Answer :


Given: Three non collinear 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,


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 cannot 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.


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