Circle easy question,

1. Feb 22, 2009

macarino

1. I have a circle equation
$$(x+a)^2+(y+b)^2=r^2$$
And suppose we start from the topmost of the circle
2.I have an angle Theta
and if Theta is larger than 0 we move counterclockwise or we will move in the opposite direction if it is positive.
, how can I find out the coordinates of a point on the circle after the move

Thanks a lot :)

2. Feb 22, 2009

HallsofIvy

You mean "negative" instead of this second "positive" don't you?

For $\theta$ between 0 and $\pi/2$, at least, You can construct a right triangle by drawing a line from (0,0) to the point and then from the point perpendicular to the right angle. The y-coordinate of the point is the "near side", the x-coordinate is the negative of the "opposite side".

3. Feb 27, 2009

macarino

I am sorry I am not a native, not understanding what you mean

In the picture below, I'd like to know what is P's coordinates given rotation angle alpha and a point O(0,a)

The circle equation is
(x+a)^2+(y+b)^2=r^2

Thanks
Regards

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4. Feb 27, 2009

HallsofIvy

You are taking $\theta$ to be 0 at the "top", (a,b+ r) and measuring it clockwise? If you were using the "standard", $\theta= 0$ to the right, (a+r, 0), and measuring counter clockwise, then you could use the parametric equations, $x= r cos(\theta)+ a$, $y= r sin(\theta)+ b$. The fact that you starting 90 degrees off that means you need to swap sine and cosine: $x= r sin(\theta)+ a$ and $y= r cos(\theta)+ a$. Now, taking $\theta= 0$ you can see that $x= r(0)+ a= a$, $y= r(1)+ b= b+ r$ as you want. Further taking $\theta= \pi/2$, we have $x= r(1)+ a= a+ r$, $y= r(0)+ b= b$ as wanted.

$x= r sin(\theta)+ a$, $y= r cos(\theta)+ b$.