What is the center 9x^2+9y^2+42x-54y+49=0 ?

The center 9x^2+9y^2+42x-54y+49=0 is (-7/3 ,3)

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center 9x^2+9y^2+42x-54y+49=0

center

(- \frac{7}{3}, 3)

Solution steps

9 x^{2} + 9 y^{2} + 42 x - 54 y + 49 = 0

Rewrite

9 x^{2} + 9 y^{2} + 42 x - 54 y + 49 = 0

in the form of the standard circle equation

(x - (- \frac{7}{3}))^{2} + (y - 3)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{3}, 3), r = 3

And the center is:

(- \frac{7}{3}, 3)

What is the center 9x^2+9y^2+42x-54y+49=0 ?
The center 9x^2+9y^2+42x-54y+49=0 is (-7/3 ,3)