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Question

If the circle x2+y22(a23a+1)x2(a25a+3)y+b2+1=0 has diameters 3x - 4y - 1 = 0 and 8x - 3y + 5 = 0, then


A
a=1,bϵR(,)
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B
a=2,bϵ[3,3]
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C
a=1,bϵ[3,3]
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D
a=1,bϵ(1,1)
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Solution

The correct option is D a=1,bϵ(1,1)
Equation of diameters are 3x - 4y - 1 = 0
8x - 3y + 5 = 0,
solving, we get centre = (-1, -1)
But centre of circle x2+y22(a23a+1)x2(a25a+3)y+b2+1=0 is (a23a+1,a25a+3)=(1,1)
a23a+2=0anda25a+4=0
a=1
Radius = 1+1(b2+1)=1b2 is defined when 1b2 > 0 bϵ(1,1)


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