What is the radius 2x^2+2y^2+2x-6y-7=0 ?

The radius 2x^2+2y^2+2x-6y-7=0 is r=sqrt(6)

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radius 2x^2+2y^2+2x-6y-7=0

radius

2 x^{2} + 2 y^{2} + 2 x - 6 y - 7 = 0

r = 6

Solution steps

2 x^{2} + 2 y^{2} + 2 x - 6 y - 7 = 0

Rewrite

2 x^{2} + 2 y^{2} + 2 x - 6 y - 7 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{3}{2}), r = 6

And the radius is:

r = 6

x^{2} + y^{2} + 10 x + 4 y + 4 = 0

3 x^{2} + 3 y^{2} - 2 x + y - 5 = 0

3 x^{2} - 12 x - 18 y + 3 y^{2} = 0

(x - 1)^{2} + (y + 4)^{2} = 9

x^{2} + y^{2} - 3 x + 4 y + 4 = 0

What is the radius 2x^2+2y^2+2x-6y-7=0 ?
The radius 2x^2+2y^2+2x-6y-7=0 is r=sqrt(6)