Contd. from my previous post
anuttarasammyak said:
The system of four equations of the circles would show one degenerate solution.
The equation of line which connects two commom points of the cirlces i and j, is
(l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0
The system of such six lines meet at a point (x,y) are recuced to three equations.
<br />
\begin{pmatrix}<br />
l_1-l_2 & m_1-m_2 & n_1-n_2 \\<br />
l_2-l_3 & m_2-m_3 & n_2-n_3 \\<br />
l_3-l_4 & m_3-m_4 & n_3-n_4 \\<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
x \\<br />
y \\<br />
1 \\<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
0 \\<br />
0 \\<br />
0 \\<br />
\end{pmatrix}<br />
This 3X3 matrix in LHS shoud have determinant zero so that there is a solution.
<br />
\begin{vmatrix}<br />
l_1-l_2 & m_1-m_2 & n_1-n_2 \\<br />
l_2-l_3 & m_2-m_3 & n_2-n_3 \\<br />
l_3-l_4 & m_3-m_4 & n_3-n_4 \\<br />
\end{vmatrix}<br />
=\sum_{i,j,k,q=1}^4 P(ijkq)l_i m_j n_k<br />
where P(ijkq)=0 when any of them equals.
P(ijkq)=1 when ijkq is even pertumation of 1234.
P(ijkq)=-1 when ijkq is odd permutation of 1234.
We expect to get coordinates of M as
<br />
\begin{pmatrix}<br />
l_1-l_4 & m_1-m_4 & n_1-n_4 \\<br />
l_2-l_4 & m_2-m_4 & n_2-n_4 \\<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
x \\<br />
y \\<br />
1 \\<br />
\end{pmatrix}<br />
=<br />
\begin{pmatrix}<br />
0 \\<br />
0 \\<br />
\end{pmatrix}<br />
or
<br />
\begin{pmatrix}<br />
x \\<br />
y \\<br />
\end{pmatrix}<br />
=\frac{-1}{(l_1-l_4 )(m_2-m_4 )-(l_2-l_4 )(m_1-m_4 )}<br />
\begin{pmatrix}<br />
m_2-m_4 & m_4-m_1 \\<br />
l_4-l_2 & l_1-l_4 \\<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
n_1-n_4 \\<br />
n_2-n_4 \\<br />
\end{pmatrix}<br />