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?
Maximum number of points in common are two.
- Suppose you are given a circle. Give a construction to find its centre.
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.
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