What is the radius x^2+4x+10y+y^2-2=0 ?

The radius x^2+4x+10y+y^2-2=0 is r=sqrt(31)

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radius x^2+4x+10y+y^2-2=0

radius

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

r = 31

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 5))^{2} = (31)^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = 31

And the radius is:

r = 31

4 x^{2} + 4 y^{2} = 25

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

x^{2} + y^{2} + 10 x - 4 y - 7 = 0

x^{2} + y^{2} = 20

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

What is the radius x^2+4x+10y+y^2-2=0 ?
The radius x^2+4x+10y+y^2-2=0 is r=sqrt(31)