What is (x+4)^2+(y+4)^2=16 ?

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

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(x+4)^2+(y+4)^2=16

(x + 4)^{2} + (y + 4)^{2} = 16

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

Solution steps

(x + 4)^{2} + (y + 4)^{2} = 16

Rewrite

(x + 4)^{2} + (y + 4)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 8 x + 4 y - 5 = 0

(x + 3)^{2} + (y - 2)^{2} = 9

9 x^{2} + 9 y^{2} = 36

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

x^{2} + (y + 3)^{2} = 9

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