Th equation of a circle circumscribing a quadrilateral whose sides in order are represented by the lines ${L_1} = 0$, ${L_2} = 0$, ${L_3} = 0$, and ${L_4} = 0$ is given by ${L_1}{L_3} + \lambda {L_2}{L_4} = 0$ provided the coefficients of ${x^2}$ and ${y^2}$ are equal and the coefficient of $xy = 0$.
Here, ${L_1}{L_3} + \lambda {L_2}{L_4} = 0$ is definitely a second degree equation and it would represent a circle if there exists a $\lambda$ for which the coefficients of ${x^2}$ and ${y^2}$ are equal and the coefficient of $xy = 0$.
I was wondering if the order mattered here and ${L_1}{L_2} + \lambda {L_3}{L_4} = 0$ would also give rise to a equation of circle if the above conditions are met.
So I took a quadrilateral that could actually be inscribed in a circle and tried it. But the equation ${L_1}{L_2} + \lambda {L_3}{L_4} = 0$ never represented a circle.
So, why should we consider the product of only the opposite sides while finding the equation of the circle circumscribing a quadrilateral?
Best Answer
Because of the way the sides are connected.
Look at the vertex where the $L_1$ and $L_2$ lines intersect. You have $L_1=L_2=0$ but $L_3\ne 0$ and $L_4\ne 0$. So you have a vertex where the product $L_3L_4$ isn't zero, and therefore a linear combination containing that product can't be uniformly zero. You get similar results at the other vertices.
But the opposite line pairs $\{L_1, L_3\}, \{L_2, L_4\}$ are never involved in a nonzero product at any vertex. So the opposite line pairs must be the ones whose products give an identically zero linear combination.