What is the radius x^2+y^2+10x+8y-4=0 ?

The radius x^2+y^2+10x+8y-4=0 is r=3sqrt(5)

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radius x^2+y^2+10x+8y-4=0

radius

x^{2} + y^{2} + 10 x + 8 y - 4 = 0

r = 35

Solution steps

x^{2} + y^{2} + 10 x + 8 y - 4 = 0

Rewrite

x^{2} + y^{2} + 10 x + 8 y - 4 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 4))^{2} = (35)^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 4), r = 35

And the radius is:

r = 35

(x - 3)^{2} + (y - 5)^{2} = 24

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

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

(x + 1)^{2} + (y - 5)^{2} = 5^{2}

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

What is the radius x^2+y^2+10x+8y-4=0 ?
The radius x^2+y^2+10x+8y-4=0 is r=3sqrt(5)