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Question

Given point are P=(1,2), Q=(7,6) is the origin. The length of the common chord of the circles with OP and OQ as diameters is

A
1
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B
2
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C
4
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D
6
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Solution

The correct option is B 6
REF.Image.
P=(1,-2) & Q = (7,6)
distance b/w PQ = (71)2+(6+2)2=62+82
=100
=10
radius of circle with centre (1,2) & point (0,0)
(x1)2+(y+2)2=r21
putting (0,0) =12+22=r21
1+4=r21=5r1=5
lll y with centre (7, 6) of pt.(0,0)
(x7)2+(y6)2=r22
(putting (0,0)) 72+(6)2=r22
49+36=r22=85r2=85
Now, the length OB i.e the common cord is given by
=(d2(r1r2)2)((r1+r2)2)d2d2
=((10)2)(585)2)((5+85)2(10)2)(10)2
(100(5(117))2)((5(1+17))2)10010
=(1005(1+17217))(5(1+17+217)100)100
=(10+107)(10710)100.(107)2(10)2100
=100×7100100=600100=6
distance / length of common chord = 6


1168799_1138851_ans_2d07a598ec5a444eb8089f6b0240d58a.jpeg

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