What is the radius x^2+y^2+6x+8y+19=0 ?

The radius x^2+y^2+6x+8y+19=0 is r=sqrt(6)

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radius x^2+y^2+6x+8y+19=0

radius

x^{2} + y^{2} + 6 x + 8 y + 19 = 0

r = 6

Solution steps

x^{2} + y^{2} + 6 x + 8 y + 19 = 0

Rewrite

x^{2} + y^{2} + 6 x + 8 y + 19 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - (- 4))^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (- 3, - 4), r = 6

And the radius is:

r = 6

x^{2} + 4 x + y^{2} - 6 y + 9 = 0

x^{2} + y^{2} - 8 x - 8 y + 28 = 0

4 x^{2} + 4 y^{2} + 40 x - 32 y = 0

x^{2} + y^{2} - 4 x + 6 y = 16

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

What is the radius x^2+y^2+6x+8y+19=0 ?
The radius x^2+y^2+6x+8y+19=0 is r=sqrt(6)