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  1. #1
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    Exclamation Complex No.

    mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
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  2. #2
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    Hello, Navesh!

    It can be solved by "eyeballing" the problem if you know a particular theorem.

    $\displaystyle \left|\frac{z - z_1}{z-z_2}\right|\: = \:k$, where $\displaystyle k \neq 1$ represents a circle. .How?
    We have: $\displaystyle |z - z_1| \:=\:k|z - z_2|$

    The distance $\displaystyle \overline{zz_1}$ is proportional to the distance $\displaystyle \overline{zz_2}$.

    By definition, the locus of $\displaystyle z$ is a Circle of Apollonius.
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  3. #3
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    Quote Originally Posted by Navesh
    mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
    I assume $\displaystyle Z_{1} \mbox{ and } Z_{2}\mbox{ are given}.$
    Let $\displaystyle Z_{1}= a +ib \mbox{ and } Z_{2} = c +id $
    Let$\displaystyle Z = x +iy $
    Now, $\displaystyle |Z - Z_{1}| = k|Z - Z_{2}|$
    This is equivalent to $\displaystyle (x-a)^2 + (y-b)^2 = k^2[(x-c)^2 + (y-d)^2]$
    It represents a circle if k is not equal to one. Do you know why?
    Last edited by malaygoel; Jun 18th 2006 at 06:16 PM.
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  4. #4
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    malaygoel you used LaTex code I am proud of you

    Use,
    Code:
    \mbox{ "type you sentences here" }
    To type things like,
    $\displaystyle Z\mbox{ and }Z'\mbox{ are given}$
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  5. #5
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    circle of appolonius

    Quote Originally Posted by Navesh
    mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
    what do u mean by circle of appolonius?
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  6. #6
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    Quote Originally Posted by Navesh
    mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
    $\displaystyle
    \mbox{thanks for helping}
    $
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  7. #7
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    Quote Originally Posted by malaygoel
    I assume $\displaystyle Z_{1} \mbox{ and } Z_{2}\mbox{ are given}.$
    Let $\displaystyle Z_{1}= a +ib \mbox{ and } Z_{2} = c +id $
    Let$\displaystyle Z = x +iy $
    Now, $\displaystyle |Z - Z_{1}| = k|Z - Z_{2}|$
    This is equivalent to $\displaystyle (x-a)^2 + (y-b)^2 = k^2[(x-c)^2 + (y-d)^2]$
    It represents a circle if k is not equal to one. Do you know why?
    The square terms will vanish if K=1

    also simplifying the eqn. we get
    $\displaystyle
    $x^2(1-k^2)+y^2(1-k^2)+2x(k^2c-a)+$$$\displaystyle $2y(k^2d-b)-k^2(c^2+d^2)=0$$
    which makes the constt. term +ve, hence makes the circle real. Am I OK?
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  8. #8
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    Hello, Navesh!

    also simplifying the eqn. we get
    $\displaystyle x^2(1-k^2)+y^2(1-k^2)+2x(k^2c-a)+$ $\displaystyle 2y(k^2d-b)-k^2(c^2+d^2) \:= \:0$ . . . not quite
    which makes the constt. term +ve, hence makes the circle real. Am I OK?

    We have: .$\displaystyle (x - a)^2+(y-b)^2\:=\:k^2[(x-c)^2+(y-d)^2]$

    Expand: .$\displaystyle x^2 - 2ax + a^2 + y^2 - 2by + b^2\:=\:k^2y^2$ $\displaystyle -\, 2k^2cy + c^2k^2 + k^2y^2 - 2k^2dy + d^2k^2$

    Rearrange: .$\displaystyle x^2 - k^2x^2 + 2ck^2x - 2ax + y^2 - k^2y^2 + 2k^2dy \,-$$\displaystyle 2by \:=\: c^2k^2 + d^2k^2 - a^2 - b^2$

    Factor: .$\displaystyle (1-k^2)x^2 + 2(k^2c - a)x + (1 - k^2)y^2 + 2(k^2d - b)$$\displaystyle y\:= \;k^2(c^2 + d^2) - (a^2 + b^2)$


    Hence: .$\displaystyle x^2 + \frac{2(k^2c - a)}{1 - k^2}x + y^2 + \frac{2(k^2d - b)}{1 - k^2}y \;=\;$ $\displaystyle \frac{k^2(c^2+d^2) - (a^2+b^2)}{1-k^2} $

    . . .Or: .$\displaystyle x^2 + \frac{2(a - k^2c)}{k^2 - 1}x + y^2 + \frac{2(b - k^2d)}{k^2 - 1}y \:=\:$ $\displaystyle \frac{(a^2 + b^2) - k^2(c^2+d^2)}{k^2 - 1} $


    And we see the restrictions on the constants.

    Obviously: $\displaystyle k \neq \pm 1$

    But equally important, the right side must be positive.

    Hence: .$\displaystyle [1]\;a^2 + b^2 > k^2(c^2 + d^2)$ and $\displaystyle |k| > 1$

    . . . or: .$\displaystyle [2]\;a^2 + b^2 < k^2(c^2 + d^2)$ and $\displaystyle |k| < 1$

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  9. #9
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    Quote Originally Posted by Soroban
    Hello, Navesh!


    We have: .$\displaystyle (x - a)^2+(y-b)^2\:=\:k^2[(x-c)^2+(y-d)^2]$

    Expand: .$\displaystyle x^2 - 2ax + a^2 + y^2 - 2by + b^2\:=\:k^2y^2$ $\displaystyle -\, 2k^2cy + c^2k^2 + k^2y^2 - 2k^2dy + d^2k^2$

    Rearrange: .$\displaystyle x^2 - k^2x^2 + 2ck^2x - 2ax + y^2 - k^2y^2 + 2k^2dy \,-$$\displaystyle 2by \:=\: c^2k^2 + d^2k^2 - a^2 - b^2$

    Factor: .$\displaystyle (1-k^2)x^2 + 2(k^2c - a)x + (1 - k^2)y^2 + 2(k^2d - b)$$\displaystyle y\:= \;k^2(c^2 + d^2) - (a^2 + b^2)$


    Hence: .$\displaystyle x^2 + \frac{2(k^2c - a)}{1 - k^2}x + y^2 + \frac{2(k^2d - b)}{1 - k^2}y \;=\;$ $\displaystyle \frac{k^2(c^2+d^2) - (a^2+b^2)}{1-k^2} $

    . . .Or: .$\displaystyle x^2 + \frac{2(a - k^2c)}{k^2 - 1}x + y^2 + \frac{2(b - k^2d)}{k^2 - 1}y \:=\:$ $\displaystyle \frac{(a^2 + b^2) - k^2(c^2+d^2)}{k^2 - 1} $


    And we see the restrictions on the constants.

    Obviously: $\displaystyle k \neq \pm 1$

    But equally important, the right side must be positive.

    Hence: .$\displaystyle [1]\;a^2 + b^2 > k^2(c^2 + d^2)$ and $\displaystyle |k| > 1$

    . . . or: .$\displaystyle [2]\;a^2 + b^2 < k^2(c^2 + d^2)$ and $\displaystyle |k| < 1$

    sorry, it was a silly mistake
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