Geometry tricks for smallest of 2 circles circumscribing points

Question

I’m trying to find which of two circles A,B,C and A,B,D has a smaller radius, where A,B,C,D are points in 2 dimensions (Cartesian), and each circle circumscribes those point triples. I’m ignoring edge cases of collinearity for now.

I can find the smaller radii easily enough by finding the circle centre for each triple (by using the perpendicular bisector of chords AB and BC, for example), and then finding each radii.

However, I’m wondering, is there’s any geometry or vector tricks to simplify the math given two of the points are shared, and I only care about finding which is smaller, not the centre or actual radius. Eventually I want to be able to find the result computationally, so the number of steps is important.

If possible it’d be really great to express answers with vector or geometric reasoning, as I find that much easier to understand than matrices and some other forms.

Thank you very much for taking a look!

Answer 1

$$ |AE| \leq |AF| \iff \angle ACB \geq \angle ADB $$

Informal or incomplete "proof":

If $C$ and $D$ are on the same side of line $AB$, then $\angle ACB = \angle ADB$ if and only if the quadrilateral is cyclic. If $D$ is outside the circumcircle of $\triangle ABC$, then $\angle ADB < \angle ACB$ and $|AF| > |AE|$. If $D$ is inside the circumcircle of $\triangle ABC$, then $\angle ADB > \angle ACB$ and $|AF| < |AE|$.

If $C$ and $D$ are opposite sides of line $AB$, then let $D'$ be the reflection of $D$ across line $AB$. $\angle AD'B = \angle ADB$, and the conclusions above apply.

So it suffices to determine the two (unsigned) angles and compare them. Or if you have the points in vector coordinates, the most efficient computer calculations are probably

$$ \begin{align*} \angle ACB &\geq \angle ADB \\ {} \iff \cos \angle ACB &\leq \cos \angle ADB \\ {} \iff \frac{\vec{CA} \cdot \vec{CB}}{|CA|\, |CB|} & \leq \frac{\vec{DA} \cdot \vec{DB}}{|DA|\, |DB|} \end{align*} $$

(Note the cosines and dot products can be negative. It might be convenient in some cases to square both sides of that last inequality to avoid the magnitude square roots, but that only helps if both have the same sign. If both are negative, squaring reverses the sense of the comparisons.)

Answer 2

Any chord is equal to the diameter of the circle, times the sine of an inscribed angle insisting on that chord: $$ AC=2r\cdot\sin\angle ABC=2r{|(A-B)\times(C-B)|\over AB\cdot BC} $$ that is: $$ 2r={AB\cdot BC\cdot CA\over|A\times B+B\times C+C\times A|}, $$ where $A\times B=x_Ay_B-y_Ax_B$ and so on.