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Thread: Converting a circle to polar coordinates, not centered at the origin

  1. #1
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    Converting a circle to polar coordinates, not centered at the origin

    Sorry, don't have a visual at the moment, but there's a circle with the center at: (1/2)a , with a radius of: (1/2)a

    It's not centered at the origin, so I'm confused as to what theta values the circle will be going from. I think it goes from 0 to 2pi, can anybody explain what theta is supposed to be? And would I be correct in assuming that r goes from 0 to a?
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    Re: Converting a circle to polar coordinates, not centered at the origin

    do you mean the center is at $\left(\dfrac a 2, ~\dfrac a 2 \right)$ ?
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    Re: Converting a circle to polar coordinates, not centered at the origin

    Quote Originally Posted by romsek View Post
    do you mean the center is at $\left(\dfrac a 2, ~\dfrac a 2 \right)$ ?
    yes

    I found the exact question in my notes, it's a different problem though: r = cos (theta) produces the kind of circle I'm talking about
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    Re: Converting a circle to polar coordinates, not centered at the origin

    Quote Originally Posted by Awesome31312 View Post
    yes

    I found the exact question in my notes, it's a different problem though: r = cos (theta) produces the kind of circle I'm talking about
    so you are all set then?
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  5. #5
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    Re: Converting a circle to polar coordinates, not centered at the origin

    Given a circle, of radius r, centered at $\displaystyle (x_0, y_0)$, so equation $\displaystyle (x- x_0)^2+ (y- y_0)^2= r^2$, the line y= mx is tangent to the circle if and only if $\displaystyle (x- x_0)^2+ (mx- x_0)^2= r^2$, a quadratic equation, has a double root. (A quadratic equation may have no solutions, one solution or two solutions. If such equation has two solutions, the line crosses the circle twice so crosses through the circle. If it has no solutions, the line completely misses the circle. If it has one solution, the line is tangent to the circle.)

    Write the equation as $\displaystyle x^2- 2x_0x+ x_0^2+ m^2x^2- 2mx_0x+ x_0^2= (1+m^2)x^2- (2+ 2m)x+ 2x_0^2= r^2$. That has one solution, and the line is tangent to the circle, if the discriminant of the equation, $\displaystyle (2+ 2m)^2- 4(1+ m^2)(2x_0^2- r^2)= 0$.

    ($\displaystyle r= cos(t\theta)$ is, specifically, the circle with center at (1/2, 0) with radius 1/2.
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    Re: Converting a circle to polar coordinates, not centered at the origin

    Well, I'm trying to figure out what theta is supposed to be, I mean the bounds of it. The guide says it goes from (-pi/2) to (pi/2), is this accurate?

    For context, I'm setting up an integral and this circle is the plane trace
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