What is x^2+y^2-x-6y+1/4 =0 ?

The solution to x^2+y^2-x-6y+1/4 =0 is Circle with (a,b)=(1/2 ,3),r=3

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x^2+y^2-x-6y+1/4 =0

x^{2} + y^{2} - x - 6 y + \frac{1}{4} = 0

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

Solution steps

x^{2} + y^{2} - x - 6 y + \frac{1}{4} = 0

Rewrite

x^{2} + y^{2} - x - 6 y + \frac{1}{4} = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 6 x - 8 y + 21 = 0

2 5^{2} = a^{2} + b^{2}

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

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

(x - 7)^{2} + (y - 4)^{2} = 25

What is x^2+y^2-x-6y+1/4 =0 ?
The solution to x^2+y^2-x-6y+1/4 =0 is Circle with (a,b)=(1/2 ,3),r=3