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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.

odesinit
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    The three points have to be all distinct from each other, but I suppose you were assuming that anyway. – David K Nov 24 '18 at 14:26
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    @DavidK And also not aligned. – user Nov 24 '18 at 14:27
  • Yes you are correct David K sorry about that. – odesinit Nov 24 '18 at 14:27
  • I'm bit confused with aligned what does that mean in this context? – odesinit Nov 24 '18 at 14:30
  • https://en.wikipedia.org/wiki/Circumscribed_circle#Straightedge_and_compass_construction – Jean-Claude Arbaut Nov 24 '18 at 14:35
  • Are you also interested in a general proof that given three not alligned points we can find exactly one circle passing throught those points or you are interested solely to the methods to find the centerof the circle? – user Nov 24 '18 at 15:26
  • @odesinit Could you please clarify if you question is about the methods to find the center or also about the existence of the circle? Thanks – user Nov 24 '18 at 16:26
  • Do you know that these points lie on a circle, or are you given three arbitrary points on the plane? – amd Nov 24 '18 at 22:37

3 Answers3

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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.

4

Yes of course, three (not aligned) points determine a circle and then we can always find its center.

See for example

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

user
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    IMO, this is only a restatement of the question. You should prove that the system of equations has always a unique solution. –  Nov 24 '18 at 14:41
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    @YvesDaoust Euclid already proved that for me more that 2000 years ago :) – user Nov 24 '18 at 14:45
  • @YvesDaoust Note also that in the link I given there are many different methods which show how to solve the system. – user Nov 24 '18 at 15:08
  • @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. – user Nov 24 '18 at 15:23
  • @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. – user Nov 24 '18 at 15:27
  • @YvesDaoust Not really kidding! The fact that the circel was already a given makes me propend for a different interpretation, but I'm ready to revise my thoughts as soon as the asker will clarify that point. – user Nov 24 '18 at 15:33
  • @YvesDaoust Maybe the asker was referring to some particular configuration for the three points on the circle, for example three not symmetric points very closer one each other and so on. I really think that the doubt was on that and not on the existence of the circle. But I'm curious to have a confirmation about that. – user Nov 24 '18 at 15:39
  • @YvesDaoust We have discussed a lot about that, I bet that the asker will clarify the point. :) – user Nov 24 '18 at 16:24
  • Yep, you are dodging :-) –  Nov 24 '18 at 16:26
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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.

the_fox
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