NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

# NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 10 Circles Ex 10.3.

Ex 10.3 Class 9 Maths Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution.
Different pairs of circles are given below:
(i)
If they intersect each other, two points are common.

(ii) If they touch each other externally or internally, only one point is common.
(iii) If they do not intersect or touch each other, no points are common.
From figures, it is obvious that these pairs may have 0 or 1 or 2 points in common. Hence, a pair of circles cannot intersect each other at more than two points.

Hence, the maximum number of common points is two.

Ex 10.3 Class 9 Maths Question 2.
Suppose you are given a circle. Give a construction to find its centre.

Solution.

Steps of Construction:

1. Take three points P, Q and R on the circle.
2. Join PQ and QR.
3. Draw MT and NS, which are respectively the perpendicular bisectors of PQ and QR and intersecting each other at a point O.

Hence, O is the centre of the circle.

Ex 10.3 Class 9 Maths Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:
We have two circles with centres O and O’, intersecting at A and B.
AB is the common chord of two circles and OO’ is the line segment joining their centres.
Let OO’ and AB intersect each other at M.

To prove that OO’ is the perpendicular bisector of AB, we join OA, OB, O’A and O’B. Now, in ∆OAO’ and ∆OBO’, we have
OA = OB    [Radii of the same circle]
O’A = O’B  [Radii of the same circle]
OO’ = OO’ [Common]
∆OAO’ ∆OBO’    [By SSS congruence rule]
1 = 2                 [C.P.C.T.]
Now, in ∆AOM and ∆BOM, we have
OA = OB    [Radii of the same circle]
OM = OM [Common]
1 = 2    [Proved above]
∆AOM = ∆BOM       [By SAS congruence rule]

3 = 4                  [C.P.C.T.]
But
3 + 4 = 180°   [Linear pair]
∴ ∠3 = 4 = 90°
AM OO’
Also, AM = BM           [C.P.C.T.]
M is the mid-point of AB.
Thus, OO’ is the perpendicular bisector of AB.

NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12