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

The radius x^2+2x+y^2+4y-6=0 is r=sqrt(11)

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

radius

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

r = 11

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 2))^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 11

And the radius is:

r = 11

x^{2} + y^{2} - 10 x - 10 y = - 1

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

x^{2} + y^{2} - 4 x + 16 y + 4 = 0

(x - 2)^{2} + (y + 3)^{2} = 16

(x - 1)^{2} + (y - 3)^{2} = 49

What is the radius x^2+2x+y^2+4y-6=0 ?
The radius x^2+2x+y^2+4y-6=0 is r=sqrt(11)