College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding: 39

Answer

$(x-2)^2+(y-3)^2=9$

Work Step by Step

RECALL: The standard equation of a circle whose center is at the point $(h, k)$ is: $(x-h)^2+(y-k)^2=r^2$ where $r$ = radius With its center at $(2, 3)$, the tentative equation of the given circle is: $(x-2)^2+(y-3)^2=r^2$ Since the circle is tangent to the $x$-axis, the point on the x-axis that is directly below the center is a point on the circle. This point is $(2, 0)$. Note that the distance from the center $(2, 3)$ to the point on the circle $(2, 0)$ is 3 units. Therefore, the standard form of the equation of the given circle is: $(x-2)^2+(y-3)^2=3^2 \\(x-2)^2+(y-3)^2=9$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.