What is the radius 9x^2+54x+9y^2-6y+28=0 ?

The radius 9x^2+54x+9y^2-6y+28=0 is r=sqrt(6)

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radius 9x^2+54x+9y^2-6y+28=0

radius

9 x^{2} + 54 x + 9 y^{2} - 6 y + 28 = 0

r = 6

Solution steps

9 x^{2} + 54 x + 9 y^{2} - 6 y + 28 = 0

Rewrite

9 x^{2} + 54 x + 9 y^{2} - 6 y + 28 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 6

x^{2} + y^{2} + 6 x + 5 y = - \frac{45}{4}

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(x - 1)^{2} + (y + 4)^{2} = 16

x^{2} + y^{2} + 2 x - 4 y - 15 = 0

x^{2} + y^{2} - 2 x - 6 y = 71

What is the radius 9x^2+54x+9y^2-6y+28=0 ?
The radius 9x^2+54x+9y^2-6y+28=0 is r=sqrt(6)