Circle : Exercise 10.3 (Mathematics NCERT Class 9th)

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

On drawing a different pairs of circles, we find that they have two points in common. Now  we shall prove that a pair of circle  cannot intersect each other at  more than two points.
Let the two  circles cut each other at three points. But  from the three  points only one circle passes. Hence, two circles if intersect each other then they will intersect at two points. These two points are the ends of their common chords. This chord can be a common chord of a number of circles which will pass through the end points of this common chord.

 Q.2    Suppose you are given a circle. Give a  construction  to find  its centre.
Sol.      Steps of Construction :

51.   Take 3  points A, B and C on the  circumference of the circle.
2.   Join  AB and BC.
3.   Draw PQ and RS, the perpendicular bisectors of AB and BC, which intersect each other at O. Then,O is the centre  of the circle.

Q.3     If two circles intersec at two  points, prove that their  centres lie on the perpendicular bisector of the common  chord.

Given : Two  circles, with  centres O and O' intersect, at two points A and B so that AB is the  common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at M.
To prove : OO' is the  perpendicular  bisector of AB.
Construction : Draw line segment OA, OB, O'A and O'B.

Proof : \Delta s OAO' and OBO' we have
OA = OB             [Radii of the same circle]
O'A = O'B             [Radii of the same circle]
and     OO' = OO'        
By  SSS criterion of  congruence, we have
\Delta \,OAO' \cong \Delta OBO'   
 \Rightarrow     \angle AOO' = \angle BOO'
 \Rightarrow     \angle AOM = \angle BOM        ... (1)
[Since \angle AOO' = \angle AOM\,and\,\angle BOM\, = \angle BOO']
In \Delta s AOM and BOM,  we have
OA = OB             [Radii of the same circle]
\angle AOM = \angle BOM        [From (1)]
and      OM = OM             [Common]
Therefore  by SAS criterion of congruence, we have
\Delta AOM \cong \Delta BOM
 \Rightarrow     AM = BM and \angle AMO = \angle BMO
But  \angle AMO + \angle BMO = 180^\circ
Therefore 2\angle AMO = 180^\circ \Rightarrow \angle AMO = 90^\circ
Thus ,     AM = BM  and \angle AMO = {\mkern 1mu} \angle BMO = 90^\circ
Hence , OO' is the perpendicular  bisector of AB.


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