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Indeterminate problem in geometry

  1. Oct 7, 2006 #1
    Hello,

    I came across this question and am wondering if anyone could help me on this...I don't even know where to start... (see attached image)


    Any help would be greatly appreciated! :biggrin:
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2006 #2

    mathman

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    It is a somewhat tedious analytic geometry problem.

    First locate Q by solving for the intersection of C1 equation and C2 equation - the latter is x^2+y^2=r^2.

    By subtracting one equation from the other you will have a linear equation in x, which is readily solvable. This x can be used to obtain y (use plus value). Notice the coordinates of P and Q both depend on r.

    Get the equation of the straight line through P and Q and find the x coordinate of the y=0 point on the line. This will also depend on r.

    Now let r get arbitrarily small.
     
    Last edited: Oct 8, 2006
  4. Oct 9, 2006 #3
    Thank you, Mathman, for answering my question. I have done what you suggested by combining the equations of the two circles to find the intersection coordinates at Q.

    C1: (x-2)^2 + y^2 = 4
    C2: x^2 + y^2 = r^2

    It comes out to
    x = (r^2)/4

    With this I can isolate r or x...either way, I now know the relation between the two variables.

    But what I seem to have trouble with is the subtracting of the two equations. Do you mean subtracting C1 from C2's equation?
    I find this odd, because it would then look like this:
    x^2 + y^2 - 4x - (x^2 - 4x + 4 + y^2 -4) = 0
    Then after everything, it just cancels out....so I know I must be plugging the x and r relation into the same equation... could you give me a little more guidance ... ,in detail, about this question?

    You say there would then be a linear equation... but ... from where?


    Please overlook my ignorance.
    Thanks!
     
  5. Oct 9, 2006 #4

    mathman

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    No!

    C1-C2 looks like: -4x+4=4-r^2. This gives the equation you used.
     
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