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Math Help - Center of a circle and the radius.

  1. #1
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    Center of a circle and the radius.

    I don't know if this is the right thread for my question, but this was just the closest thread to my mind that marked to be correct.

    So this is my question, I'm asked to find the center and the radius of the circle with this equation below, I don't know how to git rid with 2s in 2x^2+2y^2 to become x^2+y^2, if only the given is x^2+y^2, without those 2s, maybe I can solve it by myself.

    2x^2+2y^2-8x+5y-80=0

    /OFFTOPIC
    And oh, if possible can someone point me the direct equation in formulating the equation of a circle from 3 points which is, (1,-2)(5,4)(10,5). I already got the answer, I just want to make sure if it is right. Here's mine,
    ( 8.06 = sqrt(x-9)^2+(y+3)^2 )

    Thank you so much!
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  2. #2
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    Quote Originally Posted by dissidia View Post
    I don't know if this is the right thread for my question, but this was just the closest thread to my mind that marked to be correct.

    So this is my question, I'm asked to find the center and the radius of the circle with this equation below, Mr F says: I have looked below. There is no equation. Equations have an equals sign in them .... All you have posted are a series of expressions and nothing can be done with them.

    I don't know how to git rid with 2s in 2x^2+2y^2 to become x^2+y^2, if only the given is x^2+y^2, without those 2s, maybe I can solve it by myself.

    2x^2+2y^2-8x+5y-80=0

    /OFFTOPIC
    And oh, if possible can someone point me the direct equation in formulating the equation of a circle from 3 points which is, (1,-2)(5,4)(10,5). I already got the answer, I just want to make sure if it is right. Here's mine,
    ( 8.06 = sqrt(x-9)^2+(y+3)^2 )

    Thank you so much!
    Do the given points satisfy your equation? (By the way, what you have posted is a semi-circle, not a circle).
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  3. #3
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    2x^2+2y^2-8x+5y-80=0 This is it.
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  4. #4
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    Quote Originally Posted by dissidia View Post
    2x^2+2y^2-8x+5y-80=0 This is it.
    Complete the square in x and y terms:

    2(x^2 - 4x) + 2\left(y^2 + \frac{5}{2} y\right) - 80 = 0

    \Rightarrow 2([x - 2]^2 - 4) + 2\left(\left[y + \frac{5}{4}\right]^2 - \frac{25}{4}\right) - 80 = 0

    \Rightarrow 2(x - 2)^2 - 8 + 2\left(y + \frac{5}{4}\right)^2 - \frac{25}{2} - 80 = 0

    and the finishing touches for getting the standard form are left to you.
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    That was just the one i was looking for, thank you.
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  6. #6
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    That's interesting because it does NOT answer the specific question you asked "I don't know how to git rid with 2s"! I hope that you will now divide the entire equation by 2.
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