# Thread: Finding a point position on a shape after the shape was rotated

1. ## Finding a point position on a shape after the shape was rotated

Hi. First I'd like to let you know that I'm probably in the wrong section, but this is because I really, honestly, don't know where to put this question. If I knew what the problem is about, I'd have looked up on the Internet before asking.

So, I've made a picture explaining my problem : http://i.imgur.com/EkQZq.png .

If anyone can explain me how to solve this kind of problem and tell me what domain it is related to, it would be greatly appreciated.

Thanks already.

2. ## Re: Finding a point position on a shape after the shape was rotated

This is a standard problem if you allow some changes.
First lets say the center is $\displaystyle (0,0)$; that is your $\displaystyle x=0~\&~y=0$.
Then if the circle most have radius of say five.
The red point $\displaystyle (0-4,0-4)=(-4,-4)$ has well-known coordinates:
$\displaystyle \left( {4\sqrt 2 \cos \left( {\frac{{5\pi }}{4} + t} \right),4\sqrt 2 \sin \left( {\frac{{5\pi }}{4} + t} \right)} \right),\,t = 0$

If you let $\displaystyle t=\frac{-\pi}{2}$ then that rotates the point $\displaystyle (-4,-4)$ to the point $\displaystyle (-4,4)$.

Does that help?

3. ## Re: Finding a point position on a shape after the shape was rotated

Hello, Minikloon!

Another solution . . .

Say I have a circle with center $\displaystyle C(x_o,y_o).$
A point $\displaystyle P$ is at $\displaystyle (x_o\!-\!4,y_o\!-\!4).$
Rotate the circle $\displaystyle \theta$ radians counterclockwise.
Determine the new position of point $\displaystyle P.$

Code:
              * * *
*           *
*               *
*                 *

*      4  C         *
*   + - - o         *
*   :   *           *
4: *
*  o              *
* P             *
*           *
* * *

Point $\displaystyle P$ lies on the circle with center $\displaystyle C(x_o,y_o)$ and radius $\displaystyle CP = 4\sqrt{2}.$

The coordinates of $\displaystyle P$ are: .$\displaystyle \begin{Bmatrix}x &=& x_o + 4\sqrt{2}\cos\left(\theta + \frac{5\pi}{4}\right) \\ \\[-4mm] y &=& y_o + 4\sqrt{2}\sin\left(\theta + \frac{5\pi}{4}\right) \end{Bmatrix}$

4. ## Re: Finding a point position on a shape after the shape was rotated Originally Posted by Soroban Hello, Minikloon!
Another solution . . .
Another solution???
Come on, that is identical to the solution that I posted. Is it not?
Please correct me if I am wrong. How is it another solution?

5. ## Re: Finding a point position on a shape after the shape was rotated

Hello, Plato!

Don't get your shorts in a knot . . . identical?

You made some changes . . . placing the center at the origin.
. . I didn't ... I solved it using the original information.

I believe that even the touchy moderators will agree
. . that I made a significant contribution.

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