What is the center x^2+y^2=8(x+y)+4 ?

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

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center x^2+y^2=8(x+y)+4

center

x^{2} + y^{2} = 8 (x + y) + 4

(4, 4)

Solution steps

x^{2} + y^{2} = 8 (x + y) + 4

Rewrite

x^{2} + y^{2} = 8 (x + y) + 4

in the form of the standard circle equation

(x - 4)^{2} + (y - 4)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (4, 4), r = 6

And the center is:

(4, 4)

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

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

x^{2} + y^{2} + 2 x + 10 y + 18 = 0