently?
Essentially there would be this calculation to do:
$$r=\sqrt{\left(\frac{\begin{vmatrix} y_2y_3+x_1^2-r_1^2&y_1&1\\ y_3y_1+x_2^2-r_2^2&y_2&1\\ y_1y_2+x_3^2-r_3^2&y_3&1 \end{vmatrix}}{2\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}}-x_{1}\right)^{2}+\left(\frac{\begin{vmatrix} x_1&x_2x_3+y_1^2-r_1^2&1\\ x_2&x_3x_1+y_2^2-r_2^2&1\\ x_3&x_1x_2+y_3^2-r_3^2&1 \end{vmatrix}}{2\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}}-y_{1}\right)^{2}-r_{1}^{2}}$$
My approach of solving this would be the following: There must be one number $r$ such that $$ (x_C-x_1)^2+(y_C-y_1)^2-r_1^2=r^2 \\ (x_C-x_2)^2+(y_C-y_2)^2-r_2^2=r^2 \\ (x_C-x_3)^2+(y_C-y_3)^2-r_3^2=r^2 $$ simultaneously. We can rearrange this and obtain $$ x_1x_C+y_1y_C+\frac{1}{2}(r^2-x_C^2-y_C^2)=\frac{1}{2}(x_1^2+y_1^2-r_1^2) \\ x_2x_C+y_2y_C+\frac{1}{2}(r^2-x_C^2-y_C^2)=\frac{1}{2}(x_2^2+y_2^2-r_2^2) \\ x_3x_C+y_3y_C+\frac{1}{2}(r^2-x_C^2-y_C^2)=\frac{1}{2}(x_3^2+y_3^2-r_3^2) $$ With $\lambda = \frac{1}{2}(r^2-x_C^2-y_C^2)$, we get $$ \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \begin{pmatrix} x_C \\ y_C \\ \lambda \end{pmatrix} = \frac{1}{2} \begin{pmatrix} x_1^2+y_1^2-r_1^2 \\ x_2^2+y_2^2-r_2^2 \\ x_3^2+y_3^2-r_3^2 \end{pmatrix} $$ Using Cramer's rule, we can see that $$ (x_C,y_C)=\left( \frac{ \begin{vmatrix} x_1^2+y_1^2-r_1^2 & y_1 & 1 \\ x_2^2+y_2^2-r_2^2 & y_2 & 1 \\ x_3^2+y_3^2-r_3^2 & y_3 & 1 \end{vmatrix} }{ 2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} } ,\frac{ \begin{vmatrix} x_1 & x_1^2+y_1^2-r_1^2 & 1 \\ x_2 & x_2^2+y_2^2-r_2^2 & 1 \\ x_3 & x_3^2+y_3^2-r_3^2 & 1 \end{vmatrix} }{ 2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} } \right) $$ and $$ \lambda = \frac{ \begin{vmatrix} x_1 & y_1 & x_1^2+y_1^2-r_1^2 \\ x_2 & y_2 & x_2^2+y_2^2-r_2^2 \\ x_3 & y_3 & x_3^2+y_3^2-r_3^2 \end{vmatrix} }{ 2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} } $$ from which we get $$ r^2=2\lambda + x_C^2+y_C^2 = \frac{ \begin{vmatrix} x_1 & y_1 & x_1^2+y_1^2-r_1^2 \\ x_2 & y_2 & x_2^2+y_2^2-r_2^2 \\ x_3 & y_3 & x_3^2+y_3^2-r_3^2 \end{vmatrix} }{ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} } +x_C^2+y_C^2 $$ which is symmetrical with respect to the centers and radii of the circles. I do not know if this can be simplified any further. If you want one fraction of determinants, you can write it as $$ r^2= \frac{ \begin{vmatrix} x_1 & y_1 & x_1^2+y_1^2-r_1^2+x_C^2+y_C^2 \\ x_2 & y_2 & x_2^2+y_2^2-r_2^2+x_C^2+y_C^2 \\ x_3 & y_3 & x_3^2+y_3^2-r_3^2+x_C^2+y_C^2 \end{vmatrix} }{ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} } $$ Note: In order to see why $$ \begin{vmatrix} x_1^2+y_1^2-r_1^2 & y_1 & 1 \\ x_2^2+y_2^2-r_2^2 & y_2 & 1 \\ x_3^2+y_3^2-r_3^2 & y_3 & 1 \end{vmatrix} = \begin{vmatrix} x_1^2+y_2y_3-r_1^2 & y_1 & 1 \\ x_2^2+y_3y_1-r_2^2 & y_2 & 1 \\ x_3^2+y_1y_2-r_3^2 & y_3 & 1 \end{vmatrix} $$ subtract $(y_1+y_2+y_3)$ times the second column from the first column and then add $(y_2y_3+y_3y_1+y_1y_2)$ times the third column to the first column in the left matrix.