How do you do complete the square with circle in center radius form

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How do you do complete the square with circle in center radius form

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    2021-09-14T22:56:37+00:00

    we know that formula of circle in standard form

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

    where

    (h,k) is center of circle

    r is radius of circle

    Here , we can see that (x-h)^2 and (y-k)^2 are perfect squares

    so, we will change our equations in perfect square

    For example:

    x^2+y^2+16x-14y+49=0

    Firstly, we will change x terms into perfect square

    step-1:

    we will move constant term on right side

    x^2+y^2+16x-14y+49-49=0-49

    x^2+y^2+16x-14y=-49

    step-2:

    We will move all x terms altogether

    x^2+16x+y^2-14y=-49

    step-3:

    Break x terms as 2*a*b

    x^2+2\times 8\times x+y^2-14y=-49

    step-4:

    now, we can use formula

    a^2+2ab+b^2=(a+b)^2

    we will identify b

    so, we get b=8

    to make 8^2 , we will add both sides by 8^2

    x^2+2\times 8\times x+8^2+y^2-14y=-49+8^2

    now, we can write in perfect square form

    (x+8)^2+y^2-14y=15

    step-5:

    now, we can break y terms in 2ab form

    (x+8)^2+y^2-2\times 7\times y=15

    step-6:

    now, we can use formula

    a^2-2ab+b^2=(a-b)^2

    we will identify b

    so, we get b=7

    to make 7^2 , we will add both sides by 7^2

    (x+8)^2+y^2-2\times 7\times y+7^2=15+7^2

    (x+8)^2+(y-7)^2=64

    (x+8)^2+(y-7)^2=8^2

    we can see that this is in standard equation of circle form

    so, center=(-8,7)

    radius=8

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