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Math Help - coordinate circle

  1. #1
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    coordinate circle

    A circle in the standard (x,y) coordinate plane has center (3,8) and is tangent to the x-axis. The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

    The solution is (x - 3)^2 + (y - 8)^2 = 64. Why do we say the radius is 8? Why couldn't it be 3 with the above information?
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  2. #2
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    Quote Originally Posted by sarahh View Post
    A circle in the standard (x,y) coordinate plane has center (3,8) and is tangent to the x-axis. The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

    The solution is (x - 3)^2 + (y - 8)^2 = 64. Why do we say the radius is 8? Why couldn't it be 3 with the above information?
    As you may know the radius is perpendicular to the tangent in the tangent point. The tangent point is T(3, 0) because it is placed on the x-axis and therefore y_T = 0 and it must be perpendicular below the center and therefore x_T = 3.

    Now calculate the distance between these 2 points:

    r = \sqrt{(3-3)^2+(8-0)^2} = 8
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  3. #3
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    Hello, Sarah!

    Did you make a sketch?


    A circle in the standard (x,y) coordinate plane
    has center (3,8) and is tangent to the x-axis.
    The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

    The solution is (x - 3)^2 + (y - 8)^2 = 64.
    Why do we say the radius is 8?
    Why couldn't it be 3 with the above information?
    Code:
              |
              |   * * *
              *           *
            * |             *
           *  |              *
              |
          *   |   (3,8)       *
          *   |     *         *
          *   |     :         *
              |     :
           *  |     :        *
            * |     :       *
              *     :     *
        - - - + - * * * - - - - - -
              |
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