Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 37

Answer

$(x - 2)^{2}$ + $(y-4)^{2}$ = 9 $(x - 2)^{2}$ + $(y-4)^{2}$ = 26

Work Step by Step

$(x - 2)^{2}$ + $(y-4)^{2}$ = $r^{2}$ gives a circle centered at $(2,4)$ with radius r, using the general form equation for circles a) plugging in r = 3 gives $(x - 2)^{2}$ + $(y-4)^{2}$ = 9 b) the circle at $(2,4)$ passes through $(1,-1)$ if it has radius equal to the distance between the two points, so r = $\sqrt ((2-1)^{2} + (4 - (-1))^{2})$ = $\sqrt (1+25)$ = $\sqrt 26$ Plugging into the general equation gives $(x - 2)^{2} + (y-4)^{2}$ = 26
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