Find the surface that is orthogonal to the family…

Find the surface that is orthogonal to the family \[ x^2+y^2+z^2 = cx, \] where $c$ is an arbitrary parameter, and passes through the line $y=x$, $z=1$.

Step 1 · PDE and characteristic equations

As in Problem 67, eliminating $c$ gives the orthogonality PDE \[ (x^2-y^2-z^2)\,p + 2xy\,q = 2xz, \qquad \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}. \]

Step 2 · First and second integrals

As before: $z/y=c_1$ \tag{I} and $(x^2+y^2+z^2)/y=c_2$. \tag{II} General solution: \[ \frac{x^2+y^2+z^2}{y} = \Phi\!\left(\frac{z}{y}\right). \tag{$*$} \]

Step 3 · Apply the initial condition

Parametrise $y=x$, $z=1$ by $t=x$: $c_1=1/t$, $c_2=(2t^2+1)/t=2t+1/t$. With $t=1/c_1$: $c_2=2/c_1+c_1$, so $\Phi(s)=2/s+s$. Substituting $s=z/y$ into ($*$) and clearing denominators: \[ z(x^2+y^2+z^2) = 2y^2+z^2. \]

Verification

At $y=x$, $z=1$: $x^2+x^2+1=2x^2+1$. $\checkmark$ For $G=z(x^2+y^2+z^2)-2y^2-z^2$, dotting $\nabla G$ with the family's normal $(x^2-y^2-z^2,2xy,2xz)$ and using $G=0$ to substitute $z(x^2+z^2-z)=(2-z)y^2$ collapses the expression to $0$. $\checkmark$