What is the radius x^2-6x+14y-567+y^2=0 ?

The radius x^2-6x+14y-567+y^2=0 is r=25

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radius x^2-6x+14y-567+y^2=0

radius

r = 25

Solution steps

x^{2} - 6 x + 14 y - 567 + y^{2} = 0

Rewrite

x^{2} - 6 x + 14 y - 567 + y^{2} = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 7))^{2} = 2 5^{2}

Therefore the circle properties are:

(a, b) = (3, - 7), r = 25

And the radius is:

r = 25

What is the radius x^2-6x+14y-567+y^2=0 ?
The radius x^2-6x+14y-567+y^2=0 is r=25