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  1. #1
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    circle


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  2. #2
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    i just need to know how to start it, the second problem th answer for the slop cannot contain T and S.
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  3. #3
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    Quote Originally Posted by midnightwynter View Post
    i just need to know how to start it, the second problem th answer for the slop cannot contain T and S.
    I've modified the first sketch a little bit.

    If the center of the circle has the coordinates C(x_C, y_C) then the point P has the coordinates P(x_C+u , y_C + v)

    Use the right triangle with the legs u and v and the hypotenuse r . Then
    \color{red}v = r \cdot \sin(q) and
    \color{blue}u = r \cdot \cos(q)
    Attached Thumbnails Attached Thumbnails circle-krs_koord.png  
    Last edited by earboth; November 14th 2009 at 08:09 AM.
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    I've modified the 2nd sketch a little bit.

    1. Determine the coordinates of S and T. Use the attached sketch.

    2. You are supposed to know the formula to calculate the slope of a line which runs through 2 points.

    If the points are P(x_P, y_P) and Q(x_Q, y_Q) then the slope, determined by these points, is:

    m_{PQ}=\dfrac{y_Q - y_P}{x_Q - x_P}
    Attached Thumbnails Attached Thumbnails circle-diffquotient_sinfkt.png  
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  5. #5
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    Quote Originally Posted by earboth View Post
    I've modified the first sketch a little bit.

    If the center of the circle has the coordinates C(x_C, y_C) then the point P has the coordinates P(x_C+u , y_C + v)

    Use the right triangle with the legs u and v and the hypotenuse r . Then
    u = r \cdot \sin(q) and
    v = r \cdot \cos(q)
    Excuse me but I believe you have that backwards: u= r cos(q) and v= r sin(q).
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Excuse me but I believe you have that backwards: u= r cos(q) and v= r sin(q).
    Of course you are right. Thanks for spotting my mistake.

    I've "repaired" this accident.
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