What is the center 2x^2+2y^2+20x+16y+9=0 ?

The center 2x^2+2y^2+20x+16y+9=0 is (-5,-4)

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center 2x^2+2y^2+20x+16y+9=0

center

2 x^{2} + 2 y^{2} + 20 x + 16 y + 9 = 0

(- 5, - 4)

Solution steps

2 x^{2} + 2 y^{2} + 20 x + 16 y + 9 = 0

Rewrite

2 x^{2} + 2 y^{2} + 20 x + 16 y + 9 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 4))^{2} = (\frac{73}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 4), r = \frac{73}{2}

And the center is:

(- 5, - 4)

x^{2} + (y + 20)^{2} = 100

(x - 1)^{2} + (y + 2)^{2} = 25

(x + 3)^{2} + (y + 5)^{2} = 4

(x + 3)^{2} + (y + 5)^{2} = 9

x^{2} + y^{2} - 10 x + 9 y + 1 = 0

What is the center 2x^2+2y^2+20x+16y+9=0 ?
The center 2x^2+2y^2+20x+16y+9=0 is (-5,-4)