College Algebra (11th Edition)

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

Chapter 2 - Section 2.2 - Circles - 2.2 Exercises - Page 187: 45

Answer

$(x-3)^2+(y-2)^2=4$

Work Step by Step

A circle has the equation: $(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$ We are told that our circle has center $(3,2)$ (so $h=3$,$k=2$) and is tangent to the $x$-axis. If it is tangent to the $x$ axis, it must touch at $(3,0)$ (2 units below the center). Thus the radius must be $2$. Hence the equation is: $(x-3)^2+(y-2)^2=2^2$ $(x-3)^2+(y-2)^2=4$
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