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Points on a circle.

  1. Sep 15, 2010 #1
    I have a circle and I need to find a particular point along it.

    The circle has a known radius (r), center (a, b), and I know a coordinate point upon it (x1, y1), as well as an arc-length (d). From that point x1, y1, how do I find out the point x2,y2 which is arc-length d away from the point x1, y1 in either direction?
     
  2. jcsd
  3. Sep 16, 2010 #2
    Let me give you hints
    The well known relationship is
    d=r[tex]\theta[/tex]

    You know d and r, so you can find [tex]\theta[/tex]

    Now let us assume the circle is centered at origin and given point is on +ve x axis (r,0).

    So x1=r y1=0;
    Now x2= r cos[tex]\theta[/tex]
    and y2=r sin[tex]\theta[/tex]
    Also in this case x3=x2 and y3=-y2

    From above you can find relationship for general (x1,y1) and arbitrary centre of circle
     
  4. Sep 16, 2010 #3
    Well then... er... it should be simple enough. By that, the y coordinate doesn't matter at all. All that matters is the x. And, actually, basically you gave the answer right there, no hinting. When starting from any point x1,y1 (assuming center is at the origin for simplicity) then any point other than that is:

    x2 = x1 cos[tex]\theta[/tex]
    y2 = +/- (x1 sin[tex]\theta[/tex])

    Even with a center of a, b it should be simple:

    x2 = (x1 - a) cos[tex]\theta[/tex] + a
    y2 = +/- ((x1 - a) cos[tex]\theta[/tex]) + b

    My problem, I think, was that I was working it all out in degrees since my first questions, elsewhere, got answered that way.
     
    Last edited: Sep 16, 2010
  5. Sep 16, 2010 #4
    Tried this... and I think I'm doing something wrong. -_-

    Circle of origin 150,150, radius also 150. My starting point is at 100, 290 (this is inaccurate, I know, but it's MS Paint drawing a circle, so that's about as close as it'll get). Arc length is 103, so:

    [tex]\theta[/tex] = 0.68666
    cos[tex]\theta[/tex] = 0.77336
    sin[tex]\theta[/tex] = 0.63396

    so it should be:

    x2 = (100-150) x 0.77336 + 150 = 111.33
    y2 = (100-150) x 0.63396 + 150 = 118.30

    which ends up placing this point somewhere inside the circle. No idea where I'm messing up, unless the Y value matters somehow and I'm not seeing it. Geometry has never been a good subject for me. What am I doing wrong?
     
  6. Sep 16, 2010 #5

    DaveC426913

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    Gold Member

    Wouldn't be easier if the origin were 0,0?
     
  7. Sep 16, 2010 #6
    Sure. And if you can figure out how to get MS Paint to draw a complete circle around the point 0,0 which is in the upper left hand corner of the screen, I will be very impressed. I'll also use it! ^_^
     
    Last edited: Sep 16, 2010
  8. Sep 17, 2010 #7
    First of all, your equations are not correct.

    Secondly, point (100,290) does not lie on your circle since all points on the circle must be a distance of 150 units from (150,150).

    Find the distance c from (100,290) to (150,150) by:

    [tex]c = \sqrt{(100 - 150)^2 + (290 - 150)^2} \approx 148.66[/tex]


    Here is what you need to do:

    Start with a point that is on the circle, say [itex](100, 150+100\sqrt{2}) \approx(100, 291.421)[/tex]

    Find the horizontal distance between the center of the circle (a,b) and the first point on the circle (x1,y1). This is x1-a.

    Do the same for the vertical distance: y1-b

    Call the point at (a,b) point O,
    Call the point at (x1,y1), point F
    Call the point at (x2,y2), point G

    Call the point on the circle at (a+r,b) point H

    To find the angle [itex]\angle FOH[/tex] (call this [itex]\alpha[/tex]), use

    [tex]\alpha = \tan ^ {-1} \left( \frac {y1 - b}{x1 - a} \right)[/tex]

    Now, you need to find a point [itex]\theta[/tex] degrees away from point F

    [tex]\theta = \frac{d}{r}[/tex]


    There are 2 such points J & K.
    J lies on the circle an arc length of d units away from point F in the counter-clockwise direction
    K lies in on the circle an arc length of d units away in the clockwise direction.

    Using [itex]\angle JOH[/tex] (call this angle [itex]\beta[/tex]) which is equal to [itex]\alpha + \theta[/tex], find the horizontal and vertical distances from point O to point J.
    Do this by using [itex] r \cdot \cos \beta[/tex] and [itex]r \cdot \sin \beta[/tex], respectively.

    Offset these distances by the coordinates of O to get the coordinates of G; (x2, y2)
    [itex]x2 = a + r \cdot \cos \beta[/tex]
    [itex]y2 = b + r \cdot \sin \beta[/tex]

    To verify that this point lies on the circle, it must be r units away from point O, so find:

    [tex]r = \sqrt {(x2 - a)^2 + (y2 - b)^2}[/tex]


    To find point K, use [itex]\beta = \alpha - \theta[/tex] and follow the same process.

    (Note: keep all of your angles in radians)




    Therefore with circle O having center (150,150) and radius 150,
    Given point [itex](100, 150+100\sqrt{2})[/tex] and arc length d = 103

    You should come up with points J & K having the following approximate coordinates:
    Point J: (125.648, 298.010)
    Point K: (290.357, 202.916)

    Both of which lie on the circle.
     
    Last edited: Sep 17, 2010
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