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Math Help - Rearranging an equation.

  1. #1
    Member MathBlaster47's Avatar
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    Rearranging an equation.

    A question I am working on asks me to find the center and radius of a circle with the equation: x^2+y^2-10x+4y+13=0.
    The thing is, while I am pretty sure I know what to do to get my answers I'm having some trouble rearranging the equation to a workable format. I have one way that I think might work, but I've never really understood how that whole thing works.
    Here goes!

    x^2+y^2-10x+4y+13=0 \rightarrow (x^2-10x)+(y^2+4y)=-13

    Am I on the right track?
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  2. #2
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    Quote Originally Posted by MathBlaster47 View Post
    A question I am working on asks me to find the center and radius of a circle with the equation: x^2+y^2-10x+4y+13=0.
    The thing is, while I am pretty sure I know what to do to get my answers I'm having some trouble rearranging the equation to a workable format. I have one way that I think might work, but I've never really understood how that whole thing works.
    Here goes!

    x^2+y^2-10x+4y+13=0 \rightarrow (x^2-10x)+(y^2+4y)=-13

    Am I on the right track?
    You are, next complete the square for x and y

    (x+5)^2-25 + (y+2)^2 - 4 = -13

    Spoiler:
    (x+5)^2 + (y+2)^2 = 16

    Therefore we have a circle with centre (-5,-2) and radius 4
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  3. #3
    MHF Contributor
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    You need to complete the square for both x and y.


    Completing the square for

    x^2 - 10x yields x^2 - 10x + 25 = (x - 5)^2

    and completing the square for

    y^2 + 4y yields y^2 + 4y + 4 = (y + 2)^2.

    So we add 25 + 4 = 29 to both sides of your equation:

    x^2 - 10x + 25 + y^2 + 4y + 4 + 13 = 29

    and then simplify:

    (x - 5)^2 + (y + 2)^2 = 16

    Can you find the answers from this form of the equation?
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  4. #4
    Member MathBlaster47's Avatar
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    Thanks!

    Uhm...why are the numbers added always 25 and 4?
    I had thought that they were determined by the equation at hand, I realize that isn't so, but my curiosity, as they say, has been piqued.
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  5. #5
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    Quote Originally Posted by MathBlaster47 View Post
    Thanks!

    Uhm...why are the numbers added always 25 and 4?
    I had thought that they were determined by the equation at hand, I realize that isn't so, but my curiosity, as they say, has been piqued.
    They are determined by the equation at hand.

    To complete the square for x^2 + ax, you calculate

    \frac{a^2}{4} since

    x^2 + ax + \frac{a^2}{4} = \left(x + \frac{a}{2}\right)^2

    So these numbers are dependent on the coefficients of x and y.
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  6. #6
    Junior Member SuperCalculus's Avatar
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    There's a much easier way.

    You're working with a circle with equation with equation x^2 + y^2 + 2gx + 2fy + c = 0, where centre is (-g, -f) and radius is \sqrt{g^2 + f^2 - c}

    So, from the equation you gave, you can derive;
     -10x = 2gx
    and
    4y = 2fy
    Hence;

    g = -5

    f = 2

    So, plug those values into the equations

    Centre is (5, -2)

    And radius is \sqrt {25 + 4 - 13} = \sqrt {16} = 4.
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  7. #7
    Member MathBlaster47's Avatar
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    My mistake...I had been looking at an example that also used the numbers 25 and 4, so I got a smidgen turned around, thank you for setting me straight!
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  8. #8
    Member MathBlaster47's Avatar
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    Quote Originally Posted by icemanfan View Post
    They are determined by the equation at hand.

    To complete the square for x^2 + ax, you calculate

    \frac{a^2}{4} since

    x^2 + ax + \frac{a^2}{4} = \left(x + \frac{a}{2}\right)^2

    So these numbers are dependent on the coefficients of x and y.
    Just so I'm clear, since there is no loose number all I have to do is the rest of the working for completing the square?

    Quote Originally Posted by icemanfan View Post
    You need to complete the square for both x and y.


    Completing the square for

    x^2 - 10x yields x^2 - 10x + 25 = (x - 5)^2

    and completing the square for

    y^2 + 4y yields y^2 + 4y + 4 = (y + 2)^2.

    So we add 25 + 4 = 29 to both sides of your equation:

    x^2 - 10x + 25 + y^2 + 4y + 4 + 13 = 29

    and then simplify:

    (x - 5)^2 + (y + 2)^2 = 16

    Can you find the answers from this form of the equation?
    Verily, thanks!

    Quote Originally Posted by SuperCalculus View Post
    There's a much easier way.

    You're working with a circle with equation with equation x^2 + y^2 + 2gx + 2fy + c = 0, where centre is (-g, -f) and radius is \sqrt{g^2 + f^2 - c}

    So, from the equation you gave, you can derive;
     -10x = 2gx
    and
    4y = 2fy
    Hence;

    g = -5

    f = 2

    So, plug those values into the equations

    Centre is (5, -2)

    And radius is \sqrt {25 + 4 - 13} = \sqrt {16} = 4.
    Thanks for the easy method! I'll remember it in the future!


    Edit:
    I, for some reason, am feeling a little lost as to how to simplify:
    (x^2-x10+25)+(y^2+4y+4)=16
    I just have to factor out the numbers right?
    Last edited by MathBlaster47; February 19th 2010 at 06:45 PM.
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  9. #9
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    OK, I can edit. How do I delete one of my own entries? Thanks.

    I wasn't awake again, and should have looked through all the replied already done for him.
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  10. #10
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    Quote Originally Posted by Diagonal View Post
    OK, I can edit. How do I delete one of my own entries? Thanks.

    I wasn't awake again, and should have looked through all the replied already done for him.
    You can't delete. But if you report a post to a Moderator using the Report Post Tool (the small triangle found in top right corner of all posts) you can request that a post be deleted. In fact, you can report any post that you feel deserves the attention of a Moderator.
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