I have a circle like so

circle with a given radius <span class=$r$, with angle $\theta$ to the $y$-axis">

Given a rotation θ and a radius r, how do I find the coordinate (x,y)? Keep in mind, this rotation could be anywhere between 0 and 360 degrees.

For example, I have a radius of 12 and a rotation θ of 115 degrees. How would you find the point (x,y)?

  • 3
    $\begingroup$ Do you recall polar coordinates? $(r,\theta)$ [polar]=$(r\cos(\theta),r\sin(\theta))$ [cartesian]. $\endgroup$
    – Clayton
    Dec 16, 2012 at 17:30
  • $\begingroup$ Nope but that does look promising! Thanks $\endgroup$ Dec 16, 2012 at 17:33

3 Answers 3


From the picture, it seems that your circle has centre the origin, and radius $r$. The rotation appears to be clockwise. And the question appears to be about where the point $(0,r)$ at the top of the circle ends up.

The point $(0,r)$ ends up at $x=r\sin\theta$, $y=r\cos\theta$.

In general, suppose that you are rotating about the origin clockwise through an angle $\theta$. Then the point $(s,t)$ ends up at $(u,v)$ where $$u=s\cos\theta+t\sin\theta\qquad\text{and} \qquad v=-s\sin\theta+t\cos\theta.$$

  • 1
    $\begingroup$ The question is clearly stating how to find the the point (x,y), not the positive (top) point on the vertical axis. Though your math is of course correct, the answer should technically be such (as seen in my answer) $\endgroup$ Feb 7, 2018 at 20:49
  • $\begingroup$ I don't think the $\theta$ you're referring the same as the $\theta$ on the given image? $\endgroup$
    – Dini
    Mar 28, 2020 at 20:24
  • $\begingroup$ What if I'm not rotating about the origin? Is it enough to just add the origin coordinates to the resulting coordinates? $\endgroup$
    – Emil S.
    Aug 12, 2022 at 10:18

With an angle of 115° in a clockwise direction, you can find your point (x,y) as shown in your diagram with the following math:

Any point $(x,y)$ on the path of the circle is $x = r*sin(θ), y = r*cos(θ)$

thus: $(x,y) = (12*sin(115), 12*cos(115))$

So your point will roughly be $(10.876, -5.071)$ (assuming the top right quadrant is x+, y+)


The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the counter-clockwise direction.

Thus, the standard textbook parameterization is: x=cos t y=sin t

In your drawing you have a different scenario. The way it is drawn, the starting point is at the top and increasing degrees is in the clockwise direction. Thus, the standard parameterization must be modified to accommodate your situation.

Take a look at what the value of x is in your picture at the starting point and then what happens as t increases. x starts at 0 and then increases to a maximum of 1 and then returns to 0 when t = Pi.

Now you want to compare that behavior to a standard graph of sin and cos to decide which one matches that need. X=sin t behaves that way, so now you have the parameterization of x. Note, it is not x=cos t as a standard math book teaches you because in trigonometry class they typically have 0 degrees at the intersection of the x-axis and the unit circle.

Now, y in your drawing starts out at 1 and then decreases until you hit 0 and then -1 at PI. A look at the a graph of either sin or cos shows that cos behaves that way.

So... Y=cos t.

I'm posting this because it can aid someone that knows about sin and cos, but has a problem in which the 0 degree starting point is in a non-standard position and the direction of positive degrees is in the clockwise direction, not the ccw direction.

  • 1
    $\begingroup$ Thanks for contributing, but MathJax will help improve your presentation. It comes in handy when you need to write down more complex expressions. $\endgroup$
    – Toby Mak
    Jul 1, 2018 at 13:42

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