Circle of Apollonius is Circle

From ProofWiki
Jump to navigation Jump to search

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 1

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:

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


This article, or a section of it, needs explaining.
In particular: explain why the above
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.


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.

$\blacksquare$


Proof 2


Although this article appears correct, it"s inelegant. There has to be a better way of doing it.
In particular: Simplify the workings by setting one of the points at the origin and the other on the x-axis.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Improve}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


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:

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


This needs considerable tedious hard slog to complete it.
In particular: Good luck with that.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Retrieved from "https://proofwiki.org/w/index.php?title=Circle_of_Apollonius_is_Circle&oldid=601825"