# Math Help - Locus

1. ## Locus

Hi

Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point, is a circle.

How exactly do I do this?

2. Hi

First it is essential to draw the situation in order to get an idea of the way to proceed.

Let A and B be the two fixed point.
Let M be a point such as AM = 3 BM.
The aim of the exercise is to find the locus of M.

$AM = 3 BM$ is equivalent to $AM^2 = 9 BM^2$ or $\overrightarrow{AM}^2 = 9 \overrightarrow{BM}^2$

Let G be the point defined by $\overrightarrow{AG} = \frac{9}{8}\:\overrightarrow{AB}$ then $\overrightarrow{AG} - 9 \overrightarrow{BG} = \overrightarrow{0}$

$\left(\overrightarrow{AG}+\overrightarrow{GM}\righ t)^2 = 9 \left(\overrightarrow{BG}+\overrightarrow{GM}\righ t)^2$

Develop and simplify to get the answer ...

3. Originally Posted by running-gag

Let G be the point defined by $\overrightarrow{AG} = \frac{9}{8}\:\overrightarrow{AB}$ then $\overrightarrow{AG} - 9 \overrightarrow{BG} = \overrightarrow{0}$

$\left(\overrightarrow{AG}+\overrightarrow{GM}\righ t)^2 = 9 \left(\overrightarrow{BG}+\overrightarrow{GM}\righ t)^2$
Sorry I dont seem to get this bit : what do the arrows mean and where does G come in?

4. Originally Posted by xwrathbringerx
Hi

Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point, is a circle.

How exactly do I do this?
What you have here is the locus definition of a circle known as a Circle of Apollonius (also known as an Apollonian Circle). Here are some hyperlinks

Circles of Apollonius

Apollonius' Lonely Circle

Locus of Points in a Given Ratio to Two Points from Interactive Mathematics Miscellany and Puzzles

You will find a proof in the Appendix of this paper: Circle of Apallonius

5. Hello, xwrathbringerx!

Looks like you need a walk-through . . . hints don't seem to work.

Show that the locus of a point, which moves so as always to be
three times further from one fixed point than from another fixed point, is a circle.

Since this claim is made for any two points,
. . place the points on the x-axis, symmetric to the origin.
Code:
                  |     P
|     o (x,y)
| *    *
* |       *
*     |        *
- - o - - - - + - - - - o - -
(-a,0)       |       (a,0)
A         |         B
|
Let $P(x,y)$ be any point such that: . $PA \:=\:3\!\cdot\!PB$

Using the Distance Formula: . $\begin{array}{ccc}PA &=& \sqrt{(x+a)^2 + y^2} \\ \\[-3mm] PB &=& \sqrt{(x-a)^2 + y^2} \end{array}$

Then: . $PA \:=\:3\!\cdot\!PB \quad\Rightarrow\quad \sqrt{(x+a)^2 + y^2} \:=\:3\sqrt{(x-a)^2 + y^2}$

Square both sides: . $(x+a)^2 + y^2 \:=\:9\left[(x-a)^2 + y^2\right]$

Expand: . $x^2 + 2ax + a^2 + y^2 \:=\:9x^2 - 18ax + 9a^2 + 9y^2$

Simplify: . $8x^2 - 20ax + 8y^2 \:=\:-8a^2$

Divide by 8: . $x^2 - \tfrac{5}{2}ax + h^2 \:=\:-a^2$

Complete the square: . $x^2 - \tfrac{5}{2}ax + {\color{blue}\tfrac{25}{16}a^2} \:=\:-a^2 + {\color{blue}\tfrac{25}{16}a^2}$

. . And we have: . $\left(x-\tfrac{5}{4}a\right)^2 + y^2 \:=\:\tfrac{9}{16}a^2$

This is a circle with center $C\left(\tfrac{5}{4}a,\:0\right)$ and radius $r\,=\,\tfrac{3}{4}a$