What is x^2+x+y^2+8/3 y-1223/36 =0 ?

The solution to x^2+x+y^2+8/3 y-1223/36 =0 is Circle with (a,b)=(-1/2 ,-4/3),r=6

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x^2+x+y^2+8/3 y-1223/36 =0

x^{2} + x + y^{2} + \frac{8}{3} y - \frac{1223}{36} = 0

(a, b) = (- \frac{1}{2}, - \frac{4}{3}), r = 6

Solution steps

x^{2} + x + y^{2} + \frac{8}{3} y - \frac{1223}{36} = 0

Rewrite

x^{2} + x + y^{2} + \frac{8}{3} y - \frac{1223}{36} = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - (- \frac{4}{3}))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - \frac{4}{3}), r = 6

(x - 4)^{2} + (y - 2)^{2} = 36

x^{2} + y^{2} + 2 x - 10 y + 1 = 0

(x - 2)^{2} + (y + 8)^{2} = 144

(x - 0)^{2} + (y - 5)^{2} = 4^{2}

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

What is x^2+x+y^2+8/3 y-1223/36 =0 ?
The solution to x^2+x+y^2+8/3 y-1223/36 =0 is Circle with (a,b)=(-1/2 ,-4/3),r=6