Why is the distance between two circles/spheres that don't intersect minimised at points that are in the line formed by their centers? From GRE 0568



From MathematicsGRE.Com:




*

*I'm guessing the idea applies to circles also?

*Is there a way to prove this besides the following non-elegant way?



*

*Form a line between centers $C_1$ and $C_2$

*Given a point on circle/sphere 1 $(x_1,y_1)$, minimise
$$f(x_2,y_2) = (x_1-x_2)^2 + (y_1-y_2)^2$$
to get $(x_2^*,y_2^*)$


*Minimise


$$g(x_1,y_1) = (x_1-x_2^*)^2 + (y_1-y_2^*)^2$$
to get
$(x_1^*,y_1^*)$


*Show that the $(x_2^*,y_2^*)$ and $(x_1^*,y_1^*)$ are on the line.


Basically the question in your title is answered by the fact that a straight line is the shortest path between two points. If $P,Q$ are the points on your two spheres (or circles, if you are in a plane), and $C_i$ and $r_i$ are their respective centres and radii, for $i=1,2$, then $C_1-P-Q-C_2$ is a path from $C_1$ to $C_2$ of length $r_1+d(P,Q)+r_2$. Here only the middle term, the distance $d(P,Q)$ from $P$ to $Q$, depends on the choice of these points; the other two terms are constant. The shortest possible path from $C_1$ to $C_2$ is a straight line, and given that $d(C_1,C_2)\geq r_1+r_2$, this path can be obtained by choosing $P$ and $Q$ on the segment $[C_1,C_2]$. Every other choice of $P,Q$ gives a longer path, hence a larger value of $d(P,Q)$.
