What is the radius x^2+y^2-8x+12=0 ?

The radius x^2+y^2-8x+12=0 is r=2

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radius x^2+y^2-8x+12=0

radius

x^{2} + y^{2} - 8 x + 12 = 0

r = 2

Solution steps

x^{2} + y^{2} - 8 x + 12 = 0

Rewrite

x^{2} + y^{2} - 8 x + 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 0), r = 2

And the radius is:

r = 2

25 = (x + 1)^{2} + (y - 3)^{2}

(x - 2)^{2} + y^{2} = 16

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

(x + 2)^{2} + (y - 5)^{2} = 25