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Question

If the circle x2+y2+4x+22y+c=0 bisects the circumference of the circle x2+y22x+8yd=0 (c,d>0), then the maximum possible value of cd is

A
25
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B
125
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C
425
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D
625
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Solution

The correct option is D 625


Given that S1=x2+y2+4x+22y+c=0
S2=x2+y22x+8yd=0
AB is a common chord passing through C2.
Equation of AB:S1S2=0
(x2+y2+4x+22y+c)(x2+y22x+8yd)=0
6x+14y+c+d=0
The common chord passes through centre of S2 i.e., C2(1,4)
6(1)+14(4)+c+d=0
c+d=50
Since A.M.G.M.
c+d2cd
25cd
cd625

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