What is the radius x^2+y^2-4x+6y-3=0 ?

The radius x^2+y^2-4x+6y-3=0 is r=4

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radius x^2+y^2-4x+6y-3=0

radius

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

r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 3), r = 4

And the radius is:

r = 4

x^{2} + y^{2} + 6 x - 2 y - 15 = 0

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

(x + \frac{1}{2})^{2} + (y - \frac{4}{5})^{2} = \frac{9}{4}

(x + 1)^{2} + (y + 3)^{2} = 5

x^{2} + y^{2} - 4 x - 2 y + 1 = 0