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Math Help - Form a Circle Equation

  1. #1
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    Form a Circle Equation

    Could someone please double check this for me and see if I am doing it correctly.
    Here is what the equation is and I'm supposed to find is the center point for (x,y) and the radius of the circle.

    2x^2 + 2y^2 + 20x +16y + 10 =0

    2x^2 + 2y^2 +20x + 16y = -10

    Then

    2x^2/2 + 2y^2/2 +20x/2 + 16y/2 = -10/2

    Then

    x^2 + y^2 + 10x + 8y = -5

    Then

    (x^2+10x+25) + (y^2 + 8y + 16) = -5 + 25 + 16

    Then

    (x+5)^2 + (y+4)^2 = 36^2

    Which yields my answer of ....

    (5,4) with a radius of 6.

    Please let me know if I am doing these correctly... I have no way to check my answer thus no way to know if i am on the right path.
    Last edited by WhoaBlackBetty; October 7th 2008 at 11:04 AM. Reason: typo
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  2. #2
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    Quote Originally Posted by WhoaBlackBetty View Post
    Could someone please double check this for me and see if I am doing it correctly.
    Here is what the equation is and I'm supposed to find is the center point for (x,y) and the radius of the circle.

    2x^2 + 2y^2 + 20x +16y + 10 =0

    2x^2 + 2y^2 +20x + 16y = -10

    Then

    2x^2/2 + 2y^2/2 +20x/2 + 16y/2 = -10/2

    Then

    x^2 + y^2 + 10x + 8y = -5

    Then

    (x^2+10x+25) + (y^2 + 8y + 16) = -5 + 25 + 16

    Then

    (x+5)^2 + (y+4)^2 = 36 = 6^2

    Which yields my answer of ....

    (-5, -4) with a radius of 6. Don't miss the negative signs!

    Please let me know if I am doing these correctly... I have no way to check my answer thus no way to know if i am on the right path.
    ...
    Last edited by earboth; October 7th 2008 at 11:45 AM.
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  3. #3
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    Thanks, I was close. How does that turn negative? I think thats why I screwed up. Please elaborate on that... I'm not getting when it is positive or negative.
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  4. #4
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    Quote Originally Posted by WhoaBlackBetty View Post
    Thanks, I was close. How does that turn negative? I think thats why I screwed up. Please elaborate on that... I'm not getting when it is positive or negative.
    A circle is defined as the locus of points which have a constant distance to a given fixed point.

    Let be the fixed point M(h, k) (it's the midpoint of the circle) and the constant distance r (it's the radius of the circle). Now use the distance formula to calculate the distance between any point on the circle P(x, y) and the midpoint M:

    r^2 = (x-h)^2+(y-k)^2

    You see that the operation sign in the brackets must be always minus.

    In your question you have got:

    (x+5)^2 + (y+4)^2 = 6^2~\implies~(x-(-5))^2+(y-(-4))^2=6^2 and now you can see that the point M is M(-5, -4).
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  5. #5
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    Ok...Thanks a lot. So, it is since the original circle formula is negative it needs the other negative to be positive. So, whatever I pullout will be the opposite of it's sign. Thanks! That helps out a lot.
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