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: 22

Answer

$(x+1)^{2}+(y-5)^{2}=10$ .
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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 $ $( x-(-1))^{2}+(y-5)^{2}=10$ $(x+1)^{2}+(y-5)^{2}=10$ x-intercepts (y=0): $(x+1)^{2}+(0-5)^{2}=10$ $(x+1)^{2}+25=-10$ $(x+1)^{2}=-15\qquad$... $\quad$ none y-intercepts (x=0): $(0+1)^{2}+(y-5)^{2}=10$ $(y-5)^{2}=10-1$ $y-5=\pm 3$ $y=5\pm 3$ $y=2,\quad y=8\qquad (0,8),(0,2)$
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