Consider the following two surfaces: $$x^2 + y^2 + z^2 = 1 \tag{1}$$ $$x^2 + y^2 + c\phantom{^2} = z \tag{2}$$
For which values of $c$ are they (a) mutually tangent at all their points of intersections? (b) Orthogonal at all their points of intersection?
Source: Based on Shifrin's Multivariable Mathematics.
My work so far:
Solution:
(a) is met if $c < -5/4$ (trivially), $c > 1$ (trivially), or $c = -5/4$ (at $z = -1/2$).
(b) is met if $c < -5/4$ (trivially), $c > 1$ (trivially), $c = \frac{-1 - \sqrt{17}}{8}$ (at $z = \frac{-1 - \sqrt{17}}{4}$), or $c = \frac{-1 + \sqrt{17}}{8}$ (at $z = \frac{-1 + \sqrt{17}}{4}$).
Proof:
As shown in my previous question about these surfaces, these two surfaces intersect iff $c \in [-5/4, 1]$. We therefore have the trivial solution: If $c \notin [-5/4, 1]$, both conditions are met.
For $c \in [-5/4, 1]$, we look at their Jacobians: $$\begin{bmatrix}2x & 2y & 2z \end{bmatrix} \\ \begin{bmatrix}2x & 2y & -1 \end{bmatrix}.$$
If these are equal, the tangent planes are parallel. If these are equal at a point of intersection, the intersection is therefore tangential. These are equal iff $2z = -1$, so $z = -1/2$ and $c = -5/4$.
If their dot product is zero, the tangent planes are orthogonal; if the dot product is zero at a point of intersection, the intersection is orthogonal. This happens iff at the point of intersection $$4(x^2 + y^2) -2z = 0$$ so $z = 2c$. The rest follows from the quadratic equation.
Question: The case $c = 1, z = 1$ would seem geometrically to be tangential, but the Jacobians seem to show the planes not being tangential there.
From
$$ \cases{x^2 + y^2 + z^2 = 1\\ x^2 + y^2 + c = z} $$
due to symmetry we have
$$ \cases{ f(y,z) = y^2 + z^2 - 1 = 0\\ g(y,z) = y^2 -z + c = 0}\ \ \ \ \ \ (1) $$
After substituting $z = y^2+c$ into the first equation we have
$$ p(y)=y^2+(y^2+c)^2-1 = (y-r)^2(c_1 y^2+c_2 y+c_3) $$
which states that at tangency $p(y)$ should have a double root. Now equating for all $y$ we have
$$ \cases{ c^2-c_3 r^2-1=0\\ 2 c_3 r-c_2 r^2=0\\ 2 c-c_1 r^2+2 c_2 r-c_3+1=0\\ 2 c_1r-c_2=0\\ 1-c_1=0 } $$
with solutions
$$ \left[ \matrix{ c & r & c_1 & c_2 & c_3\\ -1 & 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 0 & 3 \\ -\frac 54 & -\frac{\sqrt {3}}{2}& 1 &-\sqrt{3}&\frac 34\\ -\frac 54 & \frac{\sqrt {3}}{2}& 1 &\sqrt{3}&\frac 34 } \right] $$
Now from
$$ \cases{x^2 + y^2 + z^2 = 1\\ x^2 + y^2 - \frac 54 = z}\Rightarrow x^2+y^2=\frac 34 $$
The tangency is verified along the circle $x^2+y^2=\frac 34, z = -\frac 12$ so there are infinite common tangent planes. Attached the two tangent surfaces at the black circle. At the top, the tangency for $c=1$ and at the botton for $c=-1$.
To determine the normal intersection we proceed as follows. From $(1)$
$$ \cases{ \nabla f = (y,z)\\ \nabla g = (2y,-1) } $$
so the orthogonality condition is $\nabla f\times \nabla g = 2y^2-z = 0$ and now solving for $y,z,c$
$$ \cases{ y^2 + z^2 - 1 = 0\\ y^2 -z + c = 0\\ 2y^2-z = 0 } $$
we obtain
$$ \left[ \begin{array}{ccc} c & y & z\\ \frac{1}{8} \left(\sqrt{17}-1\right) & -\frac{1}{2} \sqrt{\frac{1}{2} \left(\sqrt{17}-1\right)} & \frac{1}{4} \left(\sqrt{17}-1\right) \\ \frac{1}{8} \left(\sqrt{17}-1\right) & \frac{1}{2} \sqrt{\frac{1}{2} \left(\sqrt{17}-1\right)} & \frac{1}{4} \left(\sqrt{17}-1\right) \\ \end{array} \right] $$
now with $c = \frac{1}{8} \left(\sqrt{17}-1\right)$ we have