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Math Help - circles- geometry-coordinates

  1. #1
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    circles- geometry-coordinates

    a circle passes through the points (0,-3) and (0,3) and has radius 5

    find equations of possible solutions (two possible)

    realllyyy confused even though it's probably quite simple, even unsure where to begin

    thanks
    Last edited by lra11; December 14th 2007 at 05:34 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lra11 View Post
    a circle passes through the points (0,-3) and (0,3) and has radius 5

    find equations of possible solutions (two possible)

    realllyyy confused even though it's probably quite simple, even unsure where to begin

    thanks
    Call the center of the circle (h, k). Then the circle equation is
    (x - h)^2 + (y - k)^2 = r^2

    Thus
    (0 - h)^2 + (-3 - k)^2 = 5^2
    and
    (0 - h)^2 + (3 - k)^2 = 5^2

    Solve for h and k.

    -Dan
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  3. #3
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    Think about (0,4) & (0,-4) as possible centers.
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  4. #4
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    A circle has a center (h,k) and a radius r in the formula
    (x-h)^2 + (y-k)^2=r^2. You will end up with two equations when plugging in your information:

    1) (0-h)^2 + (-3-k)^2 = 5^2
    2) (0-h)^2 + (+3-k)^2 = 5^2

    Try to solve for h and k from using the system of two quadratics equations...
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  5. #5
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    (0,4)

    is this because 5+3=8 / 2 = 4 or am i on the wrong lines completely
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lra11 View Post
    is this because 5+3=8 / 2 = 4 or am i on the wrong lines completely
    Pet peeve warning! Danger Will Robinson! Danger! Danger!

    How can 5 + 3 = 8/2 = 4?

    Look at the outsides of this: 5 + 3 = 4 This is obviously not true! You need to write the expressions correctly, so this needs to go on two lines:
    5 + 3 = 8
    8/2 = 4

    Okay, pet peeve rant over.

    -Dan
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  7. #7
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    h^2 + k^2 + 6k + 9 = 25
    h^2 + k^2 - 6k + 9 = 25

    so

    h^2 + k^2 + 6k + 9 = 25
    -h^2 + -k^2 + 6k - 9 = -25

    12k = 0
    k = 0

    Therefore,
    h^2 + 0^2 + 6(0) + 9 = 25
    h^2 + 0^2 - 6(0) + 9 = 25

    so

    h^2 + 9 = 25

    h^2 = 25-9

    h^2=16

    h= +-4

    Centers are (4,0) and (-4,0)
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  8. #8
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    Come on! All that was needed is to note that you have a 3,4, 5 triangle.
    All that other stuff is needless.
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  9. #9
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    thanks

    thanks everyone for the help, it seems so simple now.
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  10. #10
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    Needless? Sometimes shortcuts aren't the best way to get to the answer. The longer method works whether the triangle is 3-4-5 or say...4-5-sqrt 41 or whatever.

    I'm sure they will learn and benefit from BOTH approaches...man...chill out...
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  11. #11
    Senior Member DivideBy0's Avatar
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    3-4-5 triangles are geometry.

    Solving for variables is coordinate geometry
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  12. #12
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    Yes, and given that this question was in the context of "coordinate geometry", my approach shouldn't have been an issue with Plato.

    Sheesh...
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  13. #13
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by mathdeity View Post
    Yes, and given that this question was in the context of "coordinate geometry", my approach shouldn't have been an issue with Plato.

    Sheesh...
    Well, whatever works, works, I guess. You could approach this problem from many areas of math, just like the 100+ proofs of the pythagorean theorem.
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