Question 1. Draw different pairs of circles. How many points does each pair have in common?
What is the maximum number of common points?
Solution: Let us draw different pairs of circles as shown below:
|In figure||Maximum number of common points|
Thus, two circles can have at the most two points in common.
Question 2. Suppose you are given a circle. Give a construction to find its centre.
Solution: Steps of construction
I . Take any three points on the given circle. Let these be A, B and C.
II. Join AB and BC.
III . Draw the perpendicular bisector PQ of AB.
IV. Draw the perpendicular bisector RS of BC such that it intersects PQ at O.
Thus, ‘O’ is the required centre of the given circle.
Question 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution: We have two circles with centres O and O', intersect at A and B.
∵ AB is the common chord of two circles and OO' is the line segment joining the centres of the circles.
Let OO' and AB intersect each other at M.
∴ To prove that OO' is the perpendicular bisector AB, we join OA, OB, O'A and O'B.
Now, in ΔOAO' and ΔOBO',
we have OA = OB [Radii of the same circle]
O'A= O'B [Radii of the same circle]
OO' = OO' [Common]
∴ Using the SSS criterion,
⇒ ∠1= ∠2 [c.p.c.t.]
Now, in ΔAOM and ΔBOM,
OA = OB [Radii of the same circle]
OM = OM [Common]
∠1= ∠2 [Proved]
∴ ΔOAM ≌ ΔBOM [SAS criterion]
⇒ ∠3= ∠4 [c.p.c.t.]
But ∠3 + ∠4 = 180º [Linear pair]
∴ ∠3= ∠4 = 90º each.
⇒ AM ⊥ OO'
Also AM = BM [c.p.c.t.]
⇒ M is the mid-point of AB.
Thus, OO' is the perpendicular bisector of AB.
EQUAL CHORDS AND THEIR DISTANCES FROM THE CENTRE