Question

prove that only one circle passes through three noncollinear points

Answer 1

Theorem: There is one and only one circle passing through three given non-collinear points.Given: Three non collinear points P, Q and RTo prove: There is one and only one circle passing through the points P, Q and R.Construction: Join PQ and QR.Draw perpendicular bisectors AB of PQ and CD of QR. Let the perpendicular bisectors intersect at the point O.Now join OP, OQ and OR.A circle is obtained passing through the points P, Q and R.Proof: We know that, each and every point on the perpendicular bisector of a line segment is equidistant from its ends points.Thus, OP = OQ  [Since, O lies on the perpendicular bisector of PQ]and OQ = OR.  [Since, O lies on the perpendicular bisector of QR]So,  OP = OQ = OR.Let OP = OQ = OR =  r.Now, draw a circle C(O,  r) with O as centre and  r  as radius.Then, circle C(O,  r) passes through the points P, Q and R.Next, we prove this circle is the only circle passing through the points P, Q and R.If possible, suppose there is a another circle C(O′,  t) which passes through the points P, Q, R.Then, O′ will lie on the perpendicular bisectors AB and CD.But O was the intersection point of the perpendicular bisectors AB and CD.So,  O ′ must coincide with the point O.  [Since, two lines can not intersect at more than one point]As, O′P =  t  and OP =  r; and O ′ coincides with O, we get t = r .Therefore, C(O,  r) and C(O,  t) are congruent.Thus, there is one and only one circle passing through three the given non-collinear points.

Answer 2

Statement : There is one and only one circle passing through three given noncollinear points.

Given : AB and CD are two equal chords of the circle.

OM and ON are perpendiculars from the centre at the chords AB and CD.

To prove : OM = ON.
Construction : Join OA and OC.

Proof :

In ΔAOM and ΔCON,

OA = OC . (radii of the same circle)
MA = CN . (since OM and ON are perpendicular to the chords and it bisects the chord and AM = MB, CN = ND)

∠OMA = ∠ONC = 90°
ΔAOM ≅ ΔCON (R. H. S)
OM = ON (c. p. c. t.)

Equal chords of a circle are equidistant from the centre.