find all points $(a,b,c)$ in space for which the the spheres $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}= 1$ and $x^{2}+y^{2}+z^{2}=1$ will intersect orthogonally
I used the fact that the dot products of the gradients must be zero but I got equations which yielded wrong solutions so please help.
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In general, given two spheres below
$$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^2$$ $$x^{2}+y^{2}+z^{2}=q^2$$
In order for them to intersect orthogonally, the centers of the two spheres and their intersection point form a right triangle, which means that the point $(a, b,c)$ should be at the distance
$$d=\sqrt{p^2+q^2}$$
from the origin, thus, in a sphere with the radius $d$. Given $p=q=1$ in the question, $(a,b,c)$ satisfies
$$a^2+b^2+c^2=2$$
By the rotational symmetry of the problem, whether or not the two spheres intersect orthogonally depends solely on the distance of $(a,b,c)$ from the origin. So we can simplify the problem by assuming $(a,b,c)$ is on the positive $x$-axis, so that $a > 0$ and $b = c = 0$, and then determining which values of $a$ work.
The spheres intersect when $x^2 + y^2 + z^2 = 1 = (x-a)^2 + y^2 + z^2$. Subtracting the two equations leads to $a(2x - a) = 0$, so that $x = {a \over 2}$.
They intersect orthogonally if the gradients are orthogonal at the points of intersection. So at points of intersection we require $$<2x, 2y, 2z> \cdot <2x - 2a, 2y, 2z> = 0$$ This leads to $$4(x^2 + y^2 + z^2) = 4ax$$ Since $x^2 + y^2 + z^2 = 1$ and $x = {a \over 2}$, we get that $4 = 2a^2$, so that $a = \sqrt{2}$.
We conclude that the spheres intersect orthogonally if the distance of the center $(a,b,c)$ from the origin is equal to $\sqrt{2}$.