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Question

Prove that there is one and only one circle passing through three non-collinear points.

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Solution

Given: Three non-collinear points P, Q and R
To prove: There is one and only one circle passing through the points P, Q and R.
Construction: Join PQ and QR.
Draw the 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 that passes 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 end points.
Thus, OP = OQ (∵ O lies on the perpendicular bisector of PQ)
Also, OQ = OR (∵ O lies on the perpendicular bisector of QR)
∴ OP = OQ = OR
Let OP = OQ = OR = r (Radius of a circle)
Now, draw a circle C(O, r) with O as the centre and r as the radius.
Then, circle C(O, r) passes through the points P, Q and R.
Now, we have to show that this is the only circle that passes through P, Q and R.

Let us suppose that there is an another circle C(O′, t) that passes through P, Q and R.
Then O′ will lie on the perpendicular bisectors AB and CD.
But O is the point of intersection of the perpendicular bisectors AB and CD.
∴ 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.
∴ C(O, r) and C(O, t) are congruent.
Hence, there is one and only one circle passing through three given non-collinear points.

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