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Math Help - Complex Numbers - Circles

  1. #1
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    Complex Numbers - Circles

    Determine the set of solutions for M =  z  \epsilon  C  |     |z-3| = 2|z+3|

    Hint: Try to find an equation in form of a general Circle.

    Seriously i have no idea.. im quite good in complex numbers, but this is somehow hard for me
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  2. #2
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    Hello, coobe!

    Determine the set of solutions for: . M \;=\;\bigg\{z \in C\;\bigg|\;|z-3| \,=\, 2|z+3|\,\bigg\}

    Hint: Try to find an equation in form of a general circle.
    Let P(x,y) be a solution to the equation.


    |z-3| is the distance of point P from point A(3,0).

    |z+3| is the distance of point P from point B(-3,0).


    We are told that: . d(PA) \;=\;2\cdot d(PB)

    . . Hence: . \sqrt{(x-3)^2 + y^2} \;=\;2\sqrt{(x+3)^2 + y^2}


    Square and expand: . x^2-6x+9 + y^2 \;=\;4(x^2+6x+9 + y^2)

    Simplify: . 3x^2 + 30x + 3y^2 -27 \:=\:0 \quad\Rightarrow\quad x^2 + 10x + y^2 \:=\:-9

    Complete the square: . x^2 + 10x \:{\color{red}+\: 25} + y^2 \;=\;-9 \:{\color{red}+\: 25 }


    Therefore: . (x+5)^2 + y^2 \:=\:16

    . . The solution is a circle with center (-5,0) and radius 4.


    Note: It is known as the Circle of Apollonius.
    .
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