## Thomas' Calculus 13th Edition

Published by Pearson

# Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 25

#### Answer

A circle with center at $C(0,\displaystyle \frac{3}{2})$ and radius $\displaystyle \frac{5}{2}$ .

#### Work Step by Step

Gather terms containing x and y in separate parentheses $(x^{2})+(y^{2}-3y)=4$ Complete squares: $x^{2}\quad$ is already a square $y^{2}-3y=y^{2}-2\displaystyle \cdot y\cdot\frac{3}{2}+(\frac{3}{2})^{2}-(\frac{3}{2})^{2}=(y-\frac{3}{2})^{2}-\frac{9}{4}$ The equation becomes $x^{2}+(y-\displaystyle \frac{3}{2})^{2}-\frac{9}{4}=4$ $x^{2}+(y-\displaystyle \frac{3}{2})^{2}=\frac{25}{4}$ A circle with center at $C(0,\displaystyle \frac{3}{2})$ and radius $\displaystyle \frac{5}{2}$ x-intercepts $:\quad$ $\left[\begin{array}{l} x^{2}+(0-\frac{3}{2})^{2}=\frac{25}{4}\\ x^{2}+\frac{9}{4}=\frac{25}{4}\\ x^{2}=4\\ x=\pm 2 \\ (-2,0),\quad(2,0) \end{array}\right]$ y-intercepts $:\quad \left[\begin{array}{l} 0+(y-\frac{3}{2})^{2}=\frac{25}{4}\\ y-\frac{3}{2}=\pm\frac{5}{2}\\ y=\frac{3}{2}\pm\frac{5}{2}\\ y=-1,\quad y=4\\ (0,-1),\quad(0,4) \end{array}\right]$

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