What is the radius x^2+y^2-4x+8y+5=0 ?

The radius x^2+y^2-4x+8y+5=0 is r=sqrt(15)

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radius x^2+y^2-4x+8y+5=0

radius

x^{2} + y^{2} - 4 x + 8 y + 5 = 0

r = 15

Solution steps

x^{2} + y^{2} - 4 x + 8 y + 5 = 0

Rewrite

x^{2} + y^{2} - 4 x + 8 y + 5 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = (15)^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 15

And the radius is:

r = 15

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

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

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

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

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

What is the radius x^2+y^2-4x+8y+5=0 ?
The radius x^2+y^2-4x+8y+5=0 is r=sqrt(15)