2
$\begingroup$

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?

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

$\endgroup$
  • $\begingroup$ 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 $f(s_x,s_y)>0$? $\endgroup$ – J. M. is a poor mathematician Apr 14 '12 at 14:06
  • $\begingroup$ I can see it now. But it is puzzling how people arrive at this determinant form of the equation circumcircle at first? $\endgroup$ – FiniteA Apr 14 '12 at 16:13
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.