Question

prove that ,any three points on a circle cannot be collinear

Answer 1

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.

Answer 2

Answer:

Let’s us call the centre of the circle O and the three points on the circle A, B, C

Step-by-step explanation:

Now, OA = OB = OC = radius of the circle.

If the three points A,B and C were on a straight line then they cannot be equidistant from the point O.

Hence they are not collinear.