I need to find the equation of the circular cylinder whose generators are parallel to the line $x=y=z$ whose guiding curve is the circle: $$x^2+y^2+z^2-2x-3=0, \ \ 2x+y+2z=0 $$
My approach I assumed the generator to be: $$\frac{x-a}{1}=\frac{y-b}{1}=\frac{z-c}{1}$$ for arbitrary point $P(a,b,c)$ on the cylinder. In parametric form, $(x(t),y(t),z(t))=(t+a,t+b,t+c)$. This point must satisfy the equation of the circle. on substitution, we get: $$t=\frac{-(2a+b+2c)}{5}$$ $$3t^2+t(2a+2b+2c-2)+((a-1)^2+b^2+c^2-4)=0$$ Eliminating t from above two equations, by substituting the first in second, we get: $$17a^2+18b^2+17c^2-18ab-16ac-18bc-30a+10b+20c-75=0$$ The locus of $P(a,b,c)$ gives the required equation of the cylinder.
My doubt I don't know if my answer is correct, also is it the right way to solve? Kindly suggest, if there's a more elementary way to solve. Thanks in advance.
First of all, it should be noted that this not a circular cylinder, because the axis of the cylinder is not perpendicular to the guiding circle.
The given guiding curve is the circle resulting from the intersection of the sphere
$ (x - 1)^2 + y^2 + z^2 = 4 $
and the plane
$2 x + y + 2 z = 0$
The distance between the the center of the sphere $(1, 0, 0)$ and this plane is
$ d = \dfrac{2}{\sqrt{2^2 + 1^2 + 2^2} } = \dfrac{2}{3} $
Hence, the radius of the guiding circle is
$ r = \sqrt{ 2^2 - \left(\dfrac{2}{3}\right)^2 } = \dfrac{4}{3} \sqrt{2} $
Two vectors that are orthogonal to the normal to the given plane are and of length equal to the radius are
$ u_1 = r (1, -2, 0) / \sqrt{5}$
$ u_2 = [(2, 1, 2) / 3] \times [r(1, -2, 0) /\sqrt{5} ] = r(4, 2, -5) / (3 \sqrt{5} ) $
The center of the guiding circle is at
$ C = (1, 0, 0) - d (2, 1, 2) / 3 = (1, 0, 0) - (2, 1, 2) (2/9) = 1/9 (5, -2, -4) $
Therefore, the parametric equation of the guiding circle is
$ P(t) = P_0 + \cos t \ u_1 + \sin t \ u_2 $
where $P_0 = 1/9 (5, -2, -4) $
Therefore, the parametric equation of the cylinder is
$ S(t, s) = P(t) + s (1, 1, 1) $
From which,
$ S - P_0 = \cos t u_1 + \sin t u_2 + s (1, 1, 1) $
Let $v_1, v_2$ be two vectors that are perpendicular to $(1, 1,1) $, then
$ v_1^T (S - P_0) = \cos t \ v_1^T u_1 + \sin t \ v_1^T u_2 $
and
$ v_2^T (S - P_0) = \cos t \ v_2^T u_1 + \sin t \ v_2^T u_2 $
Define
$ V = [v_1, v_2], U = [u_1, u_2] , w = [\cos t , \sin t ]^T $
Then
$ S-P_0 = U w + s [1, 1, 1]^T$
So that
$ V^T (S - P_0) = V^T U w $
So that
$ w = (V^T U)^{-1} V^T (S - P_0) $
Since $w^T w = 1$ , then the equation of the cylinder is
$ (S - P_0)^T V (V^T U)^{-T} (V^T U)^{-1} V^T (S - P_0) = 1 $
I've computed the matrix $G = V (V^T U)^{-T} (V^T U)^{-1} V^T $ and it is equal to
$ G = \begin{bmatrix} 0.19125 && -0.10125 && -0.09 \\ -0.10125 && 0.2025 && -0.10125 \\ -0.09 && -0.10125 && 0.19125 \end{bmatrix} $
Multiply the whole equation by $\dfrac{800}{9}$ we get
$( S - P_0)^T \left(\dfrac{800}{9} G \right) (S - P_0) = \dfrac{800}{9} $
where
$ \left(\dfrac{800}{9}\right) G = \begin{bmatrix} 17 && -9 && -8 \\ -9 && 18 && -9 \\ -8 && -9 && 17 \end{bmatrix} $
Note that $P_0 = 1/9 (5, -2, -4) $
$P_0$ can be replaced with $P_0 + \lambda (1, 1, 1) $
If we take $\lambda = \dfrac{4}{9} $
Then the new $P_0 = ( 1 , \dfrac{2}{9} , 0 ) $
Hence, the cartesian equation of the cylinder becomes,
$17( x - 1)^2 + 18 (y - \dfrac{2}{9})^2 + 17 z^2 - 18 (x - 1)(y - \dfrac{2}{9}) - 16 (x - 1)z - 18 (y - \dfrac{2}{9} ) z - \dfrac{800}{9} = 0$
This simplifies to,
$17 x^2 + 18 y^2 + 17 z^2 - 18 x y - 16 x z - 18 y z - 34 x - 8 y + 18 y + 4 x + 16 z + 4 z + 17 + 8/9 - 4 - \dfrac{800}{9} = 0 $
Which finally simplifies to
$ 17 x^2 + 18 y^2 + 17 z^2 - 18 x y - 16 x z - 18 y z - 30 x + 10 y + 20 z - 75 = 0 $
And this the final form of the equation of the cylinder, which as I mentioned above, is not a circular cylinder, but an elliptical cylinder.