What is x^2+(y+3)^2=16 ?

The solution to x^2+(y+3)^2=16 is Circle with (a,b)=(0,-3),r=4

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x^2+(y+3)^2=16

x^{2} + (y + 3)^{2} = 16

(a, b) = (0, - 3), r = 4

Solution steps

x^{2} + (y + 3)^{2} = 16

Rewrite

x^{2} + (y + 3)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, - 3), r = 4

137 + 6 y = - y^{2} - x^{2} - 24 x

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

x^{2} + y^{2} - 8 x - 6 y - 11 = 0

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

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

What is x^2+(y+3)^2=16 ?
The solution to x^2+(y+3)^2=16 is Circle with (a,b)=(0,-3),r=4