Question

Prove that, There is one and only one circle passing through three given non-collinear points.

Answer 1

Hint:
According to given statement, take three non collinear points on the circle i.e. A, B and C. Draw the perpendicular bisector of the points and join all the points which are required to prove the statement.

Complete step by step solution:

It is given that there are three non-collinear points in a circle that is A, B and C.
We have to prove that there is one and only one circle passing through the points A, B and C.

Construction: As we have marked three non-collinear points in circle. Now, let’s join AB and BC. Then, Draw the perpendicular bisectors of RS and PQ of the chords AB and BC respectively as shown in figure.

Let us assume PQ and RS intersect in O. Here, Join OA, OB and OC.

Let’s proof: The point O lies on the perpendicular bisector of AB.

Therefore, OA = OB

And also, O lies on the perpendicular bisector of BC.

Therefore, OB = OC

Thus, we can say from above OA = OB = OC = r (where r stands for radius of the circle).
By taking O as the centre, construct a circle of radius r \[C\left( {0,\;r} \right)\] which passes through the points A, B and C.
Let us suppose there is another circle with the centre \[O'\] and radius r, passing through the A, B and C. Then, \[O'\] will also lie on the perpendicular bisector of PQ and RS.
As we know that, the two lines cannot intersect at more than one point. So, \[O'\] must coincide with O which means \[O'\] and O lies at the same point.

Hence, there is one and only one circle passing through three non collinear points that are A, B and C.
Hence Proved.

Note:
To solve these types of questions, you must construct a figure to solve the question. Here, you can also suppose or can take assumptions for better understanding in a practical way.