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can someone check this for me?--completing the square?

its completing the square for both x & y for a circle ( (x-h)^2+(y-k)^2=r^2 )

this is the equation: x^2+y^2-2x+6y+1=0

the answer i got was (x+1)^2 (y+3) =9

my peer got (x-1)^2 (y+3) =9

i just want to know what im doing wrong, thanks SO much for the help

Update:

i know where i went wrong, thank you!

2 Answers

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  • TC
    Lv 7
    10 years ago
    Favorite Answer

    x^2 + y^2 - 2x + 6y + 1 = 0

    (x^2 - 2x) + (y^2 + 6y) = - 1

    complete sqaure of x

    (- 2/2)^2 = (- 1)^2 = 1

    complete square of y

    (6/2)^2 = (3)^2 = 9

    add (9 + 1) to both sides of equation

    (x^2 - 2x + 1) + y^2 + 6y + 9) = 9

    (x - 1)^2 + (y + 3)^2 = 9

    center is (1, - 3) with radius of 3

  • 10 years ago

    There is no need to add a constant before completing the square as the above answerers have done, you can complete the square straight away.

    Complete the square for this circle equation:

    x² + y² - 2x + 6y + 1 = 0

    x² - 2x + y² + 6y = -1

    (x - 1)² - 1 + (y + 3)² - 9 = -1

    (x - 1)² + (y + 3)² = 9

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