What is x^2+y^2-3y-4=0 ?

The solution to x^2+y^2-3y-4=0 is Circle with (a,b)=(0, 3/2),r= 5/2

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x^2+y^2-3y-4=0

x^{2} + y^{2} - 3 y - 4 = 0

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

Solution steps

x^{2} + y^{2} - 3 y - 4 = 0

Rewrite

x^{2} + y^{2} - 3 y - 4 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 9 x - 4 y - 2 = 0

3 x^{2} + 30 x + 3 y^{2} + 6 y + 51 = 0

x^{2} + y^{2} = \frac{x}{9} - \frac{y}{7} - \frac{1}{324}

x^{2} + y^{2} - 2 x + 10 y + 22 = 0

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

What is x^2+y^2-3y-4=0 ?
The solution to x^2+y^2-3y-4=0 is Circle with (a,b)=(0, 3/2),r= 5/2