## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 947: 83

#### Answer

The standard form is, ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{3}^{2}}$.

#### Work Step by Step

The standard form of the circle is, ${{\left( x-{x}' \right)}^{2}}+{{\left( y-{y}' \right)}^{2}}={{r}^{2}}$ To make the equation in standard form, we will proceed by completing the square method as follows: \begin{align} & {{x}^{2}}+{{y}^{2}}-2x+4y=4 \\ & \left( {{x}^{2}}-2\cdot x\cdot 1+{{1}^{2}} \right)+\left( {{y}^{2}}+2\cdot y\cdot 2+{{2}^{2}} \right)-5=4 \\ & {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}-5=4 \\ & {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9 \end{align} Hence, the required standard form of the circle is, ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{3}^{2}}$.

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.