<s>[INST] Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes I have come to a problem in a multivariable calculus book that I'm having trouble with.
The problem statement is :
"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{2} = 3$ and $x^{2} + (y-1)^{2} + z^{2} = 1$, the corresponding
tangent planes will be perpendicular to each other."
So I define two functions :
\begin{align}
f(x,y,z) & = (x-c)^{2} + y^{2} + z^{2} - 3 \\
g(x,y,z) & = x^{2} + (y-1)^{2} + z^{2} - 1
\end{align}
We denote the partial derivative with respect to $x$ as $f_{x}$ and so on...
We see :
\begin{align}
f_{x}(x,y,z) & = 2(x-c) = 2x - 2c\\
f_{y}(x,y,z) & = 2y\\
f_{z}(x,y,z) & = 2z
\end{align}
and :
\begin{align}
g_{x}(x,y,z) & = 2x \\
g_{y}(x,y,z) & = 2(y-1) = 2y - 2 \\
g_{z}(x,y,z) & = 2z
\end{align}
I assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. So for every intersection point (x,y,z)
we have :
\begin{equation}
\bigtriangledown f(x,y,z) \cdot \bigtriangledown g(x,y,z) = 0
\end{equation}
So :
\begin{align}
\require{cancel}
(2x - 2c, 2y, 2z) \cdot (2x, 2y-2, 2z) & = 0\\
(2x-2c)2x + (2y)(2y) - 2(2y) + 4z^{2} & = 0\\
4x^{2} - 4xc + 4y^{2} - 4y + 4z^{2} & = 0\\
4(x^{2} + y^{2} + z^{2}) - 4(xc + y) & = 0\\
\cancel{4} \left[ (x^{2}+y^{2}+z^{2}) - (xc + y) \right] & = 0\\
(x^{2}+y^{2}+z^{2}) - (xc + y) & = 0 \\
x^{2} + y^{2} + z^{2} & = xc + y
\end{align}
We also see :
\begin{align}
\require{cancel}
f(x,y,z) & = x^{2} - 2xc + c^{2} + y^{2} + z^{2} - 3 \\
         & = (x^{2} + y^{2} + z^{2}) + (c^{2} - 2xc - 3) \\
g(x,y,z) & = x^{2} + y^{2} - 2y + \cancel{1} + z^{2} - \cancel{1} \\
         & = (x^{2}+y^{2}+z^{2}) - 2y
\end{align}
and :
\begin{align}
f(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = -(c^{2}-2xc-3) = -c^{2} + 2xc + 3\\
g(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = 2y
\end{align}
So we have :
\begin{equation}
xc + y = -c^{2} + 2xc + 3 = 2y
\end{equation}
It is here that I am stuck. It seems that the goal here would be to obtain an expression for $c$ that doesn't include $x$ or $y$, but I do not know how to obtain it.
Can someone help with this ?
 [/INST] Math is beautiful in that a good problem gives you everything that you need to solve it, and nothing more.
Taking this into consideration, let's see what we're given.
We have:
$$ (x-c)^2 + y^2 + z^2 = 3 $$
$$ x^2 + (y-1)^2 + z^2 = 1 $$
So, we can define two functions:
$$ 
f(x,y,z) = (x-c)^2 + y^2 + z^2 - 3
$$
$$
g(x,y,z) = x^2 + (y-1)^2 + z^2 - 1
$$
$$
$$
Taking the partial derivatives, we have:
$$ \frac{\partial f}{\partial x} = 2x-2c  ,  \frac{\partial f}{\partial y} = 2y  ,  \frac{\partial f}{\partial z} = 2z $$
$$ \frac{\partial g}{\partial x} = 2x  ,  \frac{\partial g}{\partial y} = 2y-2  ,  \frac{\partial g}{\partial z} = 2z $$
Thus,
$$ \nabla f = (2x-2c,2y,2z) $$
$$ \nabla g = (2x,2y-2,2z) $$
You are correct to assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. Thus, we have:
$$ \nabla f(x,y,z) \cdot \nabla g(x,y,z) = 0 $$
$$ (2x-2c,2y,2z) \cdot (2x,2y-2,2z) = 0 $$
$$ 4x^2 - 4xc + 4y^2 - 4y + 4z^2 = 0 $$
Completing the square, we have:
$$ 4(x^2 - cx + \frac{1}{4} c^2) - c^2 + 4(y^2 - y + \frac{1}{4}) - 1 + 4z^2 = 0 $$
$$ 4(x - \frac{1}{2} c)^2 + 4(y - \frac{1}{2})^2 + 4z^2 = 1 + c^2 $$
Adding $ 1 + c^2 $ to both sides and simplifying, we have:
$$ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $$
$$
$$
Now we need to look back at what we're given. We know that
$$ (x-c)^2 + y^2 + z^2 = 3 $$
$$ x^2 + (y-1)^2 + z^2 = 1 $$
So, multiplying both equations by $ 2 $, we have:
$$ 2((x-c)^2 + y^2 + z^2) = 2(3)  \implies 2x^2 - 4cx + 2c^2 + 2y^2 + 2z^2 = 6 $$
$$ 2(x^2 + (y-1)^2 + z^2) = 2(1)  \implies 2x^2 + 2 - 4y + 2y^2 + 2z^2 = 2 $$
Substituting these equations into $ 4x^2 - 4cx + c^2 + 4y^2 - 4y + 2 + 4z^2 = 2c^2 $ , we can see that this simplifies to:
$$ 2c^2 + 2 = 8 $$
$$ 2c^2 = 6 $$
$$ c = \pm \sqrt3 $$
</s>