Answer:

Step-by-step explanation:

To 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.

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