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Math Help - CIRCLE equation

  1. #1
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    CIRCLE equation

    The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y?
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  2. #2
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    Quote Originally Posted by requal View Post
    The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y?
    Yes, that would be the first step.
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  3. #3
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    yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is  \pm3\sqrt{3}
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    Quote Originally Posted by requal View Post
    yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is  \pm3\sqrt{3}
    In my universe 36 - 9 = 27 (most of the time, anyway).
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    eh, thanks anyway
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  6. #6
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    Quote Originally Posted by requal View Post
    eh, thanks anyway
    I assume that means you understand where the correct answer has come from ....
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