How many such circles exist which pass though three given points in 2 dimensions?
Is it one unique circle? or possibly more than one?
Is there any proof?
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You can use Moebius transformations to prove it. And the answer is yes, if you consider the $\infty$ – user40276 Jul 05 '13 at 14:15
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1@user40276 yes to what? one unique circle? or more than one? – Salena Jul 05 '13 at 14:20
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Consider some cases: three points in general position, three points collinear but distinct, or collinear and not distinct. – dtldarek Jul 05 '13 at 14:23
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you can prove it geometrically (by drawing it). – iostream007 Jul 05 '13 at 14:34
4 Answers
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HINT:
Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$
If the circle passes through $A(a,d),B(b,e),C(p,q)$
$a^2+d^2+2g\cdot a+2f\cdot d+c=0$
$b^2+e^2+2g\cdot b+2f\cdot e+c=0$
$p^2+q^2+2g\cdot p+2f\cdot q+c=0$
So, we have three equations with three unknowns $g,f,c$
What can we conclude from here?
lab bhattacharjee
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Three equations, three variables, can have one solution if all equations are distinct, or no solution in case of contradiction, or infinite solutions in case one equation is duplicate. Which one iss it? – Salena Jul 05 '13 at 14:19
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@Salena, have a look into http://mathworld.wolfram.com/CramersRule.html. Can you recognize the co-linearity case? – lab bhattacharjee Jul 05 '13 at 18:49
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If the three points are not colinear, you can prove that there exists an unique circle by proving that it's center lies on two perpendicular bisectors of the segments connecting the tree points.[The existence is done by proving that the circle with the center at the intersection of two perpendicular bisectors and radius the distance to one of the points works; the uniqueness is done by proving that any circle passing through the three points has that center and radius].
If the three points are co linear, there is no circle (unless you consider a line as a circle of infinite radius). This can be proven by showing that if the three points are on a line in the order $A,B,C$ (i.e. $B$ between $A$ and $C$), then for any point in the plane we have
$$OB < OA \mbox{ or } OB <OC \,.$$ To Prove this, just observe that $ \angle OBA+ \angle OBC =180^o$, thus at least one of them is $\geq 90^o$. If it is $\angle OBA \geq 90^o$, then in $\Delta OBA$ the edge $AB$ is the biggest as it opposes the bigger angle.
Added If the three points are not necessarily distinct (which from how the problem is stated I don't think is the case), then it is easy to prove there are infinitely many circles.
Case 1: All three points are the same, then it is easy to construct infinitely many circles.
Case 2: Two points are the same, but third is distinct. Ignore one of the two equal points. Then any circle with the center on the Perpendicular Bisector of the segment connecting the two points, and the right radius will do.
N. S.
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@Salena Basically this is the proof of existance and uniqueness of circumcircle of a triangle... – N. S. Jul 05 '13 at 14:26
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You are missing the case where some points are not distinct, those are collinear, but there are infinitely many circles passing through them. – dtldarek Jul 05 '13 at 15:02
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Unique circle pass through given three non colinear points. To find radius of circle : Draw a trangle by joining three points. Draw line bisector of any two sides of triangle. The point of intersection point of side bisector is center point of circle passing through given three points
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When the 3 points are distinct, the others have already replied regarding a unique circle through them.
When two points coincide, the circle is tangent to a line depending on how a fixed direction point of $ \theta $ approaches x-axis.
Let the points be with coordinates $ O, P,Q $ as $ (-a,0),(0,0), \epsilon ( \cos \theta, \sin \theta ) $ respectively.
When $ \epsilon \rightarrow 0, $ $PQ$ is tangent of circle at $P$ so that
$$ \sin \theta = a/d $$ where $d$ is circle diameter.
If $ O,P,Q $ are collinear and $ \theta = 0$ or $\pi$ then your circle becomes a straight line.
Narasimham
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