What is the center 9x^2+36x+9y^2+72y+155=0 ?

The center 9x^2+36x+9y^2+72y+155=0 is (-2,-4)

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center 9x^2+36x+9y^2+72y+155=0

center

(- 2, - 4)

Solution steps

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

Rewrite

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = \frac{5}{3}

And the center is:

(- 2, - 4)

What is the center 9x^2+36x+9y^2+72y+155=0 ?
The center 9x^2+36x+9y^2+72y+155=0 is (-2,-4)