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Math Help - Cartesian equation

  1. #1
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    Cartesian equation

    I need help to find the cartesian equation for the region represented by

    | z−2+i | = 1/5 | z−8−3 i |

    The question asks me to do the following:Please put your answer in a "natural" form. If you haven't already done so, complete the square for the terms in x and complete the square for the terms in y, and hence deduce that the locus of points satisfied by the relation is associated with a circle. What is the centre of that circle?
    Any help thanks in advance .
    Last edited by skirk34; October 15th 2008 at 07:12 AM.
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  2. #2
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    Hello, skirk34!

    Find the cartesian equation for: . | z-2+i | \:= \:\tfrac{1}{5}| z-8-3 i |

    We have: . 5|z - (2-i)| \:=\:|z - (8+3i)|

    We want the points P(x,y) so that the distance from P to (8,3)
    . . is five times the distance from P to (2,-1).


    We have: . 5\sqrt{(x-2)^2 + (y+1)^2} \;=\;\sqrt{(x-8)^2 + (y-3)^2}


    Square both sides: . 25\left[(x-2)^2 + (y+1)^2\right] \;=\;(x-8)^2 + (y-3)^2

    . . which simplifies to: . 24x^2 - 84x + 24y^2 + 56y \;=\;-52


    Divide by 24: . x^2 - \frac{7}{2}x + y^2 + \frac{7}{3}y \;=\;-\frac{13}{6}


    Complete the square: . x^2 - \frac{7}{2}x + {\color{blue}\frac{49}{36}} + y^2 + \frac{7}{3}y + {\color{red}\frac{49}{36}} \;=\;-\frac{13}{6} + {\color{blue}\frac{49}{16}} + {\color{red}\frac{49}{36}}

    . . And we have: . \left(x - \frac{7}{4}\right)^2 + \left(y + \frac{7}{6}\right)^2 \;=\;\frac{325}{144}


    We have a circle . . . .center \left(\frac{7}{4},\:-\frac{7}{6}\right) and radius \frac{5\sqrt{13}}{12}



    But check my algebra and arithmetic . . . please!
    .
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  3. #3
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    Your algebra is good, thanks
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