# Length of a Chord of a circle

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.

• There always is such a chord. "Is it possible to draw" is more complicated. If the drawing is to be done with straightedge and compass, the answer is no. Feb 9, 2016 at 15:58
• To make the question more accurate you need to specify that $0<d<some value (say r)$ for example. Feb 9, 2016 at 16:02
• Related perhaps tangentially: Ptolemy's table of chords (See also my answer below.) $\qquad$ Feb 9, 2016 at 16:03
• @AndréNicolas: Constructing length x may not be possible (it's not even very well-posed), but if you have such a length, then constructing a chord of that length is trivial (you just select a point A on the circle, draw a circle of radius x centered at A, and draw a line segment from A to one of the two points where the two circles intersect). Feb 9, 2016 at 17:26
• @ruakh: You are certainly right, if $x$ is given we can construct the chord. Feb 9, 2016 at 17:46

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

Open a compass with the desired length $x$, place the needle on the circumference and draw an arc that intersects the circle. Can you imagine that you'll never meet it ?

Hint:$$\text{ chord length } = 2r\sin\bigg({\frac{\theta}{2}}\bigg)$$ where $r$ is the radius and $\theta$ is the angle subtended at the center by the chord. Note the continuity of the RHS, now use the Intermediate Value Theorem.

Yes. There are several ways to prove it.

From the axioms for Euclidean geometry: draw a circle with radius $x$ and center on your circle; see where the circles intersect. (Euclid's axioms are a little incomplete - they may not actually guarantee the intersection.)

An argument from continuity: start at one end of the diameter of your circle and move around to the other end. The length of the chord varies continuously, so takes on every value between $d$ and $0$. This argument depends on understanding enough of the structure of the real numbers.

• You actually need radius $x$ - the two intersection points won't form a diameter of the (second) circle, so they won't be distance $(x/2)+(x/2)=x$ apart. Feb 9, 2016 at 16:02
• @StevenStadnicki Fixed thanks. Feb 9, 2016 at 16:03

The answer depends on which axioms you have. If you have any axioms that give you a mesure such that for every real positive number exist a line of that lenght the answer is yes. You can see Hilbert geometry (moder versione of euclidean geometry) and his axioms about continuity and mesure

Consider a point $P$ on the circle and the diameter going through this point. Let $N$ the other end of the diameter. Let $M=N$. The angle between $\vec{PN}$ and $\vec{PM}$ is zero and the length of the chord is $d$. now if you increase this angle continuously from $0$ to $\frac{\pi}{2}$ by moving $M$ along the circle, the chord will go from $d$ to $0$ and it'll take all the values between $0$ and $d$ because the length of the chord is a continuous function of the angle.

Let $O$ be the center of the circle. Take a point $B$ at distance $b$ from O, with $0<b<r=d/2$.

Let the line thru $B,$ perpendicular to the segment $OB,$ meet the circle at points A and C.

We have $AO=CO=r=d/2.$ The triangles $ABO, CBO$ are right-angled at $B.$ So the length of the chord $AC$ is $$AB+BC=\sqrt {r^2-b^2}+\sqrt {r^2-b^2}=2\sqrt {r^2-b^2}=d\sqrt {1-(b/d)^2}.$$ The number $\sqrt {1-(b/d)^2}$ can be any value between $0$ and $1,$ depending on $b,$ so the length $AC$ can be anything between $0$ and $d$.