What is the radius (x-6)^2+(y-4)^2-64=0 ?

The radius (x-6)^2+(y-4)^2-64=0 is r=8

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radius (x-6)^2+(y-4)^2-64=0

radius

(x - 6)^{2} + (y - 4)^{2} - 64 = 0

r = 8

Solution steps

(x - 6)^{2} + (y - 4)^{2} - 64 = 0

Rewrite

(x - 6)^{2} + (y - 4)^{2} - 64 = 0

in the form of the standard circle equation

(x - 6)^{2} + (y - 4)^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (6, 4), r = 8

And the radius is:

r = 8

x^{2} + y^{2} + 12 x - 4 y - 40 = 0

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

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

4 x^{2} + 4 y^{2} = 25

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

What is the radius (x-6)^2+(y-4)^2-64=0 ?
The radius (x-6)^2+(y-4)^2-64=0 is r=8