9th Maths 10.3
NCERT Class 9th solution of Exercise 10.1
NCERT Class 9th solution of Exercise 10.2
NCERT Class 9th solution of Exercise 10.4
Exercise 10.3
Q1. Draw different pairs of circles. How many points does each pair have in common? what is the maximum number of common points?
Sol. :
Different pairs of circles
Number of common points
(a) 0 (zero)
(b) 1 (one) A,
(c) 2 (two) A,B
Answer:
The maximum number of common points is 2.
Q2. Suppose you are given a circle. Give a construction to find its centre.
Sol. :
1) Take three points A, B and C on the cricumference of the given circle.
2) Join AB and BC.
3) Draw perpendicular bisector of AB and draw perpendicular bisector of BC, which intersect at O.
4) We get centre of circle O.
Q3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chords.
Sol. :
Two circles with centre O and O′. AB is a common chord.
To Prove:
Centres O and O′ lie on the perpendicular bisector of common chord AB.
Construction:
Join OA, O′A, OB, and O′B.
Proof:
In △OAO and △OBO′.
OA=OB [Radii of the circle]
O′A=O′B [Radii of the circle]
OO′=OO′ [Common]
△OAO′≅△OBO′ [By SSS rule]
∠AOO′=∠BOO′ [ By CPCT]
∠AOM=∠BOM [By CPCT]__________(1)
In △AMO and △BMO
OA=OB [Radii of the circle]
∠AOM=∠BOM [from equation (1)]
OM=OM [Common ]
△AMO≅△BMO′ [By SAS rule ]
AM=BM[By CPCT]
∠AMO=∠BMO [By CPCT]
this gives
∠AMO+∠BMO=180∘
∠AMO+∠AMO=180∘
2∠AMO=180∘
∠AMO=180∘2
∠AMO=90∘
∠BMO=90∘
So that AM=BM
thus
OO′ is the perpendicular bisector of AB.
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