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

The radius x^2+y^2+6x-4y+3=0 is r=sqrt(10)

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

radius

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

r = 10

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 10

And the radius is:

r = 10

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

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

x^{2} + y^{2} = 3

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

x^{2} + y^{2} - 12 x + 20 y + 15 = 0

What is the radius x^2+y^2+6x-4y+3=0 ?
The radius x^2+y^2+6x-4y+3=0 is r=sqrt(10)