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

The radius x^2+y^2-8x+10y+8=0 is r=sqrt(33)

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

radius

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

r = 33

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 5))^{2} = (33)^{2}

Therefore the circle properties are:

(a, b) = (4, - 5), r = 33

And the radius is:

r = 33

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

(x + 6)^{2} + (y + 5)^{2} = 8

(x - 4)^{2} + (y + 5)^{2} = 49

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

x^{2} + y^{2} - 4 x - 7 = 0

What is the radius x^2+y^2-8x+10y+8=0 ?
The radius x^2+y^2-8x+10y+8=0 is r=sqrt(33)