University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter Appendices - Section A.3 - Lines and Circles - Exercises - Page AP-15: 21

Answer

$x^{2}+(y-2)^{2}=4$ .

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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