What is the radius x^2+6x+y^2-3y=-1/4 ?

The radius x^2+6x+y^2-3y=-1/4 is r=sqrt(11)

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radius x^2+6x+y^2-3y=-1/4

radius

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

r = 11

Solution steps

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

Rewrite

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{3}{2})^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{3}{2}), r = 11

And the radius is:

r = 11

x^{2} + y^{2} = 50

3 x^{2} + 3 y^{2} - 17 x = 0

x^{2} + y^{2} - 6 x + 5 = 0

2 x^{2} + 2 y^{2} - 9 x - 18 y - 80 = 0

5 x^{2} + 5 y^{2} + 10 x + 2 y + 3 = 0

What is the radius x^2+6x+y^2-3y=-1/4 ?
The radius x^2+6x+y^2-3y=-1/4 is r=sqrt(11)