Locus of of the third point on the circle with two fixed beacons in triangulation

Question

I was studying triangulation technique for robot localization and came across this equation (x-x12)^2 + (y-y12)^2=R12^2. Points B1 and B2 are fixed beacons on the circle and point R is the location of the robot. The co-ordinates of points R, B1, B2 are (x+iy),(x1+iy1),(x2+iy2) respectively in the complex plane. Can someone help me with the derivation of the above equation showing the locus of circle passing through R, B1 and B2, Thanks in advance

Answer 1

Since you specify all your points and equations using separate real and imaginary components, I think that thinking about this problem in terms of complex numbers is likely of little use to you.

A circle with radius $r$ and center $(x_C,y_C)$ can be described by an equation

$$r^2=(x-x_C)^2+(y-y_C)^2$$

which translates into

$$(r^2-x_C^2-y_C^2)+2x_C\,x+2y_C\,y=x^2+y^2$$

Defining $a:=r^2-x_C^2-y_C^2, b:=2x_C, c:=2y_C$ this becomes

$$a+bx+cy=x^2+y^2$$

so if you have three non-collinear points and want the circle through these, you can simply solve the linear system of equations

$$\begin{pmatrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{pmatrix} \cdot\begin{pmatrix}a\\b\\c\end{pmatrix} =\begin{pmatrix}x_1^2+y_1^2\\x_2^2+y_2^2\\x_3^2+y_3^2\end{pmatrix}$$

Then you can read the center and radius from this:

$$y_C=\frac c2\qquad x_C=\frac b2\qquad r=\sqrt{a+c_X^2+y_C^2}$$

Applied to your problem, you have three points, so you can read center (called $(x_{12},y_{12})$ in your case) and radius (called $R_{12}$) from that.