Construct a chord equal to the radius with compass and straight edge. 
A circle is given and it has a point inside the circle. How to construct a chord that passes through the point and equal to the radius of the circle with compass and straight edge?  

I know how to construct with the intersecting chords theorem. Is there any other  ways to costruct this?
 A: Well, the side length of a regular hexagon equals the circumradius, so you may simply draw an inscribed regular hexagon and rotate it until one of its sides goes through the given point. Of course this can be done only if the distance of the given point from the center of the circle is $\geq \frac{\sqrt{3}}{2}R$.

Let $P$ be the given point and $O$ the center$^{(*)}$ of the given circle $\Gamma$. You may


*

*Take any $A\in\Gamma$ and construct $B$ as one of the intersections between $\Gamma$ and a circle centered at $A$ with radius $AO$ (now $AB$ equals the radius)

*Construct $C$ as the intersection between $AB$ and a circle centered at $O$ through $P$

*Construct $D$ as one of the intersections between $\Gamma$ and a circle centered at $P$ with radius $AC$.


$DP$ meets $\Gamma$ again at $E$ and $DE$ is a solution.
$(*)$ If $O$ is not previously drawn, you may simply construct it as the intersection of the axis of two different chords. Here an alternative construction which exploits the intersecting chords theorem:



*

*Let $QR$ the diameter of $\Gamma$ through $P$, with $Q$ being closer to $P$ than $R$;

*Let $S$ be one of the intersections between $\Gamma$ and the perpendicular to $OP$ through $P$ ($PS$ is the geometric mean of $PQ$ and $PR$)

*Let $M$ be the midpoint of $OQ$ and $T$ be the intersection, closer to $S$, between the circle centered at $M$ through $O$ and the parallel to $OP$ through $S$

*Let $U$ be the projection of $T$ on $OP$: now the geometric mean of $UQ,UO$ is also the geometric mean of $PQ,PR$ and the arithmetic mean of $UQ,UO$ is half the arithmetic mean of $PQ,PR$, so...

*Draw a circle centered at $P$ with radius $UQ$ and let $V$ be one of the intersections with $\Gamma$. If $VP$ meets $\Gamma$ again at $W$, $VW$ is a chord whose length equals $OQ$.


This construction is a bit more involved than the previous one, but both approaches can be easily modified in order to solve the problem of drawing a chord through $P$ of any allowed length.
A: Let $\Omega$ be the circle.  Let $O$ be its center.  Let $r$ be its radius.  Let $\ell$ be an arbitrary length s.t. $0<\ell <2r$.  Let $P$ be arbitrary point in the plane.  Let $d$ be the distance from $O$ to $P$.  We want to construct points $A,B\in\Omega$ s.t. the chord $AB$ is of length $\ell$ and $P$ is on the line $AB$.  
Suppose that $P$ is closer to $B$ than $A$.  Note that
$$PA\cdot PB=d^2-r^2.$$
(We define $PA$ to be positive always, and $PB$ is negative if $P$ is between $A$ and $B$, otherwise $PB$ is positive.)  We have $PA-PB=\ell$.  If $PB=x$, then $PA=x+\ell$, so
$$x(x+\ell)=d^2-r^2.$$
That is
$$\left(x+\frac{\ell}{2}\right)^2=d^2-r^2+\frac{\ell^2}{4}.$$
Hence
$$x+\frac{\ell}{2}=\pm\sqrt{d^2-r^2+\frac{\ell^2}{4}}.$$
Since $PA=x+\ell\ge 0$, we see that
$$x=\sqrt{d^2-r^2+\frac{\ell^2}{4}}-\frac{\ell}{2}.$$
The points $A$ and $B$ exist iff
$$d^2\geq r^2-\frac{\ell^2}{4}.$$
This guides us to how to construct $A$ and $B$.  Let $\Gamma$ be the circle centred at $O$ with radius $OP$.  Draw an arbitrary chord $CD$ of length $\ell$ on $\Omega$.  Let the line $CD$ intersect $\Gamma$ at $E$ and $F$.  Suppose that $C$ is closer to $E$ and $D$ is closer to $F$.  Let $M$ be the midpoint of $CD$.  We see that
$$ME^2=OE^2-OM^2=OE^2-(OC^2-MC^2)=d^2-r^2+\frac{\ell^2}{4}.$$
Therefore, $$PA=x+\ell=DE.$$
Draw a circle centred at $P$ with radius $DE$.  Then this circle intersects $\Omega$ at two points $A$ and $A'$.  The line $PA$ and $PA'$ meet $\Omega$ again at $B$ and $B'$.  Then $AB=A'B'=\ell$.


