## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 347: 25

#### Answer

$(x+\frac{3}{2})^2+(y-3)=1$ Center: $(-\frac{3}{2},3)$ Radius: $r=1$

#### Work Step by Step

$4x^2+4y^2+12x-24y+41=0$ (divide the equation by 4): $x^2+y^2+3x-6y-\frac{41}{4}=0$ Notice that $3x=-2(\frac{3}{2})x$ and that $(\frac{3}{2})^2=\frac{9}{4}$ Notice that $-6y=-2(3)y$ and that $3^2=9$ $x^2+3x+\frac{9}{4}-\frac{9}{4}+y^2-6y+9-9+\frac{41}{4}=0$ $(x^2+3x+\frac{9}{4})+(y^2-6y+9)=\frac{9}{4}-\frac{41}{4}+9=\frac{9}{4}-\frac{41}{4}+\frac{36}{4}=1$ $(x+\frac{3}{2})^2+(y-3)=1$ The equation of a circle in standard form: $(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius $(x+\frac{3}{2})^2+(y-3)=1^2$ $[x-(-\frac{3}{2})]^2+(y-3)=1^2$ Center: $(-\frac{3}{2},3)$ Radius: $r=1$

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