What is the center x^2+y^2-8y+7=0 ?

The center x^2+y^2-8y+7=0 is (0,4)

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center x^2+y^2-8y+7=0

center

x^{2} + y^{2} - 8 y + 7 = 0

(0, 4)

Solution steps

x^{2} + y^{2} - 8 y + 7 = 0

Rewrite

x^{2} + y^{2} - 8 y + 7 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 4)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (0, 4), r = 3

And the center is:

(0, 4)

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

(y + 8)^{2} + (x + 2)^{2} = 72

(x - 3)^{2} + (y - (- 2))^{2} = 35

(x + 5)^{2} + (y + 6)^{2} = \frac{2}{3}

x^{2} + 5 x + y^{2} - y = \frac{5}{2}