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 $\left( 4,-1 \right)$.
Small 1566304049
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.