The given equation is a circle with center = (0, 0) and radius = 4.
We compare the given equation with the standard form of a circle[tex](x-h)^2+(y-k)^2=r^2[/tex]
So, h = 0 , k = 0, [tex]r^2=16[/tex] or r = 4.
So the given equation is a circle with center = (0, 0) and radius = 4.
The graph is attached.
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