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

(\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})

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)