What is the radius x^2+y^2-8x-8y+28=0 ?

The radius x^2+y^2-8x-8y+28=0 is r=2

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radius x^2+y^2-8x-8y+28=0

radius

r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 4), r = 2

And the radius is:

r = 2