What is the radius (x+12)^2+(y-4)^2-30=0 ?

The radius (x+12)^2+(y-4)^2-30=0 is r=sqrt(30)

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radius (x+12)^2+(y-4)^2-30=0

radius

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

r = 30

Solution steps

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

Rewrite

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

in the form of the standard circle equation

(x - (- 12))^{2} + (y - 4)^{2} = (30)^{2}

Therefore the circle properties are:

(a, b) = (- 12, 4), r = 30

And the radius is:

r = 30

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

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

x^{2} - 2 x + y^{2} - 2 y = 23

(x - 1)^{2} + (y + 3)^{2} = 4

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

What is the radius (x+12)^2+(y-4)^2-30=0 ?
The radius (x+12)^2+(y-4)^2-30=0 is r=sqrt(30)