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Math Help - Equation of unit circle, from a non-center point?

  1. #1
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    Equation of unit circle, from a non-center point?

    Hi everyone,

    This is my first post here, so please feel free to ignore/delete/relocate this post if it is in the wrong place or has been covered before.

    I was idly wondering about this the other day, and thought I would seek out the insight of those who might know.

    If we take a Unit circle, with it's center located at the origin and radius r=1, the equation of the circle is written as

    cos^2(theta) + sin^2(theta) = 1
    where theta is defined as angle POR, where P is a point on the circle, O is the origin, and R is, for the sake of simplicity, (x = 1, y = 0)

    Is there a simple way to write the equation of the same circle, with respect to a different angle, phi, defined as PQR, where Q is any random point along y=0, such as (x=-0.5, y=0)?

    I have been trying to figure it out in terms of generating triangle POQ, and invoking the laws of sines and cosines, but I inevitably going in circles (pun not intended) and wind up with something along the lines of x = x.

    Any insights would be greatly appreciated.

    Thanks,
    Matt
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mdunnbass View Post
    Hi everyone,

    This is my first post here, so please feel free to ignore/delete/relocate this post if it is in the wrong place or has been covered before.

    I was idly wondering about this the other day, and thought I would seek out the insight of those who might know.

    If we take a Unit circle, with it's center located at the origin and radius r=1, the equation of the circle is written as

    cos^2(theta) + sin^2(theta) = 1
    where theta is defined as angle POR, where P is a point on the circle, O is the origin, and R is, for the sake of simplicity, (x = 1, y = 0)

    Is there a simple way to write the equation of the same circle, with respect to a different angle, phi, defined as PQR, where Q is any random point along y=0, such as (x=-0.5, y=0)?

    I have been trying to figure it out in terms of generating triangle POQ, and invoking the laws of sines and cosines, but I inevitably going in circles (pun not intended) and wind up with something along the lines of x = x.

    Any insights would be greatly appreciated.

    Thanks,
    Matt
    when a circle is defined parametrically in this way, to change the center, you simply add the values you want to each component.

    that is, a circle with radius r centered at (h,k) can be parametrized by:

    x = r \cos \theta + h and y = r \sin \theta + k .........see if you can figure out why this works

    so the circle you want is given by: x = \cos \theta - 0.5, and y = \sin \theta

    you can check that this yields the circle (x + 0.5)^2 + y^2 = 1
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    Quote Originally Posted by Jhevon View Post
    when a circle is defined parametrically in this way, to change the center, you simply add the values you want to each component.

    that is, a circle with radius r centered at (h,k) can be parametrized by:

    x = r \cos \theta + h and y = r \sin \theta + k .........see if you can figure out why this works

    so the circle you want is given by: x = \cos \theta - 0.5, and y = \sin \theta

    you can check that this yields the circle (x + 0.5)^2 + y^2 = 1
    Wait, I thought that was for a circle where the center of the circle was not at the origin, but rather at x=-0.5
    I'm asking about a circle with its center at the origin, but the angle \phi (not \theta) is NOT residing at the origin. Hence, the radius r in this case is not constant, but rather, when \phi = 0, r=1.5. and when \phi = \pi, r=0.5
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mdunnbass View Post
    Wait, I thought that was for a circle where the center of the circle was not at the origin, but rather at x=-0.5
    indeed, it is. the center here is (-0.5, 0). i thought that is what you were asking for. my apologies for misunderstanding.

    I'm asking about a circle with its center at the origin, but the angle \phi (not \theta) is NOT residing at the origin. Hence, the radius r in this case is not constant, but rather, when \phi = 0, r=1.5. and when \phi = \pi, r=0.5
    ok, so it seems you want to define a circle in terms of an angle based at another point, say from (-0.5,0) that still describes the circle centered at the origin with radius 1? ...why? anyway, i will think about it and get back to you, if no one else has by the time i get back. i have to leave for a while.
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    Quote Originally Posted by Jhevon View Post
    ok, so it seems you want to define a circle in terms of an angle based at another point, say from (-0.5,0) that still describes the circle centered at the origin with radius 1?
    Exactly.


    ...why?
    idle curiosity.
    Last edited by mr fantastic; December 29th 2008 at 12:20 PM. Reason: Deleted potentially offensive language
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    Off-center circle equation

    I was looking for the solution to this problem to help me solve a homework problem, so I came here in hopes of finding it. After a few hours, I finally worked out a solution, but it didn't end up helping me with my homework problem.

    I started by drawing a circle of radius r and defining a new "origin" at a distance m from the center on the x-axis. Then, I selected an arbitrary point P on the circle and drew a radius from the center and a quasi-radius, rho, from the new origin, as well as dropping a perpendicular to the x-axis. The radius made an angle with the x-axis I called alpha, and rho made an angle I called beta.

    Then I found the following relations:

    r*sin(alpha) = rho*sin(beta)

    r*cos(alpha) = m + rho*cos(beta)

    Solving for rho in both cases and setting the results equal to each other, I found alpha in terms of beta, which I plugged back into the first equation. The solution is:

    rho = r*(sqrt(1-(m/r*sin(beta))^2)-m/r*cos(beta))
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