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 855: 57

Answer

Center of circle is $\left( 4,-1 \right)$ and radius is $r=2$.

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 equation of circle is ${{x}^{2}}+{{y}^{2}}-8x+2y=-13$ (equation - 2) Now add $16$ and $1$ on both the sides of (equation - 2) to complete the square twice. $\begin{align} & {{x}^{2}}+{{y}^{2}}-8x+2y+16+1=-13+16+1 \\ & \left( {{x}^{2}}-8x+16 \right)+\left( {{y}^{2}}+2y+1 \right)=4 \\ & {{\left( x-4 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( 2 \right)}^{2}} \end{align}$ Now compare the (equation – 1) with the equation${{\left( x-4 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( 2 \right)}^{2}}$. Center coordinate of circle is $\left( h=4,k=-1 \right)$. And radius of circle is $r=2$. To graph, we plot the points $\left( 4,0.414 \right)$, $\left( 4,-2.414 \right)$, $\left( 2.585,-1 \right)$, and $\left( 5.414,-1 \right)$ which are, respectively, $\sqrt2$ units above, below, left and right of $\left( 4,-1 \right)$.
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