Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 280: 66


a. $(4,5)$ b. $\sqrt 2$ c. $(x-4)^2+(y-5)^2=2$

Work Step by Step

a. Given the coordinates of the ends of a diameter: $(3,6)$ and $(5,4)$, we can find the center of the circle as $(\frac{5+3}{2},\frac{4+6}{2})$ or $(4,5)$ b. The radius is half of the diameter; we have $r=\frac{\sqrt {(5-3)^2+(4-6)^2}}{2}=\frac{\sqrt {4+4}}{2}=\sqrt 2$ c. Based on the above results, we can write the standard form of the circle’s equation as $(x-4)^2+(y-5)^2=2$
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