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Thread: Locus

  1. #1
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    Exclamation 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?
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  2. #2
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    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.

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

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

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

    Develop and simplify to get the answer ...
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  3. #3
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    Quote Originally Posted by running-gag View Post

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

    $\displaystyle \left(\overrightarrow{AG}+\overrightarrow{GM}\righ t)^2 = 9 \left(\overrightarrow{BG}+\overrightarrow{GM}\righ t)^2$
    Sorry I dont seem to get this bit : what do the arrows mean and where does G come in?
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  4. #4
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    Quote Originally Posted by xwrathbringerx View Post
    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
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  5. #5
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    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 $\displaystyle P(x,y)$ be any point such that: .$\displaystyle PA \:=\:3\!\cdot\!PB$


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

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

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

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

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

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

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

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


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

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