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