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PLZ PLZ EXPALIN :THROUGH THREE NON - COLLINEAR POINTS ONE AND ONLY ONE CIRCLE CAN PASS
MTAHS CLASS 9 CBSE
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Theorem: There is one and only one circle passing through three given 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: 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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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.
I hope it is helped you
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