What is the radius x^2+y^2-4x-7=0 ?

The radius x^2+y^2-4x-7=0 is r=sqrt(11)

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radius x^2+y^2-4x-7=0

radius

x^{2} + y^{2} - 4 x - 7 = 0

r = 11

Solution steps

x^{2} + y^{2} - 4 x - 7 = 0

Rewrite

x^{2} + y^{2} - 4 x - 7 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = 11

And the radius is:

r = 11

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

x^{2} + y^{2} - 4 x + 10 y + 20 = 0

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

(x - 3)^{2} + (y + 8)^{2} = 35

(y + 8)^{2} + (x + 2)^{2} = 72