Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set: 80

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.
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