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Practice each question, then open the answer or worked solution.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
In these two circles, no point is common.
Here, only one point, ‘P’, is common.
Even here, P is the common point.
Here, two points are common, which are P and Q.
No point is common in the above circle.
Suppose you are given a circle. Give a construction to find its centre.
The construction steps to find the centre of the circle is:
Step I: Draw a circle first.
Step II: Draw 2 chords, AB and CD, in the circle.
Step III: Draw the perpendicular bisectors of AB and CD.
Step IV: Connect the two perpendicular bisectors at a point. This intersection point of the two perpendicular bisectors 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.
It is given that two circles intersect each other at P and Q.
To prove:
OO’ is perpendicular bisector of PQ.
(i) PR = RQ
(ii) ∠PRO = ∠PRO’ = ∠QRO = ∠QRO’ = 900
Proof:
In triangles ΔPOO’ and ΔQOO’,
OP = OQ and O’P = O’Q (Since they are also the radii.)
OO’ = OO’ (It is the common side.)
So, it can be said that ΔPOO’ ≅ ΔQOO’ (SSS Congruence rule)
∴ ∠POO’ = ∠QOO’ (c.p.c.t)— (i)
Even triangles ΔPOR and ΔQOR are similar by SAS congruency.
OP = OQ (Radii)
∠POR = ∠QOR (As ∠POO’ = ∠QOO’)
OR = OR (Common arm)
So, ΔOPO’ ≅ ΔOQO’ (SAS Congruence rule)
∴ PR = QR and ∠PRO = ∠QRO (c.p.c.t) …. (ii)
As PQ is a line
∠PRO + ∠QRO = 180°
∠PRO + ∠PRO = 180° (Using (ii))
2∠PRO = 180°
∠PRO = 90°
So ∠QRO = ∠PRO = 90°
Here,
∠PRO’ = ∠QRO = 90° and ∠QRO’ = ∠PRO = 90° (Vertically opposite angles)
∠PRO = ∠QRO = ∠PRO’ = ∠QRO’ = 90°
So, OO’ is the perpendicular bisector of PQ.