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Chord of minimum length

  1. Nov 27, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    px+qy=40 is a chord of minimum length of the circle [itex](x-10)^2 + (y-20)^2 = 729 [/itex]. If the chord passes through (5,15), then [itex]p^{2013}+q^{2013}[/itex] is equal to

    2. Relevant equations

    3. The attempt at a solution

    Let chord length be L

    [itex]\frac{L}{2} = 729- \dfrac{(10p+20q-40)^2}{p^2+q^2} [/itex]

    Also
    5p+15q-40=0

    Now if I apply Lagrange's Multiplier Method using above two conditions I get some weird value of q which is a huge fraction.
     
  2. jcsd
  3. Nov 27, 2013 #2

    haruspex

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    I don't think you need to be using a Lagrange Multiplier here.
    Although it's not expressed clearly, it must be that the chord has the minimum length of all those passing through the given point. Isn't the implication clear from the geometry? Or was this given as an exercise in using Lagrange Multipliers?
     
  4. Nov 28, 2013 #3

    utkarshakash

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    From geometry I can see that minimum length will occur if the given line is parallel to Y-axis. So q=0 and p=8. The answer should then be 2^6039. But this does not match any of the options given.
     
  5. Nov 28, 2013 #4

    Mentallic

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    No, that's not correct. The minimum length would be a line parallel to the y-axis only if the point that the line passes through has a y value equal to that of the y value for the centre of the circle. That is, since the circle has centre at (10,20) then only lines that must pass through points on y=20 would be parallel to the y axis to give a minimum length chord.

    The minimum length chord passing through a point P is perpendicular to the line OP where O is the centre of the circle.
     
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