What is the radius 4x^2+4y^2-8x+4y-11=0 ?

The radius 4x^2+4y^2-8x+4y-11=0 is r=2

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radius 4x^2+4y^2-8x+4y-11=0

radius

4 x^{2} + 4 y^{2} - 8 x + 4 y - 11 = 0

r = 2

Solution steps

4 x^{2} + 4 y^{2} - 8 x + 4 y - 11 = 0

Rewrite

4 x^{2} + 4 y^{2} - 8 x + 4 y - 11 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

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

x^{2} + y^{2} - 9 x + 4 y - 181 = 0

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

What is the radius 4x^2+4y^2-8x+4y-11=0 ?
The radius 4x^2+4y^2-8x+4y-11=0 is r=2