What is the radius x^2+y^2+8x-2y-17=0 ?

The radius x^2+y^2+8x-2y-17=0 is r=sqrt(34)

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radius x^2+y^2+8x-2y-17=0

radius

x^{2} + y^{2} + 8 x - 2 y - 17 = 0

r = 34

Solution steps

x^{2} + y^{2} + 8 x - 2 y - 17 = 0

Rewrite

x^{2} + y^{2} + 8 x - 2 y - 17 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 1)^{2} = (34)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 1), r = 34

And the radius is:

r = 34

x^{2} + y^{2} - 6 x + 2 y - 15 = 0

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

(y + 4)^{2} + (x + 3)^{2} = 25

(x - 1)^{2} + y^{2} = 9

(x - 10)^{2} + y^{2} = \frac{1}{4}

What is the radius x^2+y^2+8x-2y-17=0 ?
The radius x^2+y^2+8x-2y-17=0 is r=sqrt(34)