NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.3 have been provided in accordance with NCERT syllabus and guidelines directed by CBSE. It has been designed by our expert teachers for students, to make it easier for them and prepare for the final exam. Students will be able to do their homework with the help of solutions of class 9 Maths. Also, it is helpful for students to score good marks in the exam.
Access Other Exercise Solutions of Class 9 Maths Chapter 10- Circles
Exercise 10.1 Solutions 2 Question ( 2 Short)
Exercise 10.2 Solutions 2 Question ( 2 long)
Exercise 10.4 Solutions 6 Question ( 6 long)
Exercise 10.5 Solutions 12 Questions (12 long)
Exercise 10.6 Solutions 10 Questions (10 long)
Access Answers to NCERT Class 9 Maths Chapter 10 – Circles Exercise 10.3
1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
In these two circles, no point is common.
Here, only one point “P” is common.
Even here, P is the common point.
Here, two points are common which are P and Q.
No point is common in the above circle.
2. Suppose you are given a circle. Give a construction to find its centre.
The construction steps to find the center of the circle are:
Step I: Draw a circle first.
Step II: Draw 2 chords AB and CD in the circle.
Step III: Draw the perpendicular bisectors of AB and CD.
Step IV: Connect the two perpendicular bisectors at a point. This intersection point of the two perpendicular bisectors is the centre of the circle.
3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
It is given that two circles intersect each other at P and Q.
OO’ is perpendicular bisector of PQ.
Triangle ΔPOO’ and ΔQOO’ are similar by SSS congruency since
OP = OQ and O’P = OQ (Since they are also the radii)
OO’ = OO’ (It is the common side)
So, It can be said that ΔPOO’ ΔQOO’
∴ POO’ = QOO’ — (i)
Even triangles ΔPOR and ΔQOR are similar by SAS congruency as
OP = OQ (Radii)
POR = QOR (As POO’ = QOO’)
OR = OR (Common arm)
So, ΔPOR ΔQOR
∴ PRO = QRO
Also, we know that
PRO+QRO = 180°
Hence, PRO = QRO = 180°/2 = 90°
So, OO’ is the perpendicular bisector of PQ.
This exercise has three questions based on two topics;
- The perpendicular from the Centre to a Chord
- Circle through Three Points
And under these two topics, there are two theorems defined as:
Theorem 1: The perpendicular from the centre of a circle to a chord bisects the chord.
Theorem 2: The line, drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Practice questions for Class 9 Maths Chapter 10, by reaching us at BYJU’S. Students can also reach our other learning resources to Class 9, such as notes, question papers, books, etc. provided here. The questions in Ex.10.3 are based on construction for the given condition. Students have to make use of their geometry box to solve the problems. To solve each problem mentioned in the textbook, go to the NCERT solutions and achieve an excellent score in Maths.