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Circle easy question,

  1. Feb 22, 2009 #1
    :frown:1. I have a circle equation
    [tex](x+a)^2+(y+b)^2=r^2[/tex]
    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. jcsd
  3. Feb 22, 2009 #2

    HallsofIvy

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

    For [itex]\theta[/itex] between 0 and [itex]\pi/2[/itex], 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".
     
  4. Feb 27, 2009 #3
    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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  5. Feb 27, 2009 #4

    HallsofIvy

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

    [itex]x= r sin(\theta)+ a[/itex], [itex]y= r cos(\theta)+ b[/itex].
     
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