## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

# Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 76

#### Answer

$x^{2}+y^{2}=y+\sqrt{x^{2}+y^{2}} \quad \text{or} \quad x^2+y^2-y-\sqrt{x^2+y^2}=0$

#### Work Step by Step

Multiply $r$ to both sides of the equation to obtain \begin{align*} r\cdot r &=r(\sin\theta+1) \\r^{2}&=r\sin\theta+r \end{align*} Since $r^{2}=x^{2}+y^{2}$ and $r\sin\theta=y$, then \begin{align*} r^2&=r\sin\theta+r\\ x^2+y^2&=y+\sqrt{x^2+y^2}\\ x^2+y^2-\left(y+\sqrt{x^2+y^2}\right)&=0\\ x^2+y^2-y-\sqrt{x^2+y^2}&=0\\ \end{align*}

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