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Question

If the curves ax2+4xy+2y2+x+y+5=0 and ax2+6xy+5y2+2x+3y+8=0 intersect at four concyclic points then the value of a is

A
4
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B
4
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C
6
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D
6
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Solution

The correct option is D 4
S1=ax2+4xy+2y2+x+y+5=0
S2=ax2+6xy+5y2+2x+3y+8=0
Equation of curve passing through their intersection will be S1+kS2=0
(ax2+4xy+2y2+x+y+5)+k(ax2+6xy+5y2+2x+3y+8)=0
Since the points are concyclic, the above curve should be a circle ,
Thus,Coefficient of xy=0
4+6k=0
k=23
Ceofficient of x2=Coefficient of y2
(a+ka)=2+5k
a(1+k)=2+5k
a(123)=2103
a(323)=6103
a=4

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