What is the radius 4x^2+4y^2+40x-32y=0 ?

The radius 4x^2+4y^2+40x-32y=0 is r=sqrt(41)

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radius 4x^2+4y^2+40x-32y=0

radius

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

r = 41

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 4)^{2} = (41)^{2}

Therefore the circle properties are:

(a, b) = (- 5, 4), r = 41

And the radius is:

r = 41

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

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

x^{2} + y^{2} - 2 x - 10 y + 17 = 0

(x + 5)^{2} + (y + 1)^{2} = 16

x^{2} + (y + 5)^{2} = 7

What is the radius 4x^2+4y^2+40x-32y=0 ?
The radius 4x^2+4y^2+40x-32y=0 is r=sqrt(41)