Answer
False
Work Step by Step
The given equation becomes in the form of $(x-h)^2+(y-k)^2=d$ when we complete the square.
When $d \gt 0$; it shows an equation of a circle.
When $d \lt 0$; the equations does not have a solution so there is no graph,
When $d = 0$; the graph will be a point, $(h,k)$
Thus, the given statement is false.
