## Precalculus (6th Edition) Blitzer

The given equation does not represent a circle. Its graph is a point $: \text{ }(3, 5)$.
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.