Construct circle center from chord

1

$begingroup$

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?

geometric-construction

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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!
  $endgroup$
  – lulu
  Sep 3 '16 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:54
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then yes: you're correct. Well done.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MaxW
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  

 | 
show 13 more comments

1

$begingroup$

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?

geometric-construction

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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!
  $endgroup$
  – lulu
  Sep 3 '16 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:54
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then yes: you're correct. Well done.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MaxW
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  

 | 
show 13 more comments

1

1

1

$begingroup$

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?

geometric-construction

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

$endgroup$

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?

geometric-construction

geometric-construction

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

share|cite|improve this question

share|cite|improve this question

edited Sep 3 '16 at 17:59

Lundborg

edited Sep 3 '16 at 17:59

Lundborg

edited Sep 3 '16 at 17:59

Lundborg

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

asked Sep 3 '16 at 17:49

LundborgLundborg

860416

860416

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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!
  $endgroup$
  – lulu
  Sep 3 '16 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:54
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then yes: you're correct. Well done.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MaxW
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  

 | 
show 13 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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!
  $endgroup$
  – lulu
  Sep 3 '16 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:54
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Then yes: you're correct. Well done.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – DonAntonio
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – MaxW
  Sep 3 '16 at 18:04
  
  
  

  
  
  
  

1

1

$begingroup$
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!
$endgroup$
– lulu
Sep 3 '16 at 17:50

$begingroup$
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!
$endgroup$
– lulu
Sep 3 '16 at 17:50

$begingroup$
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
$endgroup$
– DonAntonio
Sep 3 '16 at 17:54

$begingroup$
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
$endgroup$
– DonAntonio
Sep 3 '16 at 17:54

1

1

$begingroup$
Then yes: you're correct. Well done.
$endgroup$
– DonAntonio
Sep 3 '16 at 17:55

$begingroup$
Then yes: you're correct. Well done.
$endgroup$
– DonAntonio
Sep 3 '16 at 17:55

1

1

$begingroup$
@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.
$endgroup$
– DonAntonio
Sep 3 '16 at 18:04

$begingroup$
@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.
$endgroup$
– DonAntonio
Sep 3 '16 at 18:04

1

1

$begingroup$
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.
$endgroup$
– MaxW
Sep 3 '16 at 18:04

$begingroup$
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.
$endgroup$
– MaxW
Sep 3 '16 at 18:04

 | 
show 13 more comments

2 Answers
2

active

oldest

votes

0

$begingroup$

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

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1913395%2fconstruct-circle-center-from-chord%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

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

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

$endgroup$

add a comment | 

0

0

0

$begingroup$

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

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

$endgroup$

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

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

share|cite|improve this answer

share|cite|improve this answer

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

answered Sep 3 '16 at 18:06

DonAntonioDonAntonio

179k1494230

179k1494230

add a comment | 

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

$endgroup$

add a comment | 

1

1

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

$endgroup$

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.

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

share|cite|improve this answer

share|cite|improve this answer

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

answered Jan 16 at 0:17

Abhigyan MohantaAbhigyan Mohanta

111

111

add a comment | 

add a comment | 

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1913395%2fconstruct-circle-center-from-chord%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here