What is the center x^2+y^2+8y=32 ?

The center x^2+y^2+8y=32 is (0,-4)

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center x^2+y^2+8y=32

center

x^{2} + y^{2} + 8 y = 32

(0, - 4)

Solution steps

x^{2} + y^{2} + 8 y = 32

Rewrite

x^{2} + y^{2} + 8 y = 32

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 4))^{2} = (43)^{2}

Therefore the circle properties are:

(a, b) = (0, - 4), r = 43

And the center is:

(0, - 4)

x^{2} + y^{2} - 6 x + 8 y - 11 = 0

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

x^{2} + y^{2} - 12 x + 4 y = - 2

x^{2} + y^{2} + 10 x - 6 y + 15 = 0

a^{2} + b^{2} = 36