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: 64

Answer

Center of circle is $\left( 0,0 \right)$ and radius is $r=\frac{1}{2}$

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 $4{{x}^{2}}+4{{y}^{2}}=1$ (equation - 2) Multiply $\frac{1}{4}$on both the sides of equation $4{{x}^{2}}+4{{y}^{2}}=1$. $\begin{align} & 4\times \frac{1}{4}{{x}^{2}}+4\times \frac{1}{4}{{y}^{2}}=\frac{1}{4} \\ & {{x}^{2}}+{{y}^{2}}=\frac{1}{4} \end{align}$ Now compare the standard equation with equation ${{x}^{2}}+{{y}^{2}}=\frac{1}{4}$. $\begin{align} & {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=\frac{1}{4} \\ & {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}} \\ \end{align}$ Center coordinate of circle is $\left( h=0,k=0 \right)$. And radius of circle is $r=\frac{1}{2}$. Graph: To graph, we plot the points $\left( 0,0.5 \right)$, $\left( 0,-0.5 \right)$, $\left( -0.5,0 \right)$, and $\left( 0.5,0 \right)$ which are, respectively, $\frac{1}{2}$ units above, below, left and right of $\left( 0,0 \right)$.
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