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Let A, B and C be 3 noncollinear points.
Join AB and BC.
Draw the perpendicualr bisector to AB (1)
Draw the perpendicular bsector to BC (2)
Let the two perpendicular bisectors in (1) and (2) meet at O.
If X and Y are mid points of the AB and BC , then XO and YO become the perpendicular bisectors of AB and BC.
Consider the triangles OXA and OXB which are congruent , as
XA = XB . OX is perpendicular bisector to AB, by constuction.
angle OXA = angle OXB,. Right angles, by construction.
XO =XO , the common side to both triangles.
Threfore OA =OB. (3)
|||ly Taking triangles OYB and OYC we can prove that they are congruent.
So, OB=OC (4)
Therefore, from (3) and (4):
OA = OB = OC = r say.
Therefore, a circle passes through the A, B and C whose centre is O and whose radius is r =OA = OB =OC.
The circle is unique as the point O is unique and the radius r is unique.
Now if you join OZ, Z being the mid point of the third side AC, then,by the Circle theorem:The perpendicular bisector of any chord passes through the centre of the circle.
So, ZO is the line on which the centre lies, OA=OC as proved ealier, O is the centre of the circle for the chord AC also.
The circle is unique as centre O and the radius r= OA=OB=OC being unique, i.e, with a fixed centre and radius we can draw only oneand only one circle.
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