Chord with the smallest length If I have a point K inside a euclidean disc what will be chord (chord which goes through the point K) with the smallest length. I think it will be chord which is perpendicular to the diameter, which goes through the point K, but I don't know is this true. I tried to prove this with formula of length of the chord $L=2R \sin (A/2)$, where $A$ is angle between two radii drawn to the ends of chord, but the only thing I deduced from this formula is that length would be the smallest when a is the smallest. So I'm stuck with this. Is my assumption true (if so can you help with the proof) and if not what chord will have the smallest length ?
 A: I am.afraid Arthur, Winter said it all .
Just for fun:
Point $K$ given, inside a circle .
For definiteness consider the chord $KM$ , where $M$ is the centre of the circle.
$K$ divides the chord $KM$  into $2$ segments of lengths $a$ and $b.$
Inersecting Chord Theorem:
$ab = xy$ where $x,y$ are the lengths of segments formed by any chord through $K.$
$2$ equations:
1)$xy=ab=C$ (constant , given)
2)$S:= x +y $.
Want to minimize $S$ with the constraint 1).
Use 1) to eliminate $y$:
$S = x + C/x $.
AM-GM  :
$S = x+C/x \ge 2√C.$
Minimum attained for $x=C/x= √C$.
Then $y= (ab)/√C =√C$.
Hence $x=y$ , the perpendicular chord to  chord $KM
$.
Note: $KM$ is a symmetry axis, hence $x=y$ implies the chord through $K$ is perpendicular.
A: Let's say we have a unit circle around the origin $O$. A secant $\ell$ having distance $d(\ell,O)$ to $O$ is of length $$L=2\sqrt{1-d(\ell,O)^2}.$$ We can minimize this by maximizing $d(\ell,O)$. 
Now let's restrict to secants $\ell$ through $K$. We know that $d(\ell,O)$ is the length of a line segment between $O$ and $\ell$, which is perpendicular to $\ell$. The point where this segment meets $\ell$ is the closest point of $\ell$ to $O$. All other points of $\ell$ are further away from $O$. Because $\ell$ has to go through $K$, it cannot have a distance greater than $d(K,O)$ from $O$, but by choosing $K$ to be the closest point on $\ell$ to $O$, we also ensure that $d(\ell,O)$ is not smaller than $d(K,O)$. Therefore, this maximizes $d(\ell,O)=d(K,O)$ and minimizes the secants length to:
$$L_{\min}=2\sqrt{1-d(K,O)^2}.$$
Since the segment from $O$ to $K$ is a (part of the) radius of the circle, and meets $\ell$ perpendicularly in $K$, your intuition was right.
A: Given any point $P$ and any circle $C$, the power of $P$ with respect to $C$ is defined as follows:

Take a line through $P$ that intersects $C$ twice, at $A$ and $B$. The power of $P$ with respect to $C$ is $|PA|\cdot |PB|$, and is independent of what line was chosen.

(According to some conventions the power is negative when $P$ is on the inside of $C$ and positive when $P$ is outside, but we will say it is positive. Until we have to compare powers of several different points it doesn't matter.)
Given a chord through $K$, the power of $K$ with respect to the circle is length of the two parts of chord on either sides of $K$. It is the same no matter which chord you draw. You want the sum of the two parts to be minimal.
Now note that if two quantities vary so that their product is the same (such as the lengths of the two parts of the chords), their sum is the least when the quantities are equal. (Replace "their product" with "the square root of their product" and "their sum" with "their average", and you've got exactly the AM-GM inequality, which proves this statement.)
