Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 21

Answer

$x^{2}+(y-2)^{2}=4$
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Work Step by Step

The circle with center $C(h,k)$ and radus $a$ has the general equation $( x-h)^{2}+(y-k)^{2}=a^{2}.$ Here, $\qquad $ $( x-0)^{2}+(y-2)^{2}=4.$ $x^{2}+(y-2)^{2}=4$ x-intercepts (y=0): $x^{2}+(0-2)^{2}=4$ $x^{2}+4=4$ $x^{2}=0$ $ x=0\qquad$... $\quad(0,0)$ y-intercepts (x=0): $0+(y-2)^{2}=4$ $y-2=\pm 2$ $y=2\pm 2$ $y=0,\quad y=4\qquad (0,0),(0,4)$
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