Prove that the shortest distance from a line to a circle passes through the centre Let $L$ be a line and $\omega$ be a circle with centre $C$. Suppose that the shortest distance from $L$ to $\omega$ is the distance from point $P$ on $L$ to point $Q$ on $\omega$. Why does the line $PQ$ pass through $C$?
I ask this because apparently the best method to find the shortest distance from a line to a circle is to find the shortest distance from the line to the centre of the circle, then subtract the radius. But this only works if $PQ$ passes through $C$ as stated above.
 A: For any point $P$ on the line, the shortest distance from $P$ to the circle is the line that goes through the center.
(This is true regardless, but the following argument does require that $P$ is outside the circle. If $P$ is on the inside, then some point of the line is on the circle, and the distance from the line to the circle is $0$.)
To see this, consider that to get to the center from $P$, you have to first get to the circle, and then from the circle and in to the center. That last bit will always be a radius, so the shortest way from $P$ to the center must correspond to the shortest way from $P$ to the circle, plus a radius.
But the shortest way from $P$ to the center is a straight line.
A: We know that $d(p,q)\leq d(p,x)+d(x,q)$ is a property that all measure must satisfy. 
Applying the latter to your example
$$d(P,w)\leq d(P,Q)+d(Q,w)$$ where $Q$ can be any point ,and in particular a point in the circumference. (Of course that you didn't specify that the point $Q$ must be on the circumference,but for sure a point in the boundary is always going to be closer to $P$ that a point in the "interior").
 This is equivalent to 
$$d(P,w)-d(Q,w)\leq d(P,Q)$$ 
Since $Q$ is a point of the circumference the second term on the left is the radius. 
Hence the smaller distance from a point $P$ to a point $Q$ in the circle will be , as you claimed, the distance to the center minus the radius.
A: Geometrical proof:
1) Consider $2$ parallel lines $l_1,l_2$ distance $d$ .
2) A semi circle with center $O$ on $l_1$, radius $r  <d$ .
3) A line perpendicular to $l_1,l_2$ intersects the semi circle in $P$, $ l_1$ in $R$, $l_2$ in $S$.
4) $\overline {RP} +\overline {PS} = d.$
5) $\min( \overline {PS}) =$
$ d - \max(\overline {RP})= d-r$.
Now a bit tricky:
6) $O$ lies on the circle about $P_{\min}$ with radius $r$.
7) $O$ lies on line $ l_1$(construction), where distance $ l_1$ to $P_{min} =r$, i.e.
circle about $P_{min}$ with radius  $r$ intersects line l_1 in one point, touches line $ l_1 $.
8)$O$ is  point of tangency, hence $O$ lies on $SP_{min}$.
