Answer
a) $(x-3)^2+(y-4)^2=1 ; z=1$
b) $(y-4)^2+(z-1)^2=1; x=-3$
c) $(x+3)^2+(z-1)^2=1; y=4$
Work Step by Step
Definition: The equations of a circle having radius $r$ with center at $(x_0,y_0,z_0)$ are represented as:
1) when the equation of a circle lies in a plane parallel to the $xy$ plane then, we have $(x-x_0)^2+(y-y_0)^2=r^2$; $z=z_0$
2) when the equation of a circle lies in a plane parallel to the $yz$ plane then, we have $(y-y_0)^2+(z-z_0)^2=r^2$; $x=x_0$
3) when the equation of a circle lies in a plane parallel to the $xz$ plane then, we have $(x-x_0)^2+(z-z_0)^2=r^2$; $y=y_0$
Thus, we have:
a) $(x-3)^2+(y-4)^2=1 ; z=1$
b) $(y-4)^2+(z-1)^2=1; x=-3$
c) $(x+3)^2+(z-1)^2=1; y=4$
