What is the radius x^2+y^2-4x-8y+18=0 ?

The radius x^2+y^2-4x-8y+18=0 is r=sqrt(2)

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radius x^2+y^2-4x-8y+18=0

radius

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

r = 2

Solution steps

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

Rewrite

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

(x + 2)^{2} + (y - 5)^{2} = 25

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

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

What is the radius x^2+y^2-4x-8y+18=0 ?
The radius x^2+y^2-4x-8y+18=0 is r=sqrt(2)