What is the center 10x^2+10y^2-10x-y-137=0 ?

The center 10x^2+10y^2-10x-y-137=0 is (1/2 , 1/20)

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center 10x^2+10y^2-10x-y-137=0

center

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

(\frac{1}{2}, \frac{1}{20})

Solution steps

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

Rewrite

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - \frac{1}{20})^{2} = (\frac{5581}{20})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, \frac{1}{20}), r = \frac{5581}{20}

And the center is:

(\frac{1}{2}, \frac{1}{20})

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

x^{2} + y^{2} + 14 x - 16 + 64 = 0

x^{2} + 8 x + y^{2} + 4 y = 4

What is the center 10x^2+10y^2-10x-y-137=0 ?
The center 10x^2+10y^2-10x-y-137=0 is (1/2 , 1/20)