geometrylocus
I need a general formula that calculates the equidistant locus of three points $(P_x,P_y)$; in terms of the coordinates of the three points $(A_x, A_y), (B_x,B_y), (C_x,C_y)$.
Setting the distances equal yielded nothing for me. Calculating the intersection of the equidistant locus of two pairs of points depended on the x and y coordinates being different for each point, making it impractical to use.
It is an application of cross product, since
$$|\vec v \times \vec w|=|\vec v||\vec w|\sin \theta$$
and the area of triangle with sides $|\vec v|$ and $|\vec w|$ is given by
$$A=\frac12|\vec v||\vec w|\sin \theta$$
Note that it not necessary to take into account the order if we consider the absolute value.
For the calculation you should consider for example
$\vec v=(A_x-B_x,A_y-B_y,A_z-B_z)$
$\vec w=(A_x-C_x,A_y-C_y,A_z-C_z)$
$$\vec v \times \vec w=\begin{vmatrix} i&j&k\\A_x-B_x&A_y-B_y&A_z-B_z\\A_x-C_x&A_y-C_y&A_z-C_z \end{vmatrix}=...$$
Take also a look here how to calculate area of 3D triangle?
This is not too hard to see using ordinary geometry.
Let $A$, $B$ be fixed points on circle with center $O$, and $C$ any other point on the circle. In triangle $ABC$, bisecting $CB$, $AB$ at $E$, $F$, and joining $AE$, $CF$, then $G$ is the centroid of $\triangle ABC$.
In fixed triangle $AOB$, bisect $AO$ at $D$, and join $BD$, $OF$, giving $K$ the centroid of $\triangle AOB$. Finally, join $GK$.
Assuming as well known, that the centroid divides the median lines of a triangle in a $\frac{2}{1}$ ratio, then$$\frac{CG}{GF}=\frac{OK}{KF}=\frac{2}{1}$$Hence$$\frac{CF}{GF}=\frac{OF}{KF}=\frac{3}{1}$$Therefore$$GK\parallel CO$$whence$$\triangle CFO\sim\triangle GFK$$and$$\frac{CO}{GK}=\frac{3}{1}$$And since $CO$ has a fixed length for all positions of $C$, so does $GK$.
Therefore, since $K$ is fixed in position, $G$ lies always on the circumference of a circle centered on $K$.
Best Answer
In terms of complex variables, $a=A_{x}+iA_{y}$, etc. $$p= \frac{\left| \begin{array}{ccc} a & a\bar{a} & 1 \\ b & b\bar{b} & 1 \\ c & c\bar{c} & 1 \end{array} \right|} {\left| \begin{array}{ccc} a & \bar{a} & 1 \\ b & \bar{b} & 1 \\ c & \bar{c} & 1 \end{array} \right|}$$