What is x^2+y^2+2/3 x-1/2 y-13/48 =0 ?

The solution to x^2+y^2+2/3 x-1/2 y-13/48 =0 is Circle with (a,b)=(-1/3 , 1/4),r= 2/3

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x^2+y^2+2/3 x-1/2 y-13/48 =0

x^{2} + y^{2} + \frac{2}{3} x - \frac{1}{2} y - \frac{13}{48} = 0

(a, b) = (- \frac{1}{3}, \frac{1}{4}), r = \frac{2}{3}

Solution steps

x^{2} + y^{2} + \frac{2}{3} x - \frac{1}{2} y - \frac{13}{48} = 0

Rewrite

x^{2} + y^{2} + \frac{2}{3} x - \frac{1}{2} y - \frac{13}{48} = 0

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - \frac{1}{4})^{2} = (\frac{2}{3})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, \frac{1}{4}), r = \frac{2}{3}

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x^{2} + y^{2} - 8 x - 12 y + 27 = 0

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What is x^2+y^2+2/3 x-1/2 y-13/48 =0 ?
The solution to x^2+y^2+2/3 x-1/2 y-13/48 =0 is Circle with (a,b)=(-1/3 , 1/4),r= 2/3