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: $r=1$
