Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - Review Exercises - Page 354: 86a


$x^2+(y+4\sqrt 3)^2=64$

Work Step by Step

$(x-h)^2=4p(y-k)$ or, $(4-0)^2=4p(0-4)$ or, $p=-1$ Now, $x^2=-4(y-4)$ The x-coordinate of the center of the circle is: $4^2 +y^2 =8^2$ or, $y=-\sqrt {8^2-4^2}=-4\sqrt 3$ The equation of a circle is: $(x-h)^2+(y-k)^2=r^2$ or, $x^2+(y+4\sqrt 3)^2=64$
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