What is the radius x^2+y^2+4x-6y-12=0 ?

The radius x^2+y^2+4x-6y-12=0 is r=5

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radius x^2+y^2+4x-6y-12=0

radius

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

r = 5

Solution steps

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

Rewrite

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 5

And the radius is:

r = 5

x^{2} - 8 x + y^{2} + 6 y = - 9

(x + 1)^{2} + (y - 5)^{2} = 5^{2}

x^{2} + y^{2} - 13 = 0

(x + \frac{7}{2})^{2} + (y - \frac{1}{2})^{2} = 5

4 x^{2} + 4 y^{2} - 3 x + 5 y - 5 = 0