Theorem 10.5 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 10.5 class 9 Maths which is related to Chapter 10 Circles Class 9 Mathematics. After understanding theorem 10.5, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 Theorem 10.5 Class 9

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

Given

Three non-collinear points, A B and C.

To prove

There is one and only one circle passing through A, B, and C.

Construction

Join AB and BC. Draw PQ and RS as the perpendicular bisectors of AB and BC respectively.

 Let these perpendicular bisectors intersect at O. Join OA, OB, and OC.

Proof

Point O lies on the perpendicular bisector PQ of AB.

OA=OB……………. (i)

Similarly, point O lies on the perpendicular bisector RS of BC.

OB=OC………………(ii)

From equations (i) and (ii), we have

OA=OB=OC=  r

Draw a circle with center O and radius OA= r. It will also pass through points B and C.

This indicates that there is a circle passing through three points A, B, and C.

We know that two lines can intersect at one point, so we can draw only one circle with a radius of OA.

Hence, there is one and only one circle passing through A, B, and C.

 

 Related Topics

1. Theorem 10.6

2. Theorem 10.4