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Math Help - circles

  1. #1
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    circles

    Two circles are drawn in a 12-inch by 14-inch rectangle. Each circle has a diameter of 6 inches. If the circles do not extend beyond the rectangular region, what is the greatest possible distance between the centers of the two circles?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by sri340 View Post
    Two circles are drawn in a 12-inch by 14-inch rectangle. Each circle has a diameter of 6 inches. If the circles do not extend beyond the rectangular region, what is the greatest possible distance between the centers of the two circles?
    Let P=(x_1,y_1) be the center of one circle, while Q=(x_2,y_2) is the center of the other. It is clear that the distance between P and Q willl be greatest when the circles lay at opposite corners of the rectangle. On a coordinate grid, locate P and Q and then determine the distance d by

    d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
    Last edited by VonNemo19; January 30th 2010 at 04:15 PM.
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  3. #3
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    Or...

    The circles touch opposite ends of the rectangle, so their centres are 6 inches from either end.
    6+6=12 inches from side to centre combined when the centres are furthest apart. This leaves at most 2 inches from centre to centre.
    Therefore the maximum distance between the centres is 2 inches.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Or...

    The circles touch opposite ends of the rectangle, so their centres are 6 inches from either end.
    6+6=12 inches from side to centre combined when the centres are furthest apart. This leaves at most 2 inches from centre to centre.
    Therefore the maximum distance between the centres is 2 inches.
    No.
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  5. #5
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    Apologies,
    i used a radius of 6.
    Pythagoras' theorem gives \sqrt{(14-6)^2+(12-6)^2}
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Apologies,
    i used a radius of 6.
    Pythagoras' theorem gives \sqrt{(14-6)^2+(12-6)^2}
    That's okay, but I don't follow. So, you are using (14,12) as one of your points?

    Even if you used the correct radius this would yield an incorrect result. Let's wait and see if the OP is going to reply. If you like, we can talk about this via IM.

    EDIT:Oh wait. I see...it works.
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  7. #7
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    Hello, sri340!

    Did you make a sketch?
    You can almost "eyeball" the answer!



    Two circles are drawn in a 12-inch by 14-inch rectangle.
    Each circle has a diameter of 6 inches.
    If the circles do not extend beyond the rectangular region,
    what is the greatest possible distance between the centers of the two circles?
    Code:
          : - 3 - : - - - -8- - - - : - 3 - :
        - *-------o-------------------------* -
        : |   o   :   o                     | :
        : |o      :3     o                  | 3
        : |      P:                 R       | :
        : o - - - o - - - o - - - - *       | -
        : |   3     *               :       | :
        : |o          *  o          :       | :
        : |   o       o *           :       | :
       12 |       o       *         :       | 6
        : |                 *       o       | :
        : |                   * o   :   o   | :
        : |                  o  *   :      o| :
        : |                       * :Q      | :
        : |                 o       o - - - o -
        : |                         :   3   | :
        : |                  o     3:      o| 3
        : |                     o   :   o   | :
        - *-------------------------*-------* -
          : - - - - - - - 14  - - - - - - - :

    The circles have centers P and Q and radius 3.
    They are placed in opposite corners of the rectangle.

    Center P and Q with point R form a right triangle
    . . with legs: .  PR = 8,\:QR = 6

    Therefore: . PQ \,=\,10

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