## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Test - Page 800: 6

#### Answer

The polar equation is $r=-16\sin \theta$.

#### Work Step by Step

Consider the given equation ${{x}^{2}}+{{\left( y+8 \right)}^{2}}=64$ ; So, \begin{align} & {{x}^{2}}+{{\left( y+8 \right)}^{2}}=64 \\ & {{\left( r\cos \theta \right)}^{2}}+{{\left( r\cos \theta +8 \right)}^{2}}=64 \\ & {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta +16r\sin \theta +64=64 \\ & {{r}^{2}}+16r\sin \theta =0 \end{align} So, \begin{align} & {{r}^{2}}=-16r\sin \theta \\ & r=-16\sin \theta \end{align} So, the polar equation is given by $r=-16\sin \theta$. Hence, the polar equation is $r=-16\sin \theta$.

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