# prove that There is one and only one circle passing through 3 given non-collinear points Given: A, B and C are three non-collinear points.

To prove: There is one and only one circle passing through A,B and C.

Construction: Join AB and BC. Draw the perpendicular bisectors RS and PQ of the chords AB and BC respectively.

Let PQ and RS intersect in O. Joint OA, OB and OC.

Proof:

O lies on the perpendicular bisector of AB.

∴ OA = OB

Again, O lies on the perpendicular bisector BC.

∴ OB = OC

Thus, OA = OB = OC = r  (Say)

Taking O as centre, draw a circle of radius r. C(0, r) passes through A, B and C. Thus, a circle passes through the point A, B and C.

If possible, suppose there is another circle with centre O' and radius r, passing through A, B and C. Then, O' will lie on the perpendicular bisector PQ and RS.

We know that, two lines cannot intersect at more than one point, So O' must coincide with O.

Hence, there is one and only one circle passing through three non collinear points.

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given that,(i) A,B,C are non collinear point

(ii)O is the CENTER of circle

to prove; only one circle can pass through 3 non collinear points.

construction : join AB and BC  ,   draw a perpendicular bisector of AB name it MN

draw a perpendicular bisector of BC name it SR

now, we can see MN and SR meets at O

LET TAKE. tri> AMO and tri>OMB

OM=OM(common)

AM=BM(GIVEN)

,<OMA=<OMB(90)

...TRI<AMO=TRI<OMB   so,AO=BO(by CPCT)

SIMILARLY, TRI<BOS=TRI<COS(by SAS)

so, BO=OC

ALL NON-COLLINEAR POINTS JOINS O, MAKING A CIRCLE

HENCE PROVED..................

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