Circle of Apollonius is Circle

Theorem
Let $A, B$ be distinct points in the plane.

Let $\lambda \in \R_{>0}$ be a strictly positive real number.

Let $X$ be the locus of points in the plane such that:
 * $XA = \lambda \paren {XB}$

Then $X$ is in the form of a circle, known as a circle of Apollonius.


 * Circle-of-Apollonius.png

If $\lambda < 1$, then $A$ is inside the circle, and $B$ is outside.

If $\lambda > 1$, then $B$ is inside the circle, and $A$ is outside.

Proof
Let $A = \paren {x_a, y_a}, B = \paren {x_b, y_b}$.

Let $X = \paren {x, y}$.

We have:

From Equation of Circle (Cartesian): Corollary 1, this is an equation in the form:


 * $A \left({x^2 + y^2}\right) + B x + C y + D = 0$

with radius $R$ and center $\left({a, b}\right)$, where:
 * $R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
 * $\left({a, b}\right) = \left({\dfrac B {2 A}, \dfrac C {2 A} }\right)$

Here, we have:

Hence the center $\left({a, b}\right)$ can be evaluated:

Then the radius can be evaluated: