What is the radius x^2-4x+12y+y^2+14=0 ?

The radius x^2-4x+12y+y^2+14=0 is r=sqrt(26)

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radius x^2-4x+12y+y^2+14=0

radius

x^{2} - 4 x + 12 y + y^{2} + 14 = 0

r = 26

Solution steps

x^{2} - 4 x + 12 y + y^{2} + 14 = 0

Rewrite

x^{2} - 4 x + 12 y + y^{2} + 14 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 6))^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (2, - 6), r = 26

And the radius is:

r = 26

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

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

x^{2} + 10 x - 14 y + y^{2} = 26

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

2 x^{2} + 2 y^{2} + 2 x - 6 y - 7 = 0

What is the radius x^2-4x+12y+y^2+14=0 ?
The radius x^2-4x+12y+y^2+14=0 is r=sqrt(26)