What is the radius x^2+y^2-x+6y+9=0 ?

The radius x^2+y^2-x+6y+9=0 is r= 1/2

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radius x^2+y^2-x+6y+9=0

radius

x^{2} + y^{2} - x + 6 y + 9 = 0

r = \frac{1}{2}

Solution steps

x^{2} + y^{2} - x + 6 y + 9 = 0

Rewrite

x^{2} + y^{2} - x + 6 y + 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 3), r = \frac{1}{2}

And the radius is:

r = \frac{1}{2}

(x - 4)^{2} + (y + 12)^{2} = 1 7^{2}

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

x^{2} + y^{2} - 6 x - 10 y + 9 = 0

x^{2} + y^{2} + 8 x - 14 y + 1 = 0

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0