# Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 347: 26

$(x+3)^2+(y-2)^2=\frac{134}{9}$ Center: $(-3,2)$ Radius: $r=\frac{\sqrt {134}}{3}$

#### Work Step by Step

$9x^2+9y^2+54x-36y+17=0$ (divide the equation by 9): $x^2+y^2+6x-4y-\frac{17}{9}=0$ Notice that $6x=2(3)x$ and that $3^2=9$ Notice that $-4y=-2(2)y$ and that $2^2=4$ $x^2+6x+9-9+y^2-4y+4-4-\frac{17}{9}=0$ $(x^2+6x+9)-9+(y^2-4y+4)-4-\frac{17}{9}=0$ $(x+3)^2+(y-2)^2=9+4+\frac{17}{9}=\frac{81}{9}+\frac{36}{9}+\frac{17}{9}=\frac{134}{9}$ 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 $(x+3)^2+(y-2)=(\frac{\sqrt {134}}{3})^2$ $[x-(-3)]^2+(y-2)=(\frac{\sqrt {134}}{3})^2$ Center: $(-3,2)$ Radius: $r=\frac{\sqrt {134}}{3}$

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