Prove that there is one and only one circle passing through three given 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)
OQ = OR (Since 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 center and r as radius.
Then, circle C(O, r) passes through the points P, Q and R.
We have to show that this circle is the only circle passing through the points P, Q and R.
If possible, suppose there is another circle C(O′, t) which passes through the points P, Q and 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.
∴ O ′ must coincide with the point O. (Since, two lines can not intersect at more than one point)
As, O′P = t , OP = r and O ′ coincides with O,t = r
∴ C(O, r) and C(O', t) are congruent.
Hence, there is one and only one circle passing through the 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)
OQ = OR (Since 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 center and r as radius.
Then, circle C(O, r) passes through the points P, Q and R.
We have to show that this circle is the only circle passing through the points P, Q and R.
If possible, suppose there is another circle C(O′, t) which passes through the points P, Q and 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.
∴ O ′ must coincide with the point O. (Since, two lines can not intersect at more than one point)
As, O′P = t , OP = r and O ′ coincides with O,t = r
∴ C(O, r) and C(O', t) are congruent.
Hence, there is one and only one circle passing through the given non-collinear points.
