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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 18 x + 12 y + 81 = 0

x^{2} + y^{2} - 78 y - 1843 = 0

x^{2} + y^{2} - 14 x + 14 y = 132

(x^{2} + 10 x + 25) + (y^{2} - 2 y + 1) = 9

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

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