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Let $p,q,r$ and $s$ be four points on the plane. Moreover, $p,q,r$ are given in clockwise order. My book said that the following determinant is positive if and only if $s$ lies inside the circle passing through $p,q,r$. Why?

$$\det \begin{bmatrix} p_x & p_y & p_x^2+p_y^2 & 1 \\ q_x & q_y & q_x^2+q_y^2 & 1 \\ r_x & r_y & r_x^2+r_y^2 & 1 \\ s_x & s_y & s_x^2+s_y^2 & 1 \\ \end{bmatrix} $$

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If you let $s_x$ and $s_y$ be variable, and equate the mess to zero, what you have is the equation of a circle through three points (equivalently the circumcircle of the triangle formed by your given points). Now, what happens if $f(s_x,s_y)>0$? –  J. M. Apr 14 '12 at 14:06
    
I can see it now. But it is puzzling how people arrive at this determinant form of the equation circumcircle at first? –  FiniteA Apr 14 '12 at 16:13

1 Answer 1

up vote 2 down vote accepted

Following the hints from J.M., I was able to get the answer.

$$\det \begin{bmatrix} p_x & p_y & p_x^2+p_y^2 & 1 \\ q_x & q_y & q_x^2+q_y^2 & 1 \\ r_x & r_y & r_x^2+r_y^2 & 1 \\ s_x & s_y & s_x^2+s_y^2 & 1 \\ \end{bmatrix} $$

$$ =-a(s_x^2+s_y^2)-bs_x+cs_y+d\\ =-a(s_x^2+s_y^2+\frac{b}{a}s_x-\frac{c}{a}s_y+\frac{d}{a})\\ =-a((s_x+\frac{b}{2a})^2+(s_y-\frac{c}{2a})^2-\frac{b^2+a^2}{4a^2}+\frac{d}{a} )\\ =-a((s_x+\frac{b}{2a})^2+(s_y-\frac{c}{2a})^2-r^2 )\\ $$

where $a=\det \begin{bmatrix} p_x & p_y & 1 \\ q_x & q_y & 1 \\ r_x & r_y & 1 \\ \end{bmatrix}$ and $r=\frac{\sqrt{b^2+c^2-4ad}}{2a}$. Since $p,q,r$ are in clockwise order, $a>0$ and $d<0$. Therefore, the determinant in question is positiver if and only if $s$ lies inside the circumcircle.

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