Circle of Apollonius is Circle/Proof 1

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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

Circle-of-Apollonius-Construction.png

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:

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


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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.

From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, $\angle XPY$ is a right angle.

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

$\blacksquare$


Sources

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