College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Review Exercises - Page 276: 14



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Center-Radius Form to find the equation of the circle with the following characteristics: \begin{array}{l}\require{cancel} \text{center } (5,6) ,\text{ passing through } (4,9) .\end{array} $\bf{\text{Solution Details:}}$ With the given center, then $h= 5 $ and $k= 6 .$ Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ then the equation of the circle is \begin{array}{l}\require{cancel} (x-5)^2+(y-6)^2=r^2 \text{ (*)} .\end{array} Since the circle passes through the point $ (4,9) ,$ substituting the coordinates in the equation above results to \begin{array}{l}\require{cancel} (4-5)^2+(9-6)^2=r^2 \\\\ (-1)^2+(3)^2=r^2 \\\\ 1+9=r^2 \\\\ 10=r^2 .\end{array} Substituting $r^2$ in (*), the equation of the circle is \begin{array}{l}\require{cancel} (x-5)^2+(y-6)^2=10 .\end{array}
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