Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 854: 51

Answer

Center of circle is $\left( 0,0 \right)$ and radius is $r=2\sqrt{2}$

Work Step by Step

Standard equation of the circle is: ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation - 1) And the equation of the circle is ${{x}^{2}}+{{y}^{2}}=8$ (equation - 2) Now compare both the equations. $\begin{align} & {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=8 \\ & {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}} \\ \end{align}$ Center coordinate of the circle is $\left( h=0,k=0 \right)$. And radius of the circle is $r=2\sqrt{2}$. To graph, we plot the points $\left( 0,2.828 \right)$, $\left( 0,-2.828 \right)$, $\left( -2.828,0 \right)$, and $\left( 2.828,0 \right)$ which are, respectively, $2\sqrt{2}$ units above, below, left and right of $\left( 0,0 \right)$.
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