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: 63

Answer

$(x+3)^2+(y+2)^2=1$

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 $(-3, -2)$. The point on the circle that is directly to the left of the center is $(-4, -2)$. This point is 1 unit away from the center. This means that the radius is 1 unit. Since the center is at (-3, -2), we know that h=-3 and k = -2. The radius is 1 unit, so r = 1. Therefore, the equation of the given circle is: $[(x-(-3)]^2 + [(y-(-2)]^2=1^2 \\(x+3)^2+(y+2)^2=1$
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