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)} - 6 x + 2 y = - 1

x^{2} + y^{2} = 324

(x - 3)^{2} + (y + 4)^{2} = 64

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

x^{2} + y^{2} + 2 x - 3 y - 15 = 0

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