# Math Help - Calculating position on a circumference

1. ## Calculating position on a circumference

Hello all, I hope someone is able to help me get my head round this little problem.

If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?

Any help would be appreciated!

Thanks!

2. Originally Posted by gryphon5
Hello all, I hope someone is able to help me get my head round this little problem.

If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?

Any help would be appreciated!

Thanks!
If the angle is 90 degrees, the point is (350,200)?
Do your angles start at the upper axis, the "North axis", then go clockwise?
If your angles start from the normal "East axis" then go counterclockwise, then the point at 90 degrees should be (200,350).

Whatever way you have there, for arbitrary angles/degrees, you just get the components of the 150-radius that are parallel to your axes.

Let us say your angles start the usual East-axis, or positive x-axis, and then go counterclockwise.

If the angle is 90 degrees,
x = 200 +150cos(90deg) = 200 +0 = 200
y = 200 +150sin(90deg) = 200 +150 = 350
Hence, point (200,350).

If the angle is 30 degrees,
x = 200 +150cos(30deg) = 200 +129.9 = 329.9
y = 200 +150sin(30deg) = 200 +75 = 275
Hence, point (329.9,275).

If the angle is 108 degrees,
x = 200 +150cos(108deg) = 200 -46.35 = 153.65
y = 200 +150sin(108deg) = 200 +142.66 = 342.66
Hence, point (153.65,342.66).

Etc....

3. Thanks vry much for your fast response.

Sorry for not including all the information, I should have stated...

This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.

How would this change the examples you gave?

Thanks again!

4. Originally Posted by gryphon5
Thanks vry much for your fast response.

Sorry for not including all the information, I should have stated...

This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.

How would this change the examples you gave?

Thanks again!
I see.

Then here is the change.

Let us call the vertical axis as y-axis also. The horizontal axis as x-axis also. The angles start from the upper or positive y-axis, going clockwise.
Our coordinates are in the usual (x,y) ordered pair.

If the angle is 30 degrees,
y = 200 +150cos(30deg) = 200 +129.9 = 329.9
x = 200 +150sin(30deg) = 200 +75 = 275
Hence, point (275,329.9).

If the angle is 108 degrees,
y = 200 +150cos(108deg) = 200 -46.35 = 153.65
x = 200 +150sin(108deg) = 200 +142.66 = 342.66
Hence, point (342.66,342.66).

If the angle is 92.5 degrees,
y = 200 +150cos(92.5deg) =200 -6.54 = 193.46
x = 200 +150sin(92.5deg) = 200 +149.86 = 349.86
Hence, point (349.86,193.46).

If the angle is 328.4 degrees,
y = 200 +150cos(328.4deg) = 200 +127.76 = 327.76
x = 200 +150sin(328.4deg) = 200 -78.60 = 121.40
Hence, point (121.40,327.76).

In other words,
For the x-component of the radius, use sine.
For the y-component of the radius, use cosine.
It's the reverse if you're doing them in the usual manner where the angles start from the positive x-axis, going counterclockwise.

5. Hello, gryphon5!

If I have a circle that is centered at (200,200) and its radius is 150,
how do I calculate the point at any given angle?
Code:
    |         * * *
|     *           *  P
|   *               *
|  *           r  / |*
|               /   |
| *           / θ   | *
| *        O* - - - + *
| *       (h,k)     Q *
|
|  *                 *
|   *               *
|     *           *
|         * * *
|
- + - - - - - - - - - - - - - - -
|

Consider a circle with radius $r$ with center $O(h,k)$.

Point $P$ creates $\angle POQ$ with the horizontal.

In right triangle $PQO$, we have:
. . $\cos\theta = \frac{OQ}{r}\quad\Rightarrow\quad OQ = r\cos\theta$
. . $\sin\theta = \frac{PQ}{r}\quad\Rightarrow\quad PQ = r\sin\theta$

The $x$-coordinate of $P$ is: . $x \:=\:h + OQ\:=\:h + r\cos\theta$
The $y$-coordinate of $P$ is: . $y \:=\:k + PQ \:=\:k + r\sin\theta$

Therefore, point $P$ is at: . $\left(h + r\cos\theta,\:k + r\sin\theta\right)$

6. Wow, very helpful, thanks to you both that has made things much clearer

7. Sorry to re-open an old thread, but the markup seems to have gone weird, is it possible for someone to ressurect it so I can read the equations again?

Thanks,
gryphon

8. Hello!

I'll try to format all this without LaTeX . . .

Code:
    |         * * *
|     *           *  P
|   *               *
|  *           r  / |*
|               /   |
| *           / θ   | *
| *        O* - - - + *
| *       (h,k)     Q *
|
|  *                 *
|   *               *
|     *           *
|         * * *
|
- + - - - - - - - - - - - - - - -
|

Consider a circle with radius r with center O(h,k).

Point P creates /POQ with the horizontal.

In right triangle PQO, we have:
. . cosθ = OQ/r . . OQ = r·cosθ
. . sinθ = PQ/r . . . PQ = r·sinθ

The x-coordinate of P is: .x .= .h + OQ .= .h + r·cosθ
The y-coordinate of P is: .y .= .k + PQ . = .k + r·sinθ

Therefore, point P is at: .(h + r·cosθ, k + r·sinθ)