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
Let us consider $r$ as the radius of a circle and $(x_0,y_0,z_0)$ as center.
Points to be noted: i) Equation of a circle lies in a plane parallel to $xy$ plane is represented as: $(x-x_0)^2+(y-y_0)^2=r^2$; $z=z_0$
ii) Equation of a circle lies in a plane parallel to $yz$ plane is represented as:$(y-y_0)^2+(z-z_0)^2=r^2$; $x=x_0$
iii) Equation of a circle lies in a plane parallel to $xy$ plane is represented as: $(x-x_0)^2+(z-z_0)^2=r^2$; $y=y_0$
Hence, our equations are: (a) $(x-3)^2+(y-4)^2=1 ,z=1$
(b) $(y-4)^2+(z-1)^2=1,x=-3$
(c) $(x-3)^2+(z-1)^2=1, y=4$
