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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