Problems on orthogonality and tangency in 3-space. 
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*Find the set of all points $(a,b,c)$ in 3-space for which the two spheres $(x-a)^2+(y-b)^2+(z-c)^2=1$ and $x^2+y^2+z^2=1$ intersect orthogonally. 

*A cylinder whose equation is $y=f(x)$ is tangent to the surface $z^2+2xz+y=0$.
For the first one, at the set of points, $\nabla f \cdot \nabla g$ should be $0$, where $f$ and $g$ each correspond to the first two spheres. Doing this, I get $4x(x-a)+4y(y-b)+4z(z-c)=0$, which gives the function of a sphere centered at $(a/2,b/2,c/2)$. However, the answer of this problem is a sphere with center at the origin and radius $\sqrt{2}$. 
For the second one, I tried solving it by setting the first equation as $F(x,y,z)=f(x)-y$ and the second one as $G(x,y,z)=z^2+2xz+y$. But I do not know how to progress further from here.
I guess I'm not good at solving these kind of geometric problems yet. I would greatly appreciate any help.
 A: For (1), your equation is correct. You should also combine that equation with the two original equation. The points are on the intersection too. So 
$$4x(x-a)+4y(y-b)+4z(z-c)=0\implies 4x^2-4ax+4y^2-4by+4z^2-4cz=0$$
Combining this with $x^2+y^2+z^2=1$
$$4-4zx-4by-4cz=0$$
Now the first equation is $x^2-2ax+a^2+y^2-2by+b^2+z^2-2cz+c^2=1$. Combining this with equation 2 gives you $-2ax-2by-2cz=-a^2-b^2-c^2$. This and the above equation will give you what you want. 
For (2), the gradient vectors of the two surfaces are equal at the tangent point
$$(-f'(x),1,0)=(2z,1,2z+2x)$$
From here see if you can find $f(x)$.
A: Hint for the question 2.
Your cylinder has a parametric equation $(x,z) \mapsto (x, f(x), z)$. The second surface is defined by $(x,y) \mapsto (x,y,z(x,y))$ where $z(x,y)$ is defined by the implicit equation you provide.
Based on that, you can compute the normal vectors of both surfaces at a point of intersection. Those two vectors should be dependent.
(Jean-Pierre http://www.mathcounterexamples.net)
