Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.1 - Distance and Midpoint Formulas; Circles - Exercise Set: 64

Answer

$(x+1)^2+(y-1)^2=16$

Work Step by Step

RECALL: The standard form of the equation of a circle with a center at $(h, k)$ and a radius of $r$ units is: $(x-h)^2+(y-k)^2=r^2$ The given circle has its center at $(-1, 1)$. The point on the circle that is directly to the left of the center is $(-5, 1)$. This point is 4 units away from the center. This means that the radius is 4 units. Since the center is at (-1, 1), we know that h=-1 and k = 1 The radius is 4 units, so r = 4. Therefore, the equation of the given circle is: $[(x-(-1)]^2 + (y-1)^2=4^2 \\(x+1)^2+(y-1)^2=16$
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