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Given the coordinates of a single point on a circle and a length of an arc $L$, how do I find the coordinates of another point?

Or, to put in another form: I have the radius $r$, the length of the arc $L$ and $(x_1,y_1)$ the coordinates. I need to express $(x_2,y_2)$ using only $r, L, x_1$, and $y_1$.

I'm at a dead end on this.

Jean-Claude Arbaut
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Tom Klino
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1 Answers1

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Traveling along a fixed circle, is basically rotating a vector around a fixed point. For that we can use the well known rotation matrix $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$. But to succesfully do this you need to know the center of you circle first, like ccorn mentioned

gebruiker
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  • I'm afraid I don't understand. Can you describe in more details how to use the matrix? Given that the center is (0,0)? – Tom Klino May 05 '14 at 15:31
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    @Tom If you google rotation matrix you will find an overwellming amount of examples and explanations. The wikipedia one is very usefull (as usual). Further, if the center lies at the origen you only need to establish the correct value of your angle i.e. $\theta$. But you already know the full length of your circles circumference wich corresponds to $360^\circ$ and the length of your segment.... This will probably help you more than enough. – gebruiker May 05 '14 at 15:42
  • The word "overwhelming" is the key word here :-). I am looking at the wikipedia article, but it's very long and I only want to do a relative simple calculation. I would appreciate it very much if you can elaborate the answer a little so I could understand it better (it will be good for future comers too I hope) – Tom Klino May 05 '14 at 15:46
  • @Tom I think it's best if we avoid a full explanation of "how do you derive the rotation matrix" here (though it's relatively simple). The key thing you need to know is that this matrix can be multiplied with a vector $\begin{pmatrix}x\y\end{pmatrix}$ wich will result in a rotation of the vector over an angle $\theta$ in the counterclockwise direction. – gebruiker May 05 '14 at 15:53