College Algebra (11th Edition)

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

Chapter 2 - Review Exercises - Page 276: 13

Answer

$x^2+(y-3)^2=13$

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 } (0,3) ,\text{ passing through } (-2,6) .\end{array} $\bf{\text{Solution Details:}}$ With the given center, then $h= 0 $ and $k= 3 .$ 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-0)^2+(y-3)^2=r^2 \\\\ x^2+(y-3)^2=r^2 \text{ (*)} .\end{array} Since the circle passes through the point $ (-2,6) ,$ substituting the coordinates in the equation above results to \begin{array}{l}\require{cancel} (-2)^2+(6-3)^2=r^2 \\\\ (-2)^2+(3)^2=r^2 \\\\ 4+9=r^2 \\\\ 13=r^2 .\end{array} Substituting $r^2$ in (*), the equation of the circle is \begin{array}{l}\require{cancel} x^2+(y-3)^2=13 .\end{array}
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