If the circle x2+y2+4x+22y+c=0 bisects the circumference of the circle x2+y2−2x+8y−d=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 D625
Given that S1=x2+y2+4x+22y+c=0 S2=x2+y2−2x+8y−d=0 AB is a common chord passing through C2. Equation of AB:S1−S2=0 ⇒(x2+y2+4x+22y+c)−(x2+y2−2x+8y−d)=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+d2≥√cd ⇒25≥√cd ⇒cd≤625