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Math Help - Algerbil

  1. #1
    Member GAdams's Avatar
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    Algerbil

    I have to simplify this to get the equation of a circle:

    (x-4)^2 +4^2 + (x-2)^2 - 2^2 +11 = 0

    I get:

    (x-4)^2 + (x-2)^2 +16 + 4 + 11 = 0

    (x-4)^2 + (x-2)^2 = -31

    ....where am I going wrong?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by GAdams View Post
    (x-4)^2 +4^2 + (x-2)^2 - 2^2 +11 = 0
    At first, this is not a circle, it's just a parabola.

    May be you mean (x-4)^2+(y-2)^2=-23, which is clearly, an imaginary circle.
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  3. #3
    Member GAdams's Avatar
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    Quote Originally Posted by Krizalid View Post
    At first, this is not a circle, it's just a parabola.

    May be you mean (x-4)^2+(y-2)^2=-23, which is clearly, an imaginary circle.
    Here's what I started with:

    x^2 - 8x + y^2 + 4y + 11 = 0

    I completed teh square for the first couple of expressions and got to where I was in my first post.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    Here's what I started with:

    x^2 - 8x + y^2 + 4y + 11 = 0

    I completed teh square for the first couple of expressions and got to where I was in my first post.
    that's incorrect then


    x^2 - 8x + y^2 + 4y + 11 = 0

    \Rightarrow x^2 - 8x + y^2 + 4y = -11

    \Rightarrow x^2 - 8x + (-4)^2 + y^2 + 4y + 2^2 = -11 + (-4)^2 + 2^2

    \Rightarrow (x - 4)^2 + (y + 2)^2 = 3^2

    which is a circle centered at (4,-2) with radius 3
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