What is the radius x^2+y^2+2x-8y-19=0 ?

The radius x^2+y^2+2x-8y-19=0 is r=6

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radius x^2+y^2+2x-8y-19=0

radius

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

r = 6

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 4)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 1, 4), r = 6

And the radius is:

r = 6

x^{2} + y^{2} - 8 x - 13 y - 32 = 0

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

(x - 9)^{2} + y^{2} - 32 = 0

(x - 3)^{2} = 8 - (y + 3)^{2}

x^{2} + y^{2} + 10 y + 15 = 0