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Finding center and radius

  1. Jul 27, 2014 #1
    Hi,
    how do I find the center and radius from these equations? The 2 equations represent 2 different circles, by the way. I need to draw 2 circles.
     

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  2. jcsd
  3. Jul 27, 2014 #2

    phinds

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    What have you tried so far? Those equations don't mean a thing to me, but you DO have to show some effort on your own before (or in addition to) asking for help.
     
  4. Jul 27, 2014 #3
    The equations are actually equations of power transmission and power reception circle diagram.
    I am sorry for not having an attempt but I am stuck here.
     
  5. Jul 27, 2014 #4

    HallsofIvy

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    I presume that the "[itex]P_S+ jQ_S[/itex]" and "[itex]P_R+ jQ_R[/itex]" on the left of those two equations are the complex variable "z" that is to be graphed.

    An equation of the form "[itex]z= Ae^{j\theta}[/itex]", with A a real number and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], is a circle with center at 0 and radius A. An equation of the form "itex]z= B+ A{j\theta}" is a circle with center at the complex number B and radius A.

    Of course, as [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex], [itex]\theta- \pi/2[/itex] goes from [itex]-\pi/2[/itex] to [itex]3\pi/2[/itex] but the graph still covers the circle, just "starting" at a different point. The first circle has center at the point [itex]j(0.81)E_R^2/X[/itex] in the complex plane, which is [itex](0, 0.81E_R^2/X)[/itex], and radius [itex]0.9E_R^2/X[/itex]. The second has center at [itex](0, -0.81E_R^2/X)[/itex] and the same radius.
     
  6. Jul 27, 2014 #5
    How do you get [itex](0, -0.81E_R^2/X)[/itex] for the second circle? Shouldn't it be ##(0,-E_R^2/X)##? I forgot to mention that ##P_S##+##jQ_S## are indeed a complex number in the form Z= X+iY because P is the real power while Q is the reactive power.
     
    Last edited: Jul 27, 2014
  7. Jul 28, 2014 #6

    berkeman

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    You know that you *must* show your work in your posts of schoolwork questions here. Check your PMs.
     
  8. Jul 28, 2014 #7

    berkeman

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    Thread is closed. MissP.25_5 is on a temporary vacation from the PF.
     
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