Testing if a Point is in the Circumcircle of a Triangle

Question

I am currently reading about methods of Delaunay triangulation. In particular, this paper by Leach: Improving Worst-Case Optimal Delaunay Triangulation Algorithms (1992)

In it, Leach describe a method of testing whether a point $D$ is in the circumcircle of a triangle $\triangle{}ABC$. This is contrasted with the usual determinant method (see the Delunay Triangulation Wikipedia page if you want to read more on that one).

The test is described as follows

We instead use a test based on $y$-coordinates of circumcircle centres. As shown in Figure 4, for two fixed sites $A$ and $B$ the centre of the circumcircle of $A$, $B$ and a third point lies on the bisector of $A$ and $B$.

As the merge step of the algorithm always seeks the best candidate site above the topmost cross edge, a site $D$ is a better site than a site $C$ if its circumcircle centre lies lower on the bisector. The best candidate site is the one whose circumcircle centre is the lowest. Substituting $(x_A, y_A)$, $(x_B, y_B)$ and $(x_C, y_C)$ into $$(x - p)^2 + (y - q)^2 = r^2$$ and solving for $q$ gives $$q = \frac{1}{2} \cdot{} \frac{ (x_A^2 + y_A^2)(x_C - x_B) - (x_B^2 + y_B^2)(x_C - x_A) + (x_C^2 + y_C^2)(x_B - x_A) }{ (x_B - x_A)(y_C - y_A) - (x_C - x_A)(y_B - y_A) }.$$ This calculation requires 23 operations (ignoring the division by 2 and assuming common subexpression elimination). Two such calculations are needed to compare the first two candidate sites; thereafter only one is needed per candidate site, for which the cost of the InCircle test is reduced by approximately half.

I do not follow what's happening here. I understand why the circumcenters fall on the line between $A$ and $B$, but it is not immediately obvious to me:

1. Why a lower $q$ is better for the Delaunay triangulation.
2. What the method should do when the $q$ falls below the center of $AB$. Is lower still better, or is it distance from the center we really care about?

If someone could explain (1) and answer (2), this would be greatly helpful to me, thanks.

Answer 1

After some additional thought, I have worked it out.

Recalling the inscribed angle theorem, we have that

and from this, we can see that the angle of the triangle at $D$ can be ascertained from the rise above the midpoint of $AB$, $q$, and half the distance of $AB$, i.e. $$\theta = \mathrm{atan}\left(\frac{|AB|}{2q}\right)$$ and moreover, as $\frac{|AB|}{2}$ is fixed, $\theta$ will decrease with larger $q$.

This also reveals the answer to the second question: the quantity we need to consider for this method is the distance from the midpoint of $AB$.