What is the radius x^2+y^2+8x-2y+15=0 ?

The radius x^2+y^2+8x-2y+15=0 is r=sqrt(2)

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radius x^2+y^2+8x-2y+15=0

radius

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

r = 2

Solution steps

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

Rewrite

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 1), r = 2

And the radius is:

r = 2

(x + 5)^{2} + y^{2} = 4

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

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

x^{2} + y^{2} - 4 x + 8 y + 19 = 0

x^{2} + y^{2} = 125

What is the radius x^2+y^2+8x-2y+15=0 ?
The radius x^2+y^2+8x-2y+15=0 is r=sqrt(2)