What is x^2+(y+1)^2-4=0 ?

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

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x^2+(y+1)^2-4=0

x^{2} + (y + 1)^{2} - 4 = 0

(a, b) = (0, - 1), r = 2

Solution steps

x^{2} + (y + 1)^{2} - 4 = 0

Rewrite

x^{2} + (y + 1)^{2} - 4 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = 2

(x - \frac{33}{4})^{2} + y^{2} = 25

0 > (x - 1)^{2} + y^{2} - 1

x^{(2)} + y^{(2)} + 4 x - 4 y - 1 = 0

x, x^{2} = 8 - r (x)^{2}

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

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