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14 October, 18:52

Find the center and radius for the circle, given the equation: x^2+16x+y^2-6y=-12

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  1. 14 October, 19:08
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    In form

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

    center is (h, k) and radius is r

    complete the square for both

    group x and y seprately

    (x^2+16x) + (y^2-6y) = - 12

    take 1/2 of each linear coefient and square it then add positive and negative inside parentahsees

    16/2=8, 8^2=64

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

    (x^2+16x+64-64) + (y^2-6y+9-9) = - 12

    complete the square

    ((x+4) ^2-64) + ((y-3) ^2-9) = - 12

    expand

    (x+4) ^2-64 + (y-3) ^2-9=-12

    add 64+9 to both sides

    (x+4) ^2 + (y-3) ^2=61

    or

    (x - (-4)) ^2 + (y-3) ^2 = (√61) ^2

    center is (-4,3) and radius is √61
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