Centre of a circle [closed]
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I was wondering whether if given any three point ( all distinct from each other) on the circumference of circle can I always determine the centre of the a circle. If not what scenarios would this not be applicable.
geometry circle
asked Nov 24 '18 at 14:14
odesinitodesinit
346
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closed as off-topic by user21820, amWhy, Saad, José Carlos Santos, Yves Daoust Nov 24 '18 at 23:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, Saad, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
$begingroup$
I was wondering whether if given any three point ( all distinct from each other) on the circumference of circle can I always determine the centre of the a circle. If not what scenarios would this not be applicable.
geometry circle
asked Nov 24 '18 at 14:14
odesinitodesinit
346
$endgroup$
closed as off-topic by user21820, amWhy, Saad, José Carlos Santos, Yves Daoust Nov 24 '18 at 23:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, Saad, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
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– David K
Nov 24 '18 at 14:26
2
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@DavidK And also not aligned.
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– gimusi
Nov 24 '18 at 14:27
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Yes you are correct David K sorry about that.
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– odesinit
Nov 24 '18 at 14:27
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I'm bit confused with aligned what does that mean in this context?
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– odesinit
Nov 24 '18 at 14:30
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35
|
show 3 more comments
$begingroup$
I was wondering whether if given any three point ( all distinct from each other) on the circumference of circle can I always determine the centre of the a circle. If not what scenarios would this not be applicable.
geometry circle
asked Nov 24 '18 at 14:14
odesinitodesinit
346
$endgroup$
I was wondering whether if given any three point ( all distinct from each other) on the circumference of circle can I always determine the centre of the a circle. If not what scenarios would this not be applicable.
geometry circle
geometry circle
asked Nov 24 '18 at 14:14
odesinitodesinit
346
asked Nov 24 '18 at 14:14
odesinitodesinit
346
edited Nov 24 '18 at 14:28
odesinit
asked Nov 24 '18 at 14:14
odesinitodesinit
346
asked Nov 24 '18 at 14:14
odesinitodesinit
346
asked Nov 24 '18 at 14:14
odesinitodesinit
346
346
closed as off-topic by user21820, amWhy, Saad, José Carlos Santos, Yves Daoust Nov 24 '18 at 23:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, Saad, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, amWhy, Saad, José Carlos Santos, Yves Daoust Nov 24 '18 at 23:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, Saad, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
$endgroup$
– David K
Nov 24 '18 at 14:26
2
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@DavidK And also not aligned.
$endgroup$
– gimusi
Nov 24 '18 at 14:27
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Yes you are correct David K sorry about that.
$endgroup$
– odesinit
Nov 24 '18 at 14:27
$begingroup$
I'm bit confused with aligned what does that mean in this context?
$endgroup$
– odesinit
Nov 24 '18 at 14:30
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35
|
show 3 more comments
2
$begingroup$
The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
$endgroup$
– David K
Nov 24 '18 at 14:26
2
$begingroup$
@DavidK And also not aligned.
$endgroup$
– gimusi
Nov 24 '18 at 14:27
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Yes you are correct David K sorry about that.
$endgroup$
– odesinit
Nov 24 '18 at 14:27
$begingroup$
I'm bit confused with aligned what does that mean in this context?
$endgroup$
– odesinit
Nov 24 '18 at 14:30
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35
2
2
$begingroup$
The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
$endgroup$
– David K
Nov 24 '18 at 14:26
$begingroup$
The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
$endgroup$
– David K
Nov 24 '18 at 14:26
2
2
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@DavidK And also not aligned.
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– gimusi
Nov 24 '18 at 14:27
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@DavidK And also not aligned.
$endgroup$
– gimusi
Nov 24 '18 at 14:27
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Yes you are correct David K sorry about that.
$endgroup$
– odesinit
Nov 24 '18 at 14:27
$begingroup$
Yes you are correct David K sorry about that.
$endgroup$
– odesinit
Nov 24 '18 at 14:27
$begingroup$
I'm bit confused with aligned what does that mean in this context?
