# Equation of circle circumscribing the triangle formed by three lines

The question is to show that the equation of a circle circumscribing a triangle formed by the lines $a_rx+b_ry+c_r=0$ where $r=1,2,3$ is $$det[ \begin{vmatrix} \frac{a_1^2+b_1^2}{a_1x+b_1y+c_1} & a_1 & b_1\\ \frac{a_2^2+b_2^2}{a_2x+b_2y+c_2} & a_2 & b_2 \\ \frac{a_3^2+b_3^2}{a_3x+b_3y+c_3} & a_3 & b_3 \end{vmatrix}]=0$$

I tried by using that the fact that equation of circle circumscribing the sides formed by $L_1=0,L_2=0,L_3=0$ is $L_1L_2+ d L_2L_2 + e L_3L_1 =0$ subject to condition that coefficient of $xy=0$ and coefficient of $x^2=y^2$ but i cannot bring it in the form required.Any ideas?Thanks.

This uses the Simson's Theorem that the feet of the perpendiculars from any point $(x,y)$ on the circumcircle on the sides of the triangle are collinear.

The feet of the perpendicular from $(x,y)$ on $a_ix+b_iy+c_i=0$ is given by

$x_i = x - \dfrac{-a(a_ix+b_i y+c)}{a_i^2+b_i^2}, y_i = y - \dfrac{-b(a_i x+b_i y+c)}{a_i^2+b_i^2}$

Since $(x_i,y_i)$ are collinear, we have $\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$

Substituting and using transformations:

1. $C_1 \rightarrow C_1-xC_3$

2. $C_2 \rightarrow C_2-yC_3$

3. Multiplying $C_1,C_2$ by $\dfrac {a_i^2+b_i^2}{(a_ix+b_i y+c)}$ we arrive at the desired equation

Note: Please also see Derivation of feet of perpendicular

One very very brute force and tedious way to proceed would be to construct the equations to find the 3 intersections. There will be three equations and each pair will generate the coordinates of an intersection. Now for a the circle, $(x-h)^2+(y-k)^2-r^2=0$, we plug in our previously found intersection points which generates three quadratic equations in $h,k,r$. Solving this quadratic system will give $h,k,r$ in terms of the intersection points. Form the appropriate equation for the circle and then expand the determinant and compare. Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ be the intersection points:

$h = \frac{(y_1^2 + x_1^2) (y_2 - y_3) + (y_2^2 + x_2^2) (y_3 - y_1) + (y_3^2 + x_3^2) (y_1 - y_2)}{2 (x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2))}$

$k = \frac{(y_1^2 + x_1^2) (x_3 - x_2) + (y_2^2 + x_2^2) (x_1 - x_3) + (y_3^2 + x_3^2) (x_2 - x_1)}{2 (x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2))}$

$r = \frac{\sqrt{((x_1 - x_2)^2 + (y_1 - y_2)^2) ((x_1 - x_3)^2 + (y_1 - y_3)^2) ((x_2 - x_3)^2 + (y_2 - y_3)^2)}}{2 (x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2))}$

Still have to be related to the linear coefficients $a_i,b_i,c_i$. Pretty messy. However, this all looks closely related to a Vandermonde system of some sort as does your determinant.