What is the radius x^2+y^2-14x+2y+41=0 ?

The radius x^2+y^2-14x+2y+41=0 is r=3

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radius x^2+y^2-14x+2y+41=0

radius

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

r = 3

Solution steps

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

Rewrite

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 1))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (7, - 1), r = 3

And the radius is:

r = 3

x^{2} + 2 x + y^{2} + 4 y = 20

x^{2} + 10 x + y^{2} - 16 y - 11 = 0

x^{2} - 6 x + y^{2} + 8 y - 24 = 0

x^{2} + 24 x + y^{2} + 10 y = - 160

x^{2} + y^{2} - 6 x = 7