What is x^2+y^2+y-11=0 ?

The solution to x^2+y^2+y-11=0 is Circle with (a,b)=(0,-1/2),r=(3sqrt(5))/2

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x^2+y^2+y-11=0

x^{2} + y^{2} + y - 11 = 0

(a, b) = (0, - \frac{1}{2}), r = \frac{3 5}{2}

Solution steps

x^{2} + y^{2} + y - 11 = 0

Rewrite

x^{2} + y^{2} + y - 11 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{3 5}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{1}{2}), r = \frac{3 5}{2}

x^{2} - 154 x + y^{2} - 160 y = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{16} = 10

(x - 4)^{(2)} + (y - 5)^{(2)} = 49

9 - x^{2} > y^{2}

x^{2} + y^{2} - 6 x - 4 y - 18 = 0

What is x^2+y^2+y-11=0 ?
The solution to x^2+y^2+y-11=0 is Circle with (a,b)=(0,-1/2),r=(3sqrt(5))/2