$endgroup$
– odesinit
Nov 24 '18 at 14:30
$begingroup$
I'm bit confused with aligned what does that mean in this context?
$endgroup$
– odesinit
Nov 24 '18 at 14:30
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35
|
show 3 more comments
3 Answers
3
active
oldest
votes
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Yes, if the three points are indeed on a circle (i.e. not aligned) and are distinct, you can always retrieve the center.
The center is the intersection of the bisectors of the points, in pairs. The bisector of two distinct points can always be constructed, and the bisectors can only be parallel if the points are aligned.
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
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add a comment |
$begingroup$
Two points $A$, $B$ on a circle determine the chord $AB$. The perpendicular bisector of $AB$ goes through the centre of the circle. Having a third point on the circle gives you another chord, say $AC$ or $BC$. The perpendicular bisector of this second chord also goes through the centre of the circle. It follows that the centre is the point of intersection of those two perpendicular bisectors.
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
$endgroup$
add a comment |
$begingroup$
Yes of course, three (not aligned) points determine a circle and then we can always find its center.
See for example
- Equation of circle passing through 3 given points
and once we have the equation in the form $x^2+y^2+ax+by+c=0$, we can determine the center completing the square and reducing to the form $(x-x_C)^2+(y-y_C)^2=R^2$.
Refer also to the related
- Get the equation of a circle when given 3 points
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
$endgroup$
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
|
show 4 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, if the three points are indeed on a circle (i.e. not aligned) and are distinct, you can always retrieve the center.
The center is the intersection of the bisectors of the points, in pairs. The bisector of two distinct points can always be constructed, and the bisectors can only be parallel if the points are aligned.
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
$endgroup$
add a comment |
$begingroup$
Yes, if the three points are indeed on a circle (i.e. not aligned) and are distinct, you can always retrieve the center.
The center is the intersection of the bisectors of the points, in pairs. The bisector of two distinct points can always be constructed, and the bisectors can only be parallel if the points are aligned.
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
$endgroup$
add a comment |
$begingroup$
Yes, if the three points are indeed on a circle (i.e. not aligned) and are distinct, you can always retrieve the center.
The center is the intersection of the bisectors of the points, in pairs. The bisector of two distinct points can always be constructed, and the bisectors can only be parallel if the points are aligned.
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
$endgroup$
Yes, if the three points are indeed on a circle (i.e. not aligned) and are distinct, you can always retrieve the center.
The center is the intersection of the bisectors of the points, in pairs. The bisector of two distinct points can always be constructed, and the bisectors can only be parallel if the points are aligned.
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
edited Nov 24 '18 at 15:33
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
answered Nov 24 '18 at 14:31
Yves DaoustYves Daoust
128k674227
128k674227
add a comment |
add a comment |
$begingroup$
Two points $A$, $B$ on a circle determine the chord $AB$. The perpendicular bisector of $AB$ goes through the centre of the circle. Having a third point on the circle gives you another chord, say $AC$ or $BC$. The perpendicular bisector of this second chord also goes through the centre of the circle. It follows that the centre is the point of intersection of those two perpendicular bisectors.
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
$endgroup$
add a comment |
$begingroup$
Two points $A$, $B$ on a circle determine the chord $AB$. The perpendicular bisector of $AB$ goes through the centre of the circle. Having a third point on the circle gives you another chord, say $AC$ or $BC$. The perpendicular bisector of this second chord also goes through the centre of the circle. It follows that the centre is the point of intersection of those two perpendicular bisectors.
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
$endgroup$
add a comment |
$begingroup$
Two points $A$, $B$ on a circle determine the chord $AB$. The perpendicular bisector of $AB$ goes through the centre of the circle. Having a third point on the circle gives you another chord, say $AC$ or $BC$. The perpendicular bisector of this second chord also goes through the centre of the circle. It follows that the centre is the point of intersection of those two perpendicular bisectors.
