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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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