circlesgeometry
I was wondering about the possible values that the length of a chord of a circle can take. The Length of a chord is always greater than or equal to 0 and smaller than or equal to the diameter (sayd). Is it possible to draw a chord of length 'x', where x could be any real number? In other words, does the chord of circle take all the possible real values between 0andd?
Any help/comments would be much appreciated.
We can apply the Intersecting Chords Theorem.
You chord length is the length $UV$ and the segment height is the length $PX$.
The intersecting chords theorem tells us that $XP \times XQ = XU \times XV$.
Let $\ell = UV$ and $h=XP$. It follows that $UX = XV = \tfrac{1}{2}\ell$. The ICT then tells us that
$$\tfrac{1}{2}\ell \times \tfrac{1}{2}\ell = h \times XQ \, ,$$ i.e. $XQ = \tfrac{1}{4h}\ell^2$. The diameter $PQ=PX+XQ$ and $$PX + XQ = h + \frac{\ell^2}{4h}=\frac{4h^2+\ell^2}{4h}$$
The radius is then one half of this, i.e.
$$CQ = \frac{4h^2+\ell^2}{8h} \, . $$
HINT:-
Take a circle, let its radius be $r$.
Take any point in the circle. It doesn't matter what point you take for the first point.
Now as soon as you choose the second point, you have a chord. Let the two points of the chord make angle $2\theta$ at centre. So the length of the chord is $2\cdot r\cdot \sin\theta$.
Now the given constraint is $\frac{2}{3}\le2\sin\theta \le \frac{5}{6}$.
Now this can be solved easily.
Best Answer
This comes down to the intermediate value theorem.
On the circle $x^2+y^2=1$, the chord from $(1,0)$ to $(\cos\theta,\sin\theta)$ has length $2\sin\dfrac\theta2$. You can see that by means of the usual "distance formula".
As $\theta$ goes from $0$ to $\pi$ (or from $0^\circ$ to $180^\circ$ if you like), the chord goes from $0$ to $2$ and the chord is a continuous function of $\theta$. The fact that it's continuous means you can apply the intermediate value theorem and see that it assumes all intermediate values.
If you don't like transcendental functions (perhaps because proving continuity of those takes a lot of work), you can also do it like this: the point $$ \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1-t^2} \right) \tag 1 $$ goes around the circle from $(-1,0)$ back to $(-1,0)$ as $t$ goes from $-\infty$ to $+\infty$. The length of the chord from $(1,0)$ to the point in $(1)$ can also be found via the distance formula and the same kind of argument can be used.
Tangentially (no pun intended) related is this: Ptolemy's table of chords