What is the radius x^2+y^2+12x-4y-40=0 ?

The radius x^2+y^2+12x-4y-40=0 is r=4sqrt(5)

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radius x^2+y^2+12x-4y-40=0

radius

x^{2} + y^{2} + 12 x - 4 y - 40 = 0

r = 45

Solution steps

x^{2} + y^{2} + 12 x - 4 y - 40 = 0

Rewrite

x^{2} + y^{2} + 12 x - 4 y - 40 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - 2)^{2} = (45)^{2}

Therefore the circle properties are:

(a, b) = (- 6, 2), r = 45

And the radius is:

r = 45

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

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

4 x^{2} + 4 y^{2} = 25

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

x^{2} + y^{2} + 10 x - 4 y - 7 = 0

What is the radius x^2+y^2+12x-4y-40=0 ?
The radius x^2+y^2+12x-4y-40=0 is r=4sqrt(5)