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.

 A: The radical plane (3D version of radical line) of the spheres is $5x-y-z+6 = 0$. The distance of the centers of the spheres from this plane are $\dfrac{15}{\sqrt{27}}$ and $\dfrac{12}{\sqrt{27}}$. Hence the shortest distance between the spheres and the plane are $\dfrac{15}{\sqrt{27}}-1$ and $\dfrac{12}{\sqrt{27}}-2$. Since the spheres are on the opposite sides of this plane, the minimal distance is $\dfrac{15}{\sqrt{27}}+\dfrac{12}{\sqrt{27}} - 3= 3(\sqrt{3}-1) $
A: I suppose you ask for a simple proof (geometric?)  to show this? 
Let $v=\vec{O_1O_2}$, $e=\frac{v}{|v|}$ and suppose $r_1+r_2<|v|$. One circle (sphere) is included in the halfspace of $P_1$'s for which $e\cdot \vec{O_1 P_1} \leq r_1$. The other in the half space $e\cdot \vec{O_2 P_2}\geq -r_2$. The distance between $P_1$ and $P_2$ is bounded from below by
 $$e\cdot \vec{P_1P_2} = e\cdot \left(\vec{O_1O_2} +\vec{O_2P_2} - \vec{O_1P_1}\right) \geq |v|-r_2-r_1$$ (making a drawing helps). 
A: I have maybe misunderstood your question because the way you presented it was that you wanted an answer to GRE5168 question.
The answer is (E).
Proof: Let $O_1$ and $O_2$ be the centers of the spheres $S_1$ and $S_2$, with radii $R_1=1$ and $R_2=2$ resp.
The distance between the centers is 
$d=O_1O_2=\sqrt{(2+3)^2+(1-2)^2+(3-4)^2}=\sqrt{27}\approx 5.2$ which is larger than $R_1+R_2=3$. Thus $S_1$ and $S_2$ do not intersect.
Consequently, if $A_1$ and $A_2$ are the closest points on $S_1$ and $S_2$ resp., they are on line segment $[O_1O_2]$ (see remark below) with: 
$O_1A_1+A_1A_2+A_2O_2=1+\delta+2=d$. Consequently:
$$\delta=\sqrt{27}-3=3\sqrt{3}-3$$
is the (shortest) distance between $S_1$ and $S_2$.
Remark: if one among $A_1$ and $A_2$ wasn't on line segment $[O_1O_2]$, one could clearly find a closest pair, by reasoning on triangle's inequality.
A: These questions always involve the use of Pythagoras' equation 'twice', i.e., distance $d$ from centres is 
$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\tag{$\star$}$$ 
Now, if $r_1$, $r_2$ are the radii, check: if $d<r_1+r_2$ they intersect, if $d=r_1+r_2$ they touch, and if $d>r_1+r_2$ they are apart from each other.
Best to draw a diagram.
$$S_1: (x-2)^2+(y-1)^2+(z-3)^2=1$$
$$S_2: (x+3)^2+(y-2)^2+(z-4)^2=4$$
So sphere $S_1$ has radius $r_1=1$ and centre $O_1=(2,1,3)$, while sphere $S_2$ has radius $r_2=2$ and centre $O_2=(-3,2,4)$. From this the distance between the spheres is found from $(\star)$ to be $d=3\sqrt{3}>r_1+r_2=3$, and so the spheres do not intersect. Thus the minimal distance between two points on the sphere is the difference between $d$ and $r_1+r_2$, which is $3\sqrt{3}-3$, and the answer is (E). Note: the reason why the distance is minimised at points that are in the line formed by their centres is made clear by the triangle inequality.
A: 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 $d(P,Q)$.
