What is the center 9x^2+9y^2+42x+84y+65=0 ?

The center 9x^2+9y^2+42x+84y+65=0 is (-7/3 ,-14/3)

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center 9x^2+9y^2+42x+84y+65=0

center

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

(- \frac{7}{3}, - \frac{14}{3})

Solution steps

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

Rewrite

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

in the form of the standard circle equation

(x - (- \frac{7}{3}))^{2} + (y - (- \frac{14}{3}))^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{3}, - \frac{14}{3}), r = 25

And the center is:

(- \frac{7}{3}, - \frac{14}{3})

16 x^{2} + 16 y^{2} = 320

x^{2} + y^{2} - 14 x + 12 y = 36

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

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

x^{2} + y^{2} + 4 x + 20 y + 23 = 0

What is the center 9x^2+9y^2+42x+84y+65=0 ?
The center 9x^2+9y^2+42x+84y+65=0 is (-7/3 ,-14/3)