A∈ Ox , B∈ Oy , OA=OB=a . A line p moves along the circle…
$A\in Ox$, $B\in Oy$, $OA=OB=a$. A line $p$ moves along the circle with diameter $AB$, meets $Oz$ and stays parallel to $z=0$. Find the conoid described by $p$.
Explanation of Method
A conoid whose directrix is a circle on the diameter AB, standing vertically above the segment AB.
Step 1: Which circle
A horizontal line through $Oz$ can meet a circle lying in $z=0$ only at height $0$, so that reading gives no surface. For a conoid, the directrix must not lie in a plane parallel to the director plane. The circle on diameter $AB$ must therefore stand in the vertical plane through $AB$, namely $x+y=a$:
\[ \left(x-\tfrac a2\right)^2+\left(y-\tfrac a2\right)^2+z^2=\tfrac{a^2}2,\qquad x+y=a . \]
With $x+y=a$, the first equation simplifies to $x^2+y^2+z^2=a^2$.
Step 2: Generators
At height $z$, the generator joins $(0,0,z)$ to a circle point $(x_0,y_0,z)$, where
\[ x_0+y_0=a,\qquad x_0^2+y_0^2=a^2-z^2 . \]
Its points are $(sx_0,\ sy_0,\ z)$. Then $x+y=sa$, so $s=\tfrac{x+y}a$, and $x^2+y^2=s^2(a^2-z^2)$: $a^2(x^2+y^2)=(a^2-z^2)(x+y)^2$.
Checks: $A(a,0,0)$ gives $a^4=a^4$. The top of the circle, $\left(\tfrac a2,\tfrac a2,\tfrac a{\sqrt2}\right)$, gives $\tfrac{a^4}2=\tfrac{a^2}2\cdot a^2$. The $z$-axis also satisfies the equation. $\checkmark$
Result
\[\boxed{\,a^2(x^2+y^2)=(a^2-z^2)(x+y)^2\,}\]
Remark
Correction: the source puts the circle in $z=0$, tries several contradictory constructions, and ends with the cylinder $x^2+y^2=a(x+y)$. A vertical cylinder is not a conoid with director line $Oz$: its horizontal lines through $Oz$ meet it in only two points, so they do not lie on it. The answer above uses the only reading that gives a real conoid (compare Problem 324, which has the same structure).