Finding Coordinates on a Circle: Solving for Points After Rotation

[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 :)

 

[HallsofIvy]

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.

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

, how can I find out the coordinates of a point on the circle after the move

Thanks a lot :)

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".

 

[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

 

[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.