Circle of Apollonius is Circle

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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 $A = \tuple {x_a, y_a}, B = \tuple {x_b, y_b}$.

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


We have:

\(\ds XA\) \(=\) \(\ds \lambda \paren {XB}\)
\(\ds \leadsto \ \ \) \(\ds \sqrt {\paren {x - x_a}^2 + \paren {y - y_a}^2}\) \(=\) \(\ds \lambda \sqrt {\paren {x - x_b}^2 + \paren {y - y_b}^2}\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds \paren {x - x_a}^2 + \paren {y - y_a}^2\) \(=\) \(\ds \lambda^2 \paren {\paren {x - x_b}^2 + \paren {y - y_b}^2}\)
\(\ds \leadsto \ \ \) \(\ds x^2 - 2 x x_a + {x_a}^2 + y^2 - 2 y y_a + {y_a}^2\) \(=\) \(\ds \lambda^2 \paren {x^2 - 2 x x_b + {x_b}^2 + y^2 - 2 y y_b + {y_b}^2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {\lambda^2 - 1} \paren {x^2 + y^2} + 2 x \paren {\lambda^2 x_b + x_a} + 2 y \paren {\lambda^2 y_b + y_a} + {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2}\) \(=\) \(\ds 0\)

From Equation of Circle in Cartesian Plane, this is an equation in the form:

$A \paren {x^2 + y^2} + B x + C y + D = 0$

with radius $R$ and center $\tuple {a, b}$, where:

$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
$\tuple {a, b} = \tuple {\dfrac B {2 A}, \dfrac C {2 A} }$


Here, we have:

\(\ds A\) \(=\) \(\ds \lambda^2 - 1\)
\(\ds B\) \(=\) \(\ds 2 \paren {\lambda^2 x_b + x_a}\)
\(\ds C\) \(=\) \(\ds 2 \paren {\lambda^2 y_b + y_a}\)
\(\ds D\) \(=\) \(\ds {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2}\)


Hence the center $\tuple {a, b}$ can be evaluated:

\(\ds a\) \(=\) \(\ds \dfrac B {2 A}\)
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {\lambda^2 x_b + x_a} } {2 \paren {\lambda^2 - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {\lambda^2 x_b + x_a} {\lambda^2 - 1}\)


\(\ds b\) \(=\) \(\ds \dfrac C {2 A}\)
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {\lambda^2 y_b + y_a} } {2 \paren {\lambda^2 - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {\lambda^2 y_b + y_a} {\lambda^2 - 1}\)


Then the radius can be evaluated:

\(\ds R\) \(=\) \(\ds \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \paren {\lambda^2 - 1} } \sqrt {4 \paren {\lambda^2 x_b + x_a}^2 + 4 \paren {\lambda^2 y_b + y_a}^2 - 4 \paren {\lambda^2 - 1} \paren { {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2} } }\)