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

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. (Doh)

I started by drawing a circle of radius $\displaystyle r$ and defining a new "origin" at a distance $\displaystyle m$ 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, $\displaystyle rho$, 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 $\displaystyle alpha$, and $\displaystyle rho$ made an angle I called $\displaystyle beta$.

Then I found the following relations:

$\displaystyle r*sin(alpha) = rho*sin(beta)$

$\displaystyle r*cos(alpha) = m + rho*cos(beta)$

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:

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