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Two rods of lengths ‘a’ and ‘b’ slide along coordinate axes, such that their ends are concyclic. Locus of centre of circle is

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a
4(x2+y2)=a2+b2
b
4(x2+y2)=a2-b2
c
4(x2-y2)=a2-b2
d
xy=ab

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detailed solution

Correct option is C

Let the circle be x2+y2+2gx+2fy+c=0

Length of intercept on x-axis is =2g2-c

and on y-axis  is =2f2-c

Given that  a=2g2-c b=2f2-c Now a2-b2=4(g2-c)-4(f2-c) Now locus of centre (-g,-f) is  a2-b2=4(x2-c)-4(y2-c)             =4(x2-y2) a2-b2=4(x2-y2)

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