What is the radius 4x^2+4y^2+8x-y=0 ?

The radius 4x^2+4y^2+8x-y=0 is r=(sqrt(65))/8

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radius 4x^2+4y^2+8x-y=0

radius

4 x^{2} + 4 y^{2} + 8 x - y = 0

r = \frac{65}{8}

Solution steps

4 x^{2} + 4 y^{2} + 8 x - y = 0

Rewrite

4 x^{2} + 4 y^{2} + 8 x - y = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{1}{8})^{2} = (\frac{65}{8})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{1}{8}), r = \frac{65}{8}

And the radius is:

r = \frac{65}{8}

(x - 2)^{2} + (y - 3)^{2} = 64

x^{2} + (y + 2)^{2} = (20^{2})

x^{2} + y^{2} - 10 x - 8 y + 32 = 0

x^{2} + y^{2} + 6 x + 8 y + 19 = 0

x^{2} + 4 x + y^{2} - 6 y + 9 = 0

What is the radius 4x^2+4y^2+8x-y=0 ?
The radius 4x^2+4y^2+8x-y=0 is r=(sqrt(65))/8