## Thomas' Calculus 13th Edition

(a) $x^2+(y-2)^2=4 ,z=0$ (b) $(y-2)^2+z^2=4,x=0$ (c) $x^2+z^2=4, y=2$
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) $x^2+(y-2)^2=4 ,z=0$ (b) $(y-2)^2+z^2=4,x=0$ (c) $x^2+z^2=4, y=2$