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

• Aug 15th 2011, 02:17 PM
Minikloon
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.

• Aug 15th 2011, 02:49 PM
Plato
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 $(0,0)$; that is your $x=0~\&~y=0$.
Then if the circle most have radius of say five.
The ‘red’ point $(0-4,0-4)=(-4,-4)$ has well-known coordinates:
$\left( {4\sqrt 2 \cos \left( {\frac{{5\pi }}{4} + t} \right),4\sqrt 2 \sin \left( {\frac{{5\pi }}{4} + t} \right)} \right),\,t = 0$

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

Does that help?
• Aug 15th 2011, 04:27 PM
Soroban
Re: Finding a point position on a shape after the shape was rotated
Hello, Minikloon!

Another solution . . .

Quote:

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

Code:

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

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

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

• Aug 15th 2011, 04:59 PM
Plato
Re: Finding a point position on a shape after the shape was rotated
Quote:

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?
• Aug 16th 2011, 05:01 AM
Soroban
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.