For previous exercise – Circles – Exercise 10.2 – Class IX

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

Solution:

Maximum number of points in common are two.

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

Solution:

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.

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

Solution:

**Given :** 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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