What is the radius 4x^2+4y^2-4x+12y-6=0 ?

The radius 4x^2+4y^2-4x+12y-6=0 is r=2

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radius 4x^2+4y^2-4x+12y-6=0

radius

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

r = 2

Solution steps

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

Rewrite

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{3}{2}))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = 2

And the radius is:

r = 2

x^{2} + y^{2} = 30

x^{2} + y^{2} = 34

x^{2} + y^{2} - 10 x + 6 y + 18 = 0

2 x^{2} + 2 y^{2} + 10 x = 2 y + 7

x^{2} + y^{2} - 16 y + 39 = 0

What is the radius 4x^2+4y^2-4x+12y-6=0 ?
The radius 4x^2+4y^2-4x+12y-6=0 is r=2