Circles – Exercise 10.3 – Class IX

For previous exercise – Circles – Exercise 10.2 – Class IX


  1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

Circles - Exercise 10.3 – Class IX

Maximum number of points in common are two.


  1. Suppose you are given a circle. Give a construction to find its centre.

Solution:

10.3.2

Take points A, B and C on the circumference of the circle.

Join AB and BC

Draw PQ and RS, perpendicular to the bisector of AB and BC, which intersect each other at O. Then, O is the centre of the circle.

 


  1. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Circles - Exercise 10.3 – Class IXGiven : AB is the common chord of two intersecting circles (O, r) and (O′, r′).

To Prove : Centres of both circles lie on the perpendicular bisector of chord AB, i.e., AB is bisected at right angle by OO′.

Construction : Join AO, BO, AO′ and BO′.

Proof : In ∆AOO′ and ∆BOO′

AO = OB (Radii of the circle (O, r)

AO′ = BO′ (Radii of the circle (O′, r′))

OO′ = OO′ (Common)

∴ ∆AOO′ ≅ ∆BOO′ (SSS congruency)

⇒ ∠AOO′ = ∠BOO′ (CPCT)

Now in ∆AOC and ∆BOC

∠AOC = ∠BOC (∠AOO′ = ∠BOO′)

AO = BO (Radii of the circle (O, r))

OC = OC (Common)

∴ ∆AOC ≅ ∆BOC (SAS congruency)

⇒ AC = BC and ∠ACO = ∠BCO …(i) (CPCT)

⇒ ∠ACO + ∠BCO = 180° ..(ii) (Linear pair)

⇒ ∠ACO = ∠BCO = 90° (From (i) and (ii))

Hence, OO’ lie on the perpendicular bisector of AB


For next exercise – Circles – Exercise 10.4 – Class IX


 

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