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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

4 x^{2} + 4 y^{2} = 128

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

x^{2} + y^{2} \geq 5

x^{2} + y^{2} - 6 x + 6 y + 14 = 0

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

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