What is the center x^2+y^2+2x+4y-20=0 ?

The center x^2+y^2+2x+4y-20=0 is (-1,-2)

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center x^2+y^2+2x+4y-20=0

center

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

(- 1, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 1, - 2)

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

x^{2} + y^{2} = - 4 x + 3 y

(x + 4)^{2} + (y - 10)^{2} = 274

x^{2} + y^{2} - 10 x + 12 y + 25 = 0

y^{2} + x^{2} - 12 x + 18 y + 116 = 0

What is the center x^2+y^2+2x+4y-20=0 ?
The center x^2+y^2+2x+4y-20=0 is (-1,-2)