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Theorem – 5 There is one and only one circle passing through three given non-collinear points.

Given : - A, B and C are three non-collinear points.

To Prove :One and only one circle will pass through A, B and C.

Construction : AB and BC are joined. PQ and RS are drawn perpendicular bisector of AB and BC which intersect at O

Proof :

Statements	Reason
1. O lies on the perpendicular bisector of AB.	1. PQ is perpendicular bisector of AB, every point on PQ is equidistant
2. Every point equidistant from B and C lies on RS which is the perpendicular under bisector of BC OB = OC	2. Every point of the perpendicular bisector is equidistant from B and C
3. OA = OB = OC O is equidistant from A, B and C.	3. From (1) and (2)

Hence, if a circle is drawn with centre O and radius OA will pass through B and C.

Since O is the only one point on the bisector of AB and BC, hence only one circle will pass through A, B and C.