# 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.

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 1

Let $P$ be an arbitrary point such that $\dfrac {AP} {PB} = \lambda$.

Let $\angle APB$ be bisected internally and externally to intersect $AB$ at $X$ and $Y$ respectively.

Then:

$\dfrac {AX} {XB} = \dfrac {AP} {PB} = \lambda$

and:

$\dfrac {AY} {YB} = \dfrac {AP} {PB} = \lambda$

Thus $X$ and $Y$ are the points which divide $AB$ internally and externally in the given ratio.

Therefore, for a given $A$, $B$ and $\lambda$, $X$ and $Y$ are fixed.

Hence by Thales' Theorem, $P$ lies on the circumference of a circle of which $XY$ is a diameter.

$\blacksquare$

## Proof 2

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:

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