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Is it possible to construct the center of a circle from a single chord?

If I construct the perpendicular bisector of the chord and then construct the perpendicular bisector of that, wouldn't the exact point of intersection be the center of the circle?

This is what I imagined.

Lundborg
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    No....the chord may not pass through the center. Of course your bisector passes through the center but you can't tell where. Two chords in general position gets the job done! – lulu Sep 03 '16 at 17:50
  • If you meant: "if I construct the perp. bisector of the cord, form a cord with this perp. bisector ....etc.", then yes: that perp. bisector is going to be a diameter and thus bisecting it you get the circle's center. +1 – DonAntonio Sep 03 '16 at 17:54
  • @DonAntonio that was what I meant yes. – Lundborg Sep 03 '16 at 17:55
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    Then yes: you're correct. Well done. – DonAntonio Sep 03 '16 at 17:55
  • I don't understand. The perpendicular bisector is an (infinite) line. It is not a diameter. – lulu Sep 03 '16 at 17:56
  • @lulu please see the attached picture – Lundborg Sep 03 '16 at 17:59
  • @DonAntonio So there is no chance, with any chord, that this procedure goes wrong? – Lundborg Sep 03 '16 at 17:59
  • Just to give an explicit example: in the Cartesian Plane let $P=(-1,0)$ and $Q=(1,0)$. Then the perpendicular bisector of $PQ$ is the $y$-axis. Those two points lie on infinitely many circles. They lie on $x^2+y^2=1$ for example, but also on $x^2+(y-a)^2= 1+a^2$ for all $a$. – lulu Sep 03 '16 at 18:00
  • @lulu but you have been given a circle already. – Lundborg Sep 03 '16 at 18:01
  • I don't understand that picture at all. One chord does not determine a circle uniquely, as my last comment illustrates. – lulu Sep 03 '16 at 18:01
  • @lulu I'm not trying to determine a circle, I'm trying to determine a center by construction. – Lundborg Sep 03 '16 at 18:01
  • What do you mean you've been given a circle? You never said that. How are you given the circle? If I have other points on it, then I can make several chords. All you need is two non-parallel chords. Then you can intersect the two perpendicular bisectors. Three points is enough. – lulu Sep 03 '16 at 18:01
  • @lulu and A.G this is a task in geometric construction as the tag suggests. Imagine someone has drawn a circle and a chord, I would like to determine where the center of the circle is. No algebra is involved in this. – Lundborg Sep 03 '16 at 18:03
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    @Neutronic As far as I can see it, no. We have the basic theorem in Euclidean geometry: when we have a cord in a circle (= a line segment joining two distinct points of a circle), then the cord's perpendicular bisector passes through the circles center (the other direction is also true, so this is an iff theorem). Thus, as I noted in my first comment, if the cord's perp. bisec. is made into a cord (meaning: take the segment of the perp. bis. joining two different points on the circle), it is always a diameter. – DonAntonio Sep 03 '16 at 18:04
  • As I say: if you have three distinct points $P,Q,R$ on the circle then form the two chords $PQ$ and $PR$. Draw the two perpendicular bisectors...they intersect at the center of the circle. – lulu Sep 03 '16 at 18:04
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    Whoa... The OP has a circle, then draws a chord, then the perpendicular bisector of that chord to get a chord which is a diameter of the circle. Now bisecting the diameter yields the center of the circle. – MaxW Sep 03 '16 at 18:04
  • @MaxW Exactly so. That's precisely how I understood it from the beginning. – DonAntonio Sep 03 '16 at 18:05
  • @DonAntonio Thank you. Please add an answer with you comments so I can accept.

    MaxW thats how I understood Don's first answer aswell but thank you!

     – Lundborg Sep 03 '16 at 18:06

2 Answers2

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Yes definitely. This is because the perpendicular bisector of the chords is the diameter of the circle, and it's bisector will definitely be the centre.

Abhigyan Mohanta
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We have the basic theorem in Euclidean geometry: when we have a cord in a circle (= a line segment joining two distinct points of a circle), then the cord's perpendicular bisector passes through the circles center (the other direction is also true, so this is an iff theorem). Thus, as I noted in my first comment, if the cord's perp. bisec. is made into a cord (meaning: take the segment of the perp. bis. joining two different points on the circle), it is always a diameter

DonAntonio
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