What is the center 36x^2+36y^2-12x-27y-8=0 ?

The center 36x^2+36y^2-12x-27y-8=0 is (1/6 , 3/8)

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center 36x^2+36y^2-12x-27y-8=0

center

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

(\frac{1}{6}, \frac{3}{8})

Solution steps

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

Rewrite

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

in the form of the standard circle equation

(x - \frac{1}{6})^{2} + (y - \frac{3}{8})^{2} = (\frac{5}{8})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{6}, \frac{3}{8}), r = \frac{5}{8}

And the center is:

(\frac{1}{6}, \frac{3}{8})

(x - 3)^{2} + (y + 1)^{2} = 16

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

x^{2} + y^{2} + 2 x - 4 y - 31 = 0

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

(x - 1)^{2} + (y + 3)^{2} = 4

What is the center 36x^2+36y^2-12x-27y-8=0 ?
The center 36x^2+36y^2-12x-27y-8=0 is (1/6 , 3/8)