Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.



Given : Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common chord.


To Find : Length of common chord PQ


OPO' = 90° [Tangent at a point on the circle is perpendicular to the radius through point of contact ]


So OPO is a right-angled triangle at P


Using Pythagoras in OPO', we have


(OO')2= (O'P)2+ (OP)2


(OO')2 = (4)2 + (3)2


(OO')2 = 25


OO' = 5 cm


Let ON = x cm and NO' = 5 - x cm


In right angled triangle ONP


(ON)2+ (PN)2= (OP)2


x2 + (PN)2 = (3)2


(PN)2= 9 - x2 [1]


In right angled triangle O'NP


(O'N)2 + (PN)2= (O'P)2


(5 - x)2 + (PN)2 = (4)2


25 - 10x + x2 + (PN)2 = 16


(PN)2 = -x2+ 10x - 9[2]


From [1] and [2]


9 - x2 = -x2 + 10x - 9


10x = 18


x = 1.8


From (1) we have


(PN)2 = 9 - (1.8)2


=9 - 3.24 = 5.76


PN = 2.4 cm


PQ = 2PN = 2(2.4) = 4.8 cm


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