What is the center 12x^2+12y^2-18x+4y-5=0 ?

The center 12x^2+12y^2-18x+4y-5=0 is (3/4 ,-1/6)

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center 12x^2+12y^2-18x+4y-5=0

center

12 x^{2} + 12 y^{2} - 18 x + 4 y - 5 = 0

(\frac{3}{4}, - \frac{1}{6})

Solution steps

12 x^{2} + 12 y^{2} - 18 x + 4 y - 5 = 0

Rewrite

12 x^{2} + 12 y^{2} - 18 x + 4 y - 5 = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{1}{6}))^{2} = (\frac{145}{12})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{1}{6}), r = \frac{145}{12}

And the center is:

(\frac{3}{4}, - \frac{1}{6})

(x - 3)^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} - 10 x + 9 y + 1 x = 0

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

3 (x - 5)^{2} + 3 (y + 9)^{2} = 48

x^{2} + y^{2} + 8 y = 32

What is the center 12x^2+12y^2-18x+4y-5=0 ?
The center 12x^2+12y^2-18x+4y-5=0 is (3/4 ,-1/6)