Finding equation and centre of circle through 3 points using matrices Based on answer given here: Get the equation of a circle when given 3 points.  We can find equation of circle through points $(1,1), (2,4), (5,3)$ by taking:
$\left|\begin{array}{cccc}
      x^2+y^2&x&y&1\\
      1^2+1^2&1&1&1\\
      2^2+4^2&2&4&1\\
      5^2+3^2&5&3&1\\
    \end{array}\right|=0$
Can anyone explain why this works?  I know generally what determinant equal zero means but can't see why doing this works.  

This comes from the general equation of a circle
$$x^2+y^2+Ax+By+E=0$$ which gives since $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are points in the circle
$$x^2+y^2+Ax+By+E=0\\x_1^2+y_1^2+Ax_1+By_1+E=0\\x_2^2+y_2^2+Ax_2+By_2+E=0\\x_3^2+y_3^2+Ax_3+By_3+E=0$$ This is equivalent to 
$$\left (\begin{matrix}x^2+y^2&x&y&1\\x_1^2+y_1^2&x_1&y_1&1\\x_2^2+y_2^2&x_2&y_2&1\\x_3^2+y_3^2&x_3&y_3&1\end{matrix}\right)*\left(\begin{matrix}1\\A\\B\\C\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\\0\end{matrix}\right)$$
Because, clearly, there is not unique solution ( in particular the zero matrix  in the RHS is solution) the determinant must be equal to zero.
