To prove that "one and only one circle can pass through three non-collinear points"
Non-collinear points:-
Non-collinear means simply ‘does not lies on a common line’, so it is a property of a set of points (at least two) of a geometry.
We join these three points with line. After joining these points when we draw a perpendicular bisector to each of these lines.
When perpendicular bisectors are drawn they cut each other and from that point the length of each line is equal. This line is the radius of the circle,hence a circle drawn on these non-collinear points.
When there are three non-collinear points (not lying on a single line), then only one circle can be drawn on which all these points lie. But, if we try to draw any other circle on these points then it would also be of the same radius.
Its derivation is given below,
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 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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