What is the radius 4x^2+4y^2-x+2y-11=0 ?

The radius 4x^2+4y^2-x+2y-11=0 is r=(sqrt(181))/8

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radius 4x^2+4y^2-x+2y-11=0

radius

4 x^{2} + 4 y^{2} - x + 2 y - 11 = 0

r = \frac{181}{8}

Solution steps

4 x^{2} + 4 y^{2} - x + 2 y - 11 = 0

Rewrite

4 x^{2} + 4 y^{2} - x + 2 y - 11 = 0

in the form of the standard circle equation

(x - \frac{1}{8})^{2} + (y - (- \frac{1}{4}))^{2} = (\frac{181}{8})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{8}, - \frac{1}{4}), r = \frac{181}{8}

And the radius is:

r = \frac{181}{8}

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

x^{2} + y^{2} + 2 y - 15 = 0

x^{2} + y^{2} = 144

(x - \frac{6}{5})^{2} + (y + \frac{1}{3})^{2} = \frac{49}{9}

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

What is the radius 4x^2+4y^2-x+2y-11=0 ?
The radius 4x^2+4y^2-x+2y-11=0 is r=(sqrt(181))/8