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: 23

Answer

$(x+\sqrt{3})^{2}+(y+2)^{2}=4$ .
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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 $ $( x-(-\sqrt{3}))^{2}+(y-(-2))^{2}=2^{2}$ $(x+\sqrt{3})^{2}+(y+2)^{2}=4$ x-intercepts (y=0): $(x+\sqrt{3})^{2}+(0+2)^{2}=4$ $(x+\sqrt{3})^{2}+4=4$ $(x+\sqrt{3})^{2}=0$ $ x=-\sqrt{3}\qquad$... $\quad (-\sqrt{3},0) $ y-intercepts (x=0): $(0+\sqrt{3})^{2}+(y+2)^{2}=4$ $3+(y+2)^{2}=4$ $(y+2)^{2}=1$ $y+2=\pm 1$ $y=-2\pm 1$ $y=-1,\quad y=-3\qquad (0,-1),(0,-3)$
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