What is the center x^2+y^2+4x+4y+7=0 ?

The center x^2+y^2+4x+4y+7=0 is (-2,-2)

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center x^2+y^2+4x+4y+7=0

center

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

(- 2, - 2)

Solution steps

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

Rewrite

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 2))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 2), r = 1

And the center is:

(- 2, - 2)

x^{2} + y^{2} - 10 x - 16 y + 53 = 0

x^{2} + y^{2} - 12 x + 8 y - 5 = 0

(x - 5)^{2} + (y - 3)^{2} = 49

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

(x + 2)^{2} + (y + 3)^{2} = 9