Find the cylindrical surface through x^2+y^2=4x, z=0 with…

Find the cylindrical surface through $x^2+y^2=4x,\ z=0$ with generators parallel to $\vec p=\{1,1,1\}$.

Explanation of Method

The cylinder through a circle in xOy with generators parallel to the diagonal direction.

The generator through $M$ meets $z=0$ at $(x-z,\ y-z,\ 0)$, so

\[ (x-z)^2+(y-z)^2=4(x-z). \]

\[x^2+y^2+2z^2-2xz-2yz-4x+4z=0\]

Check: $(4,0,0)+\{1,1,1\}=(5,1,1)$ gives $25+1+2-10-2-20+4=0$. $\checkmark$

$x^2+y^2=4x$

The oblique cylinder: generators parallel to $\{1,1,1\}$ through the circle $(x-2)^2+y^2=4,\ z=0$.

\[\boxed{\,x^2+y^2+2z^2-2xz-2yz-4x+4z=0\,}\]