**Given:** Three non-collinear points, P, Q and R.

**To prove:** Only one circle passing through points P, Q, and R.

**Construction:** Join line segments PQ and QR and draw perpendicular bisectors MN and ST on PQ and RQ respectively.

**Proof:**

As O lies on MN, the perpendicular bisector of PQ,

In △OXP and △OXQ, we have

OX=OX [Common]

∠OXP =∠OXQ [both are same right angles]

XP = XQ [MN is the perpendicular bisector]

So,△OXP≅△OXQ [By SAS]

therefore, OP=OQ [C.P.C.T.]

Similarly, △OYQ≅△OYR

and OQ=OR [CPCT]

therefore, OP = OQ = OR = r

Suppose O as the centre and r as the radius, draw a circle C(O, r) which will pass through P, Q and R.

If possible, suppose there is another circle C(O’,r’) passing through P, Q and R. Then O’ will lie on the perpendicular bisector MN and QR.

Since two lines cannot intersect at more than one point, O’ must coincide with O.

Hence, C(O’, r’) = C(O,r)

therefore, there is one and only one circle passing through the three non-collinear points P, Q and R. Proved

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