Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises - Page 200: 9


one point only

Work Step by Step

RECALL: The circle $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and a radius of $r$ units. The given equation has its "center" at $(0, 0)$ and has $r=0$. Since the radius is zero, then the graph must only be a point - $(0, 0)$, Thus, the given equation has only one point on its graph.
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