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.( Their tangent planes should be perpendicular at each point of intersection.)
If we consider the equations of the two spheres as level surfaces and take the gradients the we get – $$\nabla f_1 = 2(x-a)i+2(y-b)j+2(z-c)k;$$ $$\nabla f_2 = 2xi+2yj+2zk.$$
Since the two spheres intersect orthogonally at $(x,y,z)$, we must have $\nabla f_1 \cdot \nabla f_2=0$. This gives $$x(x-a)+y(y-a)+z(z-a)=0.$$
Can anyone suggest how to proceed from here to obtain the set of points (a,b,c)?
Best Answer
If $(x,y,z)$ belongs to the sphere $x^2+y^2+z^2=1$, then we have $$\nabla f_1\cdot \nabla f_2=0 \iff x(x-a)+y(y-b)+z(z-c)=0 \\ \iff x^2+y^2+z^2-(ax+by+cz)=0 \\ \iff ax+by+cz =1$$ On the other hand, if $(x,y,z)$ also belongs to the sphere $(x-a)^2+(y-b)^2+(z-c)^2=1$, we have $$x^2+y^2+z^2-2(ax+by+cz)+a^2+b^2+c^2=1$$ And we know $x^2+y^2+z^2=1=ax+by+cz$ so $$1-2(1)+a^2+b^2+c^2=1 \\ \text{i.e.}\;\; a^2+b^2+c^2=2$$ Therefore, spheres $x^2+y^2+z^2=1$ and $(x-a)^2+(y-b)^2+(z-c)^2=1$ intersect orthogonally iff $(a,b,c)$ lies in the sphere $x^2+y^2+z^2=2$.