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What is the equation of this curve?

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Let C be the circle with center (a/2, 0) and radius a/2. Let L be the line with equation x=a, find the polar equation of the curve produced in the following manner: FOr every angle [tex]\theta[/tex], -[tex]\pi/2[/tex] <[tex]\theta[/tex] < [tex]\pi/2[/tex], consider the ray from the origin that makes the angle [tex]\theta[/tex] with the positive x axis. This ray will intersect C at a point A and L at a point B. Let O be the origin. Then the point P on the ray is on the curve if line segment OP = line segment AB.

    http://img341.imageshack.us/img341/8910/cireq.jpg [Broken]

    If I trace it out, which I did in my example, it begins to look like a centroid. However I don't know how to verify this. I'm not sure exactly how to systematically approach this problem.
    My idea is to break the equation of the resulting curve into two parts: first the length of the segment from the origin to when it hits the boundary of the circle, then the distance from the circle to the vertical line; the resulting curve being the difference between the former and the latter. I believe the distance of the line segment from O to A is simply cos[tex]\theta[/tex]. I'm not sure how I'd express the rest of it though.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 14, 2010 #2

    LCKurtz

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    Hints:

    What is the equation of the line in polar coordinates: r1 = ?? in terms of θ?

    What is the equation of the circle in polar coordinates: r2 = ?? in terms of θ?

    Then your curve, in polar equations is r = r1 - r2.

    The curve approaches x = a as a vertical asymptote.
     
  4. Mar 14, 2010 #3
    Darn, it's been too long since I've learned about polar coordinates. I know that the equation of the circle in polar coordinates is r=acos[tex]\theta[/tex]. Some research is needed to find a straight line!

    Thanks very much for your assistance!
     
    Last edited: Mar 14, 2010
  5. Mar 14, 2010 #4
    Okay, I looked through my old calculus books and have relearned that the equation of a vertical line is a=rcos(theta)

    Resulting curve: acos(theta)-asec(theta)
     
  6. Mar 14, 2010 #5

    LCKurtz

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    Yes, but you mean r = a sec(theta) - a cos(theta)
     
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