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Homework Help: Circles on the complex plane

  1. Jul 18, 2003 #1
    In general, |z - zo|=r, where z_o is a fixed point and r is a positive number, represents a circle centered at z_o and with radius r. |z - z1|=k|z - z2|, where z_1 and z_2 are fixed points, also apparently represents a circle, except maybe in the case where k=1. Then we have a line, or a circle of infinite radius. So to find the radius of the circle for |z - z1|=k|z - z2|, I could try to rewrite the equation to fit |z - zo|=r. I did this, and I got a frightening answer. I shall attempt to show it here. The work is much too long and too tedious to write here in full form but I'll explain briefly. x is the x component of z, y is the y component of z, x_1 is the x component of z_1, y_1...y component of z_1, etc.
    Square both sides, distribute the k^2, collect x's and y's on the left, complete the square to get (x-somthing)^2+(y-same thing)^2=some big mess
    simplify the right hand side, rewrite in terms of z1 and z2 as much as possible, take the sqrt of each side,

    Anyone who feels like trying this problem could verify my result/ show me a better way of doing it?
  2. jcsd
  3. Jul 18, 2003 #2
    Hi StephenPrivitera,
    the problem is surely symmetrical wrt. the line Z1 Z2.
    So let's look at this line, and the 2 points A, B where the circle intersects it (points=capital, distances=small):


    Let |Z1 - Z2| = d.
    From the drawing, we see:
    I. d = ka + a
    II. b + d = kb
    III. 2r = a+b.

    Three unknowns: a, b, r.
    Three equations: Perfect!
    Last edited: Jul 18, 2003
  4. Jul 19, 2003 #3
    Really there are four unknowns, a,b,r, and d. If you substitute in for d, then there are two equations and three unknowns. You can solve for r in terms of b or a. I did it for a and got,

    Also, how do you know that the diameter lies on the line z1z2?
  5. Jul 19, 2003 #4
    You know the value of d, because you know Z1 and Z2. I defined d = |Z1 - Z2|.

    Concerning the symmetry: It's clear that the circle can have only 2 points in common with the line Z1 Z2. Let's assume the center C is not on Z1 Z2. Now take the mirror image C' of C wrt. to the line Z1 Z2. The circle centered at C', and going through A, B is obviously another possible solution. Now, since you stated that the circle is defined by the given equation, this is a contradiction. Thus, C is on Z1 Z2.
  6. Jul 20, 2003 #5
    good point, so rather than eliminating d, i should eliminate a and b;

    I'm still upset the other way didn't work... There was a (k2-1)-1 factor there. Maybe the numerator can simplify somehow to dk. I'll try it again. Thanks for the help.
    Last edited: Jul 20, 2003
  7. Jul 20, 2003 #6
    Glad I could help you :wink:.
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