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 `triangleOAO` and `triangleOBO'`.
`OA=OB` [Radii of the circle]
`O'A=O'B` [Radii of the circle]
`OO'=OO'` [Common]
`triangleOAO'congtriangleOBO'` [By `SSS` rule]
`angleAOO'=angleBOO'` [ By CPCT]
`angleAOM=angleBOM` [By CPCT]__________(1)
In `triangleAMO` and `triangleBMO`
`OA=OB` [Radii of the circle]
`angleAOM=angleBOM` [from equation (1)]
`OM=OM` [Common ]
`triangleAMOcongtriangleBMO'` [By `SAS` rule ]
`AM=BM`[By CPCT]
`angleAMO=angleBMO` [By CPCT]
this gives
`angleAMO+angleBMO=180^circ`
`angleAMO+angleAMO=180^circ`
`2angleAMO=180^circ`
`angleAMO=180^circ/2`
`angleAMO=90^circ`
`angleBMO=90^circ`
So that `AM=BM`
thus
`OO'` is the perpendicular bisector of `AB`.
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