What is the radius x^2+y^2-8x+4y+16=0 ?

The radius x^2+y^2-8x+4y+16=0 is r=2

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radius x^2+y^2-8x+4y+16=0

radius

x^{2} + y^{2} - 8 x + 4 y + 16 = 0

r = 2

Solution steps

x^{2} + y^{2} - 8 x + 4 y + 16 = 0

Rewrite

x^{2} + y^{2} - 8 x + 4 y + 16 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

74 = (24 - x)^{2} + (22 - y)^{2}

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

(x + 4)^{2} + (y + 3)^{2} = 16

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

x^{2} + y^{2} + 6 x - 2 y - 54 = 0