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In various algorithms such as the construction of the Voronoi diagram, the incircle test allows to check if four points are cocircular. It can be written as

$$\Delta=\begin{vmatrix} x_a&y_a&x_a^2+y_a^2&1\\ x_b&y_b&x_b^2+y_b^2&1\\ x_c&y_c&x_c^2+y_c^2&1\\ x_d&y_d&x_d^2+y_d^2&1\\ \end{vmatrix}=0.$$

It can be interpreted as the coplanarity test of the four points lifted to the paraboloid $z=x^2+y^2.$

I would like to relate the value of the determinant (when not zero) to the distance of one point to the circle formed by the other three, which should be an expression with a $\Delta$ factor. I know how to find the center and radius of the circle, but I am hoping that there is a more direct way.

Update:

I have worked it out analytically, but I am still looking for a more geometric approach.

1 Answers 1

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The hard way:

WLOG, the point $a$ is at the origin, and the equation of the circle $abc$ reduces to

$$2xx_o+2yy_o=x^2+y^2=z.$$

The center $(x_o, y_o)$ is found by solving

$$\begin{cases}2x_bx_o+2y_by_o=z_b,\\ 2x_cx_o+2y_cy_o=z_c.\end{cases}$$

Then the radius of the circle is $r=\sqrt{z_o}$ and the requested distance

$$\delta=\sqrt{(x_d-x_o)^2+(y_d-y_o)^2}-\sqrt{z_o},$$

which we can write as

$$\delta=\frac{z_d-2x_ox_d-2y_oy_d}{\sqrt{(x_d-x_o)^2+(y_d-y_o)^2}+\sqrt{z_o}}.$$

Remains to express the numerator in terms of $\Delta$. Using Cramer,

$$z_d-2x_ox_d-2y_oy_d=\frac{\begin{vmatrix}x_b&y_b\\x_c&y_c\end{vmatrix}z_d-\begin{vmatrix}z_b&y_b\\z_c&y_c\end{vmatrix}x_d-\begin{vmatrix}x_b&z_b\\x_c&z_c\end{vmatrix}y_d}{\begin{vmatrix}x_b&y_b\\x_c&y_c\end{vmatrix}}=\frac{\Delta}{\begin{vmatrix}x_b&y_b\\x_c&y_c\end{vmatrix}}.$$

Finally,

$$\delta=\frac\Delta{\begin{vmatrix}x_b&y_b\\x_c&y_c\end{vmatrix}\left(\sqrt{(x_d-x_o)^2+(y_d-y_o)^2}+\sqrt{z_o}\right)}$$ which can be approximated as

$$\delta\approx\frac\Delta{\begin{vmatrix}x_b&y_b\\x_c&y_c\end{vmatrix}2r}=\frac{\Delta}{\sqrt{\begin{vmatrix}z_b&y_b\\z_c&y_c\end{vmatrix}^2+\begin{vmatrix}x_b&z_b\\x_c&z_c\end{vmatrix}^2}}.$$