# General Formula for Equidistant Locus of Three Points

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.

• For points not lying on the same line, think about circle and its center. – Anurag A Jan 27 '16 at 4:03
• May you elaborate, this just seems like the definition? – user307727 Jan 27 '16 at 4:05
• Am I correct in assuming that your three points $A,B$ and $C$ are fixed? If yes, then the point that can be equidistant from three non collinear points can just be the center of a circle passing through those three points. – Anurag A Jan 27 '16 at 4:08
• I understand this, but I need to know how to calculate the center of this circle. – user307727 Jan 27 '16 at 4:11

In terms of complex variables, $a=A_{x}+iA_{y}$, etc. $$p= \frac{\left| \begin{array}{ccc} a & a\bar{a} & 1 \\ b & b\bar{b} & 1 \\ c & c\bar{c} & 1 \end{array} \right|} {\left| \begin{array}{ccc} a & \bar{a} & 1 \\ b & \bar{b} & 1 \\ c & \bar{c} & 1 \end{array} \right|}$$
As said, you need to find the coordinates $(X,Y)$ of the center and the radius $R$ of the circle. So, let us write the equations $$(X-x_A)^2+(Y-y_A)^2=R^2 \tag 1$$ $$(X-x_B)^2+(Y-y_B)^2=R^2 \tag 2$$ $$(X-x_C)^2+(Y-y_C)^2=R^2 \tag 3$$ Subtract $(1)$ from $(2)$ and $(1)$ from $(3)$. So, you get $$2(x_A-x_B)X+2(y_A-y_B)Y=(x_A^2+y_A^2)-(x_B^2+y_B^2) \tag 4$$ $$2(x_A-x_C)X+2(y_A-y_C)Y=(x_A^2+y_A^2)-(x_C^2+y_C^2) \tag 5$$ which are simple to solve for $X,Y$. When this is done, use $(1)$ to get $R^2$.