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

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

(- \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)

(x - (- 5))^{2} + (y - (7))^{2} = (11)^{2}

(x + 4)^{2} + (y + 1)^{2} = 9

(x + 3)^{2} + (y - 2)^{2} = 25

(x + 3)^{2} + (y - 2)^{2} = 13

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

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)