Prove that there is only one Circle passing through 3 non colinear points
To prove the said statement, consider three non collinear points A, B and C lying on a plane. Let the perpendicular bisectors of AB and BC intersect at O.
To prove that a circle passes through these three points A, B and C
Join OA, OB and OC
In ∆OMA and ∆OMB
OM = OM (common)
∠OMA = ∠OMB = 90° (OM ⊥ AB)
AM = MB (by construction)
∴ ∆OMA ≅ ∆OMB (SAS congruence)
⇒ OA = OB … (1)
Similarly by taking congruence between the triangles ONB and ONC, it is obtained
OB = OC … (2)
From (1) and (2)
OA = OB = OC
This shows that on taking O as centre and radius as OA or OB or OC a circle can be drawn which passes through A, B and C. This circle is called the circum circle of triangle ABC.
Now, it is required to prove that this circle is unique.
It is known that two lines may intersect maximum at one point.
So, the perpendicular bisectors of AB and BC intersect maximum at one point
Since they have already intersected at a point O so, there is no chance of intersection of these two lines more
So, the circum circle is unique
This shows that the circle is unique.