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 856: 82


(a) Center of the circle is $\left( 0,-1727.25 \right)$ (b) $r=1728.75\text{ cm}$

Work Step by Step

(a) For the center of the circle, Use distance formula, $\begin{align} & \sqrt{{{\left( 72-0 \right)}^{2}}+{{\left( 0-k \right)}^{2}}}=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 1.5-k \right)}^{2}}} \\ & \sqrt{\left( 5184+{{k}^{2}} \right)}=\sqrt{\left( 2.25+{{k}^{2}}-3k \right)} \end{align}$ Squaring on both the sides of $\sqrt{\left( 5184+{{k}^{2}} \right)}=\sqrt{\left( 2.25+{{k}^{2}}-3k \right)}$. $\begin{align} & 5184+{{k}^{2}}=2.25+{{k}^{2}}-3k \\ & 5184-2.25={{k}^{2}}-{{k}^{2}}-3k \\ & 5181.75=-3k \\ & \frac{5181.75}{3}=-k \end{align}$ Simplify More, $k\approx -1727.25$ Thus, the center of the circle is $\left( 0,-1727.25 \right)$ (b) For the radius, We will use the distance formula between center of circle $\left( 0,-1727.25 \right)$ to the any point on board $\left( 72,0 \right)\text{ and }\left( 0,1.5 \right)$ We will use$\left( 0,1.5 \right)$. Put center of circle $\left( 0,-1727.25 \right)$ and point $\left( 0,1.5 \right)$ in the standard equation of a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$. $\begin{align} & r=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( -1727.25-1.5 \right)}^{2}}} \\ & r=1728.75 \\ \end{align}$ Thus, the radius $r=1728.75\text{ cm}$.
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.