Meaning of this 4x4 determinant 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?
$$\det
  \begin{bmatrix}
    p_x & p_y & p_x^2+p_y^2 & 1 \\
    q_x & q_y & q_x^2+q_y^2 & 1 \\
    r_x & r_y & r_x^2+r_y^2 & 1 \\
    s_x & s_y & s_x^2+s_y^2 & 1 \\
  \end{bmatrix}
$$

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 $d&lt0$. Therefore, the determinant in question is positiver if and only if $s$ lies inside the circumcircle.
