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Finding the (x,y) coordinates of arbitrary circles

  1. Feb 3, 2007 #1
    I have just started curiously thinking about this. The trig functions cos and sin give the (x,y) coordinates of the unit circle. How would i go about using the trig functions to finding (x,y) coordinates of an arbitrary circle?

    What im saying is, the cos and sin only work for the circle x^2 + y^2 = 1

    I would like a general method for finding the (x,y) coordinates of the circle
    (x-h)^2 + (y-k)^2 = r^2
    I knwo this would involve simply transforming the original function of of a f(x)=sin(x) or f(x)=cos(x) to some form of f(x)= a * sin(bx+c) + d and f(x)= a * cos(bx+c) + d
    However, i cant reason out what the values of a,b,c,d would be to transform a trig function to give the (x,y) coordinates of any arbitrary circle.
  2. jcsd
  3. Feb 3, 2007 #2
    Radius is a simple fix, consider the pythagorean identity

    sin2t + cos2t = 1
    Now let
    x = cos(t) and
    y = sin(t)

    What do you get? Can you see how to modify this and get a circle of arbitrary radius?

    Actually an arbitrary center is fairly easy as well, start with the equation of the circle and from there you should be able to see a similar substitution.
  4. Feb 3, 2007 #3
    well, actually i had bigger trouble with the circles with an arbitrary center.
    For a radius, just simply modify the amplitude to the value of the radius.

    but i still cant think of how to modify the functions for an arbitrary center
  5. Feb 3, 2007 #4
    In terms of x and y, how does a circle with arbitrary center at say (h,k) differ from a circle with a center at the origin? What do both equations look like? What changes?
  6. Feb 3, 2007 #5


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    firstly, there is something missing in the question.... you said "trig fns cos and sin give the (x,y) coordinates of the unit circle"....?? How do they do that? They only give you the coordinates of a certain point on the unit circle if you have specified the angle! Now, once you have an angle and a radius (which is 1 in this case), you have effectively written everything in terms polar coordinates. Sin and cos are functions that helps you to tranform from one to the other...

    ok, regarding your arbitrary circle business, it appears that you want a rather long-handed way to specify (x,y).... anyway... still doable I guess... assuming that you know the equation of the circle (then again why would u want to go all the way to express things in sin and cos?) in standard form
    (x-h)^2 + (y-k)^2 = r^2, then simply ignore h and k for the moment and express your coordinates in terms of cos and sin of an angle then shift your answer by h and k respectively for the x and y coords.

    Again this is just polar coordinates in disguise, this time you have also translated your origin relative to what you begin with.
  7. Feb 3, 2007 #6


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    If the center of the circle is (a, b), just shift the values. If [itex](x-a)^2+ (y-b)^2= R^2[/itex] is the circle, for angle [itex]\theta[/itex], measured counter clockwise from the point (a+R,b), [itex]x= a+ Rcos\theta [/itex], [itex]y= b+ Rsin(\theta )[/itex].
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