Answer
The equation is the point of $(0;0)$
Work Step by Step
The equation of a circle with the center of $(0;0)$ and the radius of $r$ is:
$x^2+y^2=r^2$
As $r^2=0$ the radius of the circle equals to 0.
This means that the circle actually is only a point.
The equation is the point of $(0;0)$.
(Another approach can be:
The sum of two squares can only be 0 if both of the terms equal to 0.
This means, that x and y can only be 0, which means the point of $(0;0)$)
