What is the center x^2+y^2+8y=2x+20 ?

The center x^2+y^2+8y=2x+20 is (1,-4)

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center x^2+y^2+8y=2x+20

center

x^{2} + y^{2} + 8 y = 2 x + 20

(1, - 4)

Solution steps

x^{2} + y^{2} + 8 y = 2 x + 20

Rewrite

x^{2} + y^{2} + 8 y = 2 x + 20

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 4))^{2} = (37)^{2}

Therefore the circle properties are:

(a, b) = (1, - 4), r = 37

And the center is:

(1, - 4)

x^{2} - 2 x + y^{2} - 2 y = 23

13^{2} = (x - (- 4))^{2} + (y - (- 3))^{2}

(x + 3)^{2} + y^{2} = 25

13.8 + x^{2} + y^{2} = 17

[x + 5]^{2} + [y + 1]^{2} = 9