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Circle through Three Points

Theorem 3 : One and only one circle can be drawn through three non-collinear points.
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 : Draw line segments PQ and QR. Draw perpendicular bisectors MN and ST of PQ and RQ respectively. Since P, Q, R are not collinear, MN is not parallel to ST and will intersect, at the point O. Join OP, OQ and OR (Fig,).


Proof :
As O lies on MN, the perpendicular bisector of PQ,

   OP = OQ
Similarly, QO = RO

  OP = OQ = OR = r(say)

Taking O as the centre and r as the radius, draw a circle C(O, r) which will pass through P, Q and R. This proves that, there is a circle passing through the points P, Q and R. If possible, suppose there is another circle C(O', r') passing through P, Q and R. Then 'O' will lie on the perpendicular bisector MN of PQ and ST of QR. Since two lines cannot intersect at more than one point, O' must coincide with O. Since OP = r, O'P = r'.
We have, r = 'r
Hence, C(O', r') = C(O, r)
If follows that, there is a unique circle passing through the three vertices P, Q, R of
This circle is called the circum-circle of the triangle PQR and its centre is called cirum-centre of

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