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

To prove: A circle passes through these three points P, Q and R, and such circle is one and only one.

Construction: Join PQ and QR. Their perpendicular bisectors AL and BM intersect at O. 

Join OP, OQ and OR. 

Proof: ∵ Point O is at the perpendicular bisector of chord PQ. 

∴ OP = OQ …(i) 

Similarly O is at the perpendicular bisector of chord QR 

=> OQ = OR …(ii) 

From Eqn. (i) & (ii) 

OP = OQ = OR = r (Let) 

Now taking O as a centre and r radius if we draw a circle it will pass through all three points P, Q and R, i.e., P, Q and R exist on the circumference of the circle. 

Now let another circle be (O’, s) which pass through points P, Q and R and perpendicular bisectors of PQ and QR i.e., AL and BM passes through the centre O’. 

But the intersection point of AL and BM is O. i.e. 

O’ and O coincide each other or O and O’ are the same point 

∴ OP = r and OP’ = s and O and O’ are coincide => r = s. 

=> C (O, r) = C (O’, s) 

=> There is one and only one circle through which three non-collinear points P, Q and R pass. 

=> There is one and only one circle passing through three given non-collinear points.