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
$endgroup$
Two points $A$, $B$ on a circle determine the chord $AB$. The perpendicular bisector of $AB$ goes through the centre of the circle. Having a third point on the circle gives you another chord, say $AC$ or $BC$. The perpendicular bisector of this second chord also goes through the centre of the circle. It follows that the centre is the point of intersection of those two perpendicular bisectors.
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
answered Nov 24 '18 at 14:27
the_foxthe_fox
2,89021537
2,89021537
add a comment |
add a comment |
$begingroup$
Yes of course, three (not aligned) points determine a circle and then we can always find its center.
See for example
- Equation of circle passing through 3 given points
and once we have the equation in the form $x^2+y^2+ax+by+c=0$, we can determine the center completing the square and reducing to the form $(x-x_C)^2+(y-y_C)^2=R^2$.
Refer also to the related
- Get the equation of a circle when given 3 points
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
$endgroup$
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
|
show 4 more comments
$begingroup$
Yes of course, three (not aligned) points determine a circle and then we can always find its center.
See for example
- Equation of circle passing through 3 given points
and once we have the equation in the form $x^2+y^2+ax+by+c=0$, we can determine the center completing the square and reducing to the form $(x-x_C)^2+(y-y_C)^2=R^2$.
Refer also to the related
- Get the equation of a circle when given 3 points
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
$endgroup$
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
|
show 4 more comments
$begingroup$
Yes of course, three (not aligned) points determine a circle and then we can always find its center.
See for example
- Equation of circle passing through 3 given points
and once we have the equation in the form $x^2+y^2+ax+by+c=0$, we can determine the center completing the square and reducing to the form $(x-x_C)^2+(y-y_C)^2=R^2$.
Refer also to the related
- Get the equation of a circle when given 3 points
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
$endgroup$
Yes of course, three (not aligned) points determine a circle and then we can always find its center.
See for example
- Equation of circle passing through 3 given points
and once we have the equation in the form $x^2+y^2+ax+by+c=0$, we can determine the center completing the square and reducing to the form $(x-x_C)^2+(y-y_C)^2=R^2$.
Refer also to the related
- Get the equation of a circle when given 3 points
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
edited Nov 24 '18 at 14:26
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
answered Nov 24 '18 at 14:21
gimusigimusi
92.8k84494
92.8k84494
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
|
show 4 more comments
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
1
1
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
$begingroup$
IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution.
$endgroup$
– Yves Daoust
Nov 24 '18 at 14:41
1
1
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Euclid already proved that for me more that 2000 years ago :)
$endgroup$
– gimusi
Nov 24 '18 at 14:45
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system.
$endgroup$
– gimusi
Nov 24 '18 at 15:08
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust By the same approaches we can generalize the result. Anyway since the question was on how to find the center, I was focused to give some hint on that and not for a general proof for a well known result by Euclidean geometry. I can ask if the asker is also interested in that proof.
$endgroup$
– gimusi
Nov 24 '18 at 15:23
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
$begingroup$
@YvesDaoust Note that he is stating "given any three point ( all distinct from each other) on the circumference of circle..." then the existence of the circle is out of the dicussion. That was my interpretation of the OP. I've just asked for a specific clarification about that to the asker.
$endgroup$
– gimusi
Nov 24 '18 at 15:27
|
show 4 more comments
2
$begingroup$
The three points have to be all distinct from each other, but I suppose you were assuming that anyway.
$endgroup$
– David K
Nov 24 '18 at 14:26
2
$begingroup$
@DavidK And also not aligned.
$endgroup$
– gimusi
Nov 24 '18 at 14:27
$begingroup$
Yes you are correct David K sorry about that.
$endgroup$
– odesinit
Nov 24 '18 at 14:27
$begingroup$
I'm bit confused with aligned what does that mean in this context?
$endgroup$
– odesinit
Nov 24 '18 at 14:30
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Jean-Claude Arbaut
Nov 24 '18 at 14:35