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Math Help - polar coordinates

  1. #1
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    polar coordinates

    A circle C has center at the origin and radius 3. Another circle K has a diameter with one end at the origin and the other end at the point (0, 13). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let (r,\theta) be the polar coordinates of P, chosen so that  r is positive and 0 \leq \theta \leq \pi/2

    Find  r and \theta
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  2. #2
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    Quote Originally Posted by viet View Post
    A circle C has center at the origin and radius 3. Another circle K has a diameter with one end at the origin and the other end at the point (0, 13). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let (r,\theta) be the polar coordinates of P, chosen so that  r is positive and 0 \leq \theta \leq \pi/2

    Find  r and \theta
    So you are looking for the intersections of the circles
    x^2 + y^2 = 9
    and
    x^2 + \left (y - \frac{13}{2} \right )^2 = \left ( \frac{13}{2} \right ) ^2

    From the first equation x^2 = 9 - y^2. Inserting this into the second equation gives:
    (9 - y^2) + \left (y - \frac{13}{2} \right ) ^2 = \frac{169}{4}

    Solve this for y. I get:
    y = \frac{9}{13}

    There are two x values for this y value:
    x = \pm \sqrt{9 - \left ( \frac{9}{13} \right ) ^2} = \pm \frac{12}{13} \sqrt{10}

    Thus the point of intersection in QI is
    \left ( \frac{12}{13} \sqrt{10}, \frac{9}{13} \right )

    I leave putting this into polar coordinates to you.

    -Dan
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  3. #3
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    Hello, viet!

    A circle C has center at the origin and radius 3.
    Another circle K has a diameter with one end at the origin and the other end at the point (0, 13).
    The circles C and K intersect in two points.

    Let P be the point of intersection of C and K which lies in the first quadrant.
    Let (r,\,\theta) be the polar coordinates of P, chosen so that r is positive and 0 \leq \theta \leq \pi/2

    Find r and \theta
    If you are familiar with polar equations of circle, it's easier.


    Circle C has the equation: . r\:=\:3

    Circle K has the equation: . r \:=\:13\sin\theta

    They intersect where: . 13\sin\theta \:=\:3

    Hence, we have: . \sin\theta \:=\:\frac{3}{13}\quad\Rightarrow\quad\theta \:=\:\sin^{-1}\left(\frac{3}{13}\right) \:=\:0.232868178 \:\approx\:0.233


    Therefore, point P is: . 3,\,0.233)

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