College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 6 - Section 6.4 - Summary of the Conic Sections - 6.4 Exercises - Page 618: 1

Answer

circle

Work Step by Step

We can see that both x and y are on the power of two with the same coefficients. This given equation can be easily transformed to the general equation of circle whose center is at the origin and a radius $r$: $x^2+y^2=R^2$ Here, it is: $x^2+y^2=12^2$ Therefore, it is a circle with the center of $(0,0)$, and with the radius of 12.
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