What is x^2+y^2+3y=12 ?

The solution to x^2+y^2+3y=12 is Circle with (a,b)=(0,-3/2),r=(sqrt(57))/2

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x^2+y^2+3y=12

x^{2} + y^{2} + 3 y = 12

(a, b) = (0, - \frac{3}{2}), r = \frac{57}{2}

Solution steps

x^{2} + y^{2} + 3 y = 12

Rewrite

x^{2} + y^{2} + 3 y = 12

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{57}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{3}{2}), r = \frac{57}{2}

(x - 3)^{2} + (y - 1)^{2} = 13

x^{2} + (y - 5)^{2} = 6

x^{2} + y^{2} - 8 x + 5 y - 3 = 0

4 x^{2} + 4 y^{2} - 10 x - 12 y = 20

16 x^{2} + 96 x + 16 y^{2} - 40 y - 315 + 315 = 0 + 315

What is x^2+y^2+3y=12 ?
The solution to x^2+y^2+3y=12 is Circle with (a,b)=(0,-3/2),r=(sqrt(57))/2