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

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

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

center

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

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

Solution steps

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

Rewrite

9 x^{2} - 42 x + 9 y^{2} + 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})

x^{2} + y^{2} - 8 y + 7 = 0

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

(y + 8)^{2} + (x + 2)^{2} = 72

(x - 3)^{2} + (y - (- 2))^{2} = 35

(x + 5)^{2} + (y + 6)^{2} = \frac{2}{3}

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