What is the center x^2+z^2+4x-4z=-4 ?

The center x^2+z^2+4x-4z=-4 is (-2,2)

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center x^2+z^2+4x-4z=-4

center

x^{2} + z^{2} + 4 x - 4 z = - 4

(- 2, 2)

Solution steps

x^{2} + z^{2} + 4 x - 4 z = - 4

Rewrite

x^{2} + z^{2} + 4 x - 4 z = - 4

in the form of the standard circle equation

(x - (- 2))^{2} + (z - 2)^{2} = 2^{2}

Therefore the circle properties are:

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

And the center is:

(- 2, 2)

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(x - 1)^{2} + (y - 5)^{2} = 17

(x - 2)^{2} + (y + 3)^{2} = \frac{16}{9}

x^{2} + y^{2} - 64 = 0

x^{2} + (y - 2)^{2} = 36