Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Cumulative Review Exercises - Page 435: 18

Answer

The graph is shown below:
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Work Step by Step

Since the equation is of a circle, make the equation into the general form of a circle $\left( x-{{h}^{2}} \right)+{{\left( y-k \right)}^{2}}={{r}^{2}}$. The center of the circle is expressed by the coordinates $\left( h,k \right)$ and r represents the radius of the circle. Simplify: $\begin{align} & {{x}^{2}}+{{y}^{2}}-2x+4y=4 \\ & {{x}^{2}}-2x+1+{{y}^{2}}+4y=5 \\ & {{\left( x-1 \right)}^{2}}+{{y}^{2}}+4y=5 \end{align}$ Also, $\begin{align} & {{\left( x-1 \right)}^{2}}+{{y}^{2}}+4y+4=5+4 \\ & {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9 \end{align}$ Thus, the center is $\left( 1,-2 \right)$ and the radius is 3. Equation ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$ can be expressed as $y=\sqrt{9-{{\left( x-1 \right)}^{2}}}-2$. Substitute $x=1$ in the above equation: $\begin{align} & y=\sqrt{9-{{\left( 1-1 \right)}^{2}}}-2 \\ & =\pm 3-2 \\ & =1,-5 \end{align}$ At $x=4$ , $\begin{align} & y=\sqrt{9-{{\left( 4-1 \right)}^{2}}}-2 \\ & =\sqrt{9-9}-2 \\ & =-2 \end{align}$ At $x=-2$ , $\begin{align} & y=\sqrt{9-{{\left( -2-1 \right)}^{2}}}-2 \\ & =\sqrt{9-{{\left( -3 \right)}^{2}}}-2 \\ & =\sqrt{9-9}-2 \\ & =-2 \end{align}$ Thus, the points $\left( 1,1 \right),\left( 1,-5 \right),\left( 4,-2 \right)$ and $\left( -2,-2 \right)$ satisfy the $y=\sqrt{9-{{\left( x-1 \right)}^{2}}}-2$.
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