Proving spheres are orthogonal

Question

Given two spheres in $\mathbb{R}^3$:

$x^2+y^2+z^2=2ax; \ \ \ x^2+y^2+z^2 = 2by$

and $a,b>0$, and $\gamma$ the intersection of the spheres, show that for any $p_0 \in \gamma$, the spheres are orthogonal at $p_0$.

I am not sure I fully understand the problem but here is what I tried:

First of all assuming $p_0$ is some point s.t each sphere can be represented as a function $z_1, z_2 : \mathbb{R}^2 \to \mathbb{R}$ in a neighborhood of $p_0$ (if it isn't we can use another variable), what I think we want to show is that the tangent spaces to $z_1, z_2$ are orthogonal at this point, i.e $\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle=0.$ But when I calculate this I get that it does not equal 0:

$z_1 = \sqrt{2ax-x^2-y^2}, z_2 = \sqrt{2by-x^2-y^2}$

$\nabla z_1 = (\frac{2a-2x}{2 \sqrt{2ax-x^2-y^2}}, \frac{-2y}{2 \sqrt{2ax-x^2-y^2}})$

$\nabla z_2 = (\frac{-2x}{2 \sqrt{2by-x^2-y^2}}, \frac{2b-2y}{2 \sqrt{2by-x^2-y^2}})$

$\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle \neq 0.$

Can someone explain where is my mistake?

Answer 1

First solution:

Here is a quick analytic solution first. Let $p_0=(x_0,y_0,z_0)$ be a point on both spheres.

Then the tangent plane in $p_0$ to the one or the other sphere has the equation: $$ \begin{aligned} 2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2a(x-x_0) &=0\ ,\\ 2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2b(y-y_0) &=0\ , \end{aligned} $$ so that two (unnormalized) normal vectors to the two planes are $(x_0-a,y_0,z_0)$ and $(x_0, y_0-b,z_0)$. Their scalar product is: $$ x_0(x_0-a)+y_0(y_0-b)+z_0^2=\frac 12\Big( \ (x_0^2+y_0^2+z_0^2-2ax_0)+(x_0^2+y_0^2+z_0^2-2by_0)\ \Big)=0\ . $$

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Second solution:

Trying to go as the OP goes is also possible, even if complicated. The locally defined function $z_1$ is only a part of the parametrization $$ (x,y)\to(x,y,z_1(x,y))\ . $$ Letting $x$ variate around $x_0$, and keeping $y=y_0$ we obtain a tangent vector $v_x$ to the sphere, similarly keeping $x=x_0$ and letting $y$ variate around $y_0$ we get an other tangent vector $v_y$. These two vectors, and the normal vector $n_1=v_x\times v_y$, are: $$ \begin{aligned} v_x &= (1,0,\nabla_x z_1(x_0,y_0)) =\left(1, 0, \frac{a-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\right)\ ,\\ v_y &= (0,1,\nabla_y z_1(x_0,y_0)) =\left(0, 1, \frac{0-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\right)\ ,\\ n_1 &= v_x\times v_y = \left( \ -\frac{a-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\ , \ -\frac{0-y_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\ , \ 1\ \right) \\ &\sim \left(\ x_0-a\ ,\ y_0\ ,\ \sqrt{2ax_0-x_0^2-y_0^2}\ \right) \\ &=(\ x_0-a\ ,\ y_0\ ,\ z_0\ )\ . \end{aligned} $$ The computation of the corresponding normal vector using the second parametrization delivers analogously the value $$ (\ x_0\ ,\ y_0-b\ ,\ z_0\ )\ . $$ As in the first solution, these vectors are orthogonal.

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Third solution:

A geometrical solution is as follows.

The first sphere is a sphere $(S_a)$ centered in $(a,0,0)$ with radius $a$.

The second sphere is a sphere $S_b$ centered in $(0,b,0)$ with radius $b$.

The inversion with center $O=(0,0,0)$ and power $2ab$ transforms $S_a$ in the plane with equation $x=b$, and $S_b$ in the plane with equation $y=a$. Obviously, these planes intersect orthogonally, the inversion is an orthogonal transformation, we apply it again, so $S_a$ and $S_b$ intersect orthogonally.

Answer 2

The gradients are not applied properly.

Let $f(x,y,z)=x^2+y^2+z^2-2ax$ and $g(x,y,z)=x^2+y^2+z^2-2by$.
Then the spheres are given by $f(x,y,z)=0$ respectively $g(x,y,z)=0$.

The gradients of $f$ and $g$ are orthogonal to their tangent spaces. So to verify if the tangent spaces are orthogonal, we need that for every $(x,y,z)$ on both spheres that $\langle \nabla f(x,y,z),\nabla g(x,y,z)\rangle=0$.

Answer 3

Let$$f(x,y,z)=(x-a)^2+y^2+z^2\quad\text{and}\quad g(x,y,z)=x^2+(y-b)^2+z^2.$$Then your spheres are the surfaces$$\{(x,y,z)\in\Bbb R^3\mid f(x,y,z)=a^2\}\quad\text{and}\quad\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=b^2\}.$$If $p=(x_0,y_0,z_0)\in\gamma$ then the tangent plane to the first sphere at $p$ is orthogonal to$$\nabla f(x_0,y_0,z_0)=\bigl(2(x_0-a),2y_0,2z_0\bigr)$$and the tangent plane to the second sphere at $p$ is orthogonal to$$\nabla g(x_0,y_0,z_0)=\bigl(2x_0,2(y_0-b),2z_0\bigr).$$And we have\begin{align}\bigl\langle\nabla f(x_0,y_0,z_0),\nabla g(x_0,y_0,z_0)\bigr\rangle&=\bigl\langle\bigl(2(x_0-a),2y_0,2z_0\bigr),\bigl(2x_0,2(y_0-b),2z_0\bigr)\bigr\rangle\\&=4x_0^{\,2}+4y_0^{\,2}+4z_0^{\,2}-2ax_0-2by_0\\&=2(x_0^{\,2}+y_0^{\,2}+z_0^{\,2}-ax_0)+2(x_0^{\,2}+y_0^{\,2}+z_0^{\,2}-by_0)\\&=0+0\text{ (since $p\in\gamma$)}\\&=0.\end{align}And, since the vectors $\nabla f(x_0,y_0,z_0)$ and $\nabla g(x_0,y_0,z_0)$ are orthogonal, then so are tangent planes at $p$, since one of them is orthogonal to the first vector, whereas the other one is orthogonal to the second vector.