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Find the locus of center of circle

  1. Jan 5, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A variable circle cuts x and y axes so that intercepts are of given length k1 and k2. Find the locus of center of circle

    2. Relevant equations

    3. The attempt at a solution
    There must be four intercepts but only two are given.
     
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  3. Jan 5, 2013 #2

    tiny-tim

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    hi utkarshakash! :smile:
    yes, but you only need two :wink:

    get on with it! :smile:

    hint: K1K2 is a chord of the circle
     
  4. Jan 5, 2013 #3

    phinds

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    There is a way of interpreting this problem so that it makes sense and you get a single circle as the answer, but I'm dubious as to whether or not that is the intended interpretation of the problem.

    Other than that interpreation (and, can you see the condition that allows/forces it?) then I believe you are right that to get a single circle as the answer, you need all four intercepts (actually, three will do, and two will always give you two circles as the answer if you assume that the given intercepts are just 2 of the three or 4 that exist)
     
  5. Jan 6, 2013 #4

    utkarshakash

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    I already know that. I assume the general equation of the circle and then proceed ahead. I am thinking of drawing perpendiculars from the centre on two chords. But will this help?
     
  6. Jan 6, 2013 #5

    tiny-tim

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    hi utkarshakash! :wink:
    you only have one chord, but yes, do draw the bisector! :smile:

    (i'm assuming it mean that the x intercept is always (k1,0), and the y intercept is always (0,k2))
     
  7. Jan 6, 2013 #6

    utkarshakash

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    I am not taking it that way. I was thinking k1 and k2 to be two separate chords.
     
  8. Jan 6, 2013 #7

    tiny-tim

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    in that case, how could there be four intercepts? :confused:
    ok, then assume the centre is at (x,y) and that the radius is r,

    then find the two lengths of the intercepts with the axes, put one equal to k1, the other equal to k2, and solve for x and y …

    what do you get? :smile:
     
  9. Jan 6, 2013 #8

    haruspex

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    The only way I can make sense of the question is to take k1 and k2 to be the distances between intercepts on the respective axes, not the coordinates of the intercepts. That way you do need all four.
    utkarshakash, is that what you meant?
     
  10. Jan 7, 2013 #9

    utkarshakash

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    Yes, exactly what I meant.:approve: But tiny-tim is taking it the other way. He is assuming k1 and k2 to be the coordinates of the intercepts on x and y axes. Which one is correct?
     
  11. Jan 7, 2013 #10

    tiny-tim

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    no, i've taken it either way!

    for your way …
     
  12. Jan 7, 2013 #11

    utkarshakash

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    I just made a mistake in understanding the question. Actually k1k2 is a chord of the circle as you said and not that they are 2 separate chords.
     
  13. Jan 7, 2013 #12

    tiny-tim

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    ok, in that case the first thing is to draw the bisector of the chord :wink:
     
  14. Jan 8, 2013 #13

    utkarshakash

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    By drawing the bisector and joining the center and (0,k2) I get a right triangle. But my idea is to calculate the distance between centre and the given two points and equate them. But that way I get the wrong answer.
     
  15. Jan 8, 2013 #14

    tiny-tim

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    i'm not following this :redface:

    if it's on the bisector, why do you need to check that? :confused:
     
  16. Jan 9, 2013 #15

    utkarshakash

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    Ok so what do you want me to do after drawing the bisectors?
     
  17. Jan 9, 2013 #16

    tiny-tim

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    (bisectors? i thought there was only one chord now? :confused:)

    well, where will the centre be in relation to the bisector?
     
  18. Jan 9, 2013 #17

    utkarshakash

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    Sorry it was a typo. I can find the equation of line passing through the mid point of chord and perpendicular to it. Next I can satisfy the centre in the equation. Am I correct?
     
  19. Jan 9, 2013 #18

    tiny-tim

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    do you mean that the centre will lie on the line, and so the centre will satisfy the equation?

    yes :smile:
     
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