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.