Answer
The given equation does not represent a circle.
Its graph is a point $: \text{ }(3, 5)$.
Work Step by Step
RECALL:
The standard form of the equation of a circle is given as
$(x-h)^2+(y-k)^2 = r^2$
where
$(h, k)$ is the center
$r$ is the length of the radius (distance of each point on the circle from the center)
This means that in the given equation, the center is at $(3, 5)$ and the radius is 0.
The radius of a circle cannot be zero as it would mean that the distance of the points on the circle from the center is 0.
Having $r=0$ means that the graph is actually a POINT (the center itself), not a circle.
