What is the center y^2+4x-20-2y=-x^2 ?

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

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center y^2+4x-20-2y=-x^2

center

y^{2} + 4 x - 20 - 2 y = - x^{2}

(- 2, 1)

Solution steps

y^{2} + 4 x - 20 - 2 y = - x^{2}

Rewrite

y^{2} + 4 x - 20 - 2 y = - x^{2}

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 1), r = 5

And the center is:

(- 2, 1)

(x - 1)^{2} + (y - 3)^{2} = 16

(x + 3)^{2} + (y - 3)^{2} = 8

x^{2} + y^{2} \leq 1

x^{2} + y^{2} - 4 x - 8 y - 16 = 0

y^{2} + x^{2} \leq 4