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

• Dec 29th 2008, 10:49 AM
mdunnbass
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
• Dec 29th 2008, 11:15 AM
Jhevon
Quote:

Originally Posted by mdunnbass
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 $(x + 0.5)^2 + y^2 = 1$
• Dec 29th 2008, 11:23 AM
mdunnbass
Quote:

Originally Posted by Jhevon
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
• Dec 29th 2008, 11:28 AM
Jhevon
Quote:

Originally Posted by mdunnbass
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.

Quote:

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.
• Dec 29th 2008, 11:31 AM
mdunnbass
Quote:

Originally Posted by Jhevon
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.

Quote:

...why?
idle curiosity.
(Evilgrin)
• Mar 21st 2009, 04:45 PM
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 $r$ 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:

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