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Thread: Distance.

  1. #1

    Thread Starter
    Fanatic Member prog_tom's Avatar
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    Talking Distance.

    In the diagram attached,

    Find the distance from P to AD.

    Notice the curved line from D to B, it's actually a circle, with radius 4, i didn't draw it good. And P is also on a circle with radius 2. There are two circles in the square. AD is 4 units.

    I tried to set midpoint M to be (0,0), and P to be (1/4,sqrt(3)/4) since it's not a unit circle, it has a radius of two. And i tried to suppose there was a point T on line AD, which is (-2,sqrt(3)/4). So finding the distance should be easy,

    TP = sqrt((x2-x1)^2+(y2-y1)^2)

    = sqrt((9/4)^2)
    = 9/4

    However I think I may be incorrect. Can anyone help me confirming the answer? Thanks.
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  2. #2
    vbuggy krtxmrtz's Avatar
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    Why not directly calculate where the circles have their intersection? The equation of a circle is,

    (x - a)2 + (y - b)2 = r2

    Your circles have these values:

    r = 4, a = 0, b = 4
    and
    r = 2, a = 2, b = 0

    Then,

    16 = x2 + (y - 4)2
    4 = (x - 2)2 + y2

    If you work this out you arrive at the equations

    x2 + y2 - 8y = 0
    x2 - 4x + y2 = 0

    To easily get to the solution, add and substract 2xy to/from each one of them:

    x2 + y2 +2xy - 2xy - 8y = 0
    x2 - 4x + y2 +2xy - 2xy = 0

    This can be rearranged as

    (x - y)2 + 2xy -8y = 0
    (x - y)2 + 2xy -4y = 0

    Now substract one from another and you get

    4x - 8y = 0 or x = 2y

    Substitute this back into one of the 2 original equations and finally:

    4y2 + y2 -8y = 0
    y = 8/5 = 1.6 and x = 16/5 = 3.2

  3. #3

    Thread Starter
    Fanatic Member prog_tom's Avatar
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    what are a, and b?

    prog_tom
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  4. #4
    vbuggy krtxmrtz's Avatar
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    Originally posted by prog_tom
    what are a, and b?
    Sorry, forgot about them at the last minute...
    They are the x and y coordinates of the center of the circle.

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