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

The radius 10x^2+10y^2-10x-y-137=0 is r=(sqrt(5581))/(20)

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

radius

r = \frac{5581}{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 radius is:

r = \frac{5581}{20}

What is the radius 10x^2+10y^2-10x-y-137=0 ?
The radius 10x^2+10y^2-10x-y-137=0 is r=(sqrt(5581))/(20)