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Two rods of lengths \mathrm{a \: \: and\: \: b} slide along the \mathrm{x}-axis and \mathrm{y}-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is
 

Option: 1

\mathrm{4\left(x^2+y^2\right)=a^2+b^2}

 


Option: 2

\mathrm{x^2+y^2=a^2+b^2}
 


Option: 3

\mathrm{4\left(x^2-y^2\right)=a^2-b^2}
 


Option: 4

\mathrm{x^2-y^2=a^2-b^2}


Answers (1)

best_answer

Let \mathrm{C(h, k)}be the centre of the circle passing through the end points of the rod \mathrm{A B} and \mathrm{PQ} of lengths \mathrm{a \: and\: b} respectively, \mathrm{CL\: and\: CM} be perpendiculars from \mathrm{C} on \mathrm{A B \: \: and \: \: P Q} respectively.
\mathrm{ \therefore A L=\frac{a}{2} \text { and } B M=\frac{b}{2} }
then \mathrm{ C A=C P } (radii of the same circle)

\mathrm{ \Rightarrow \quad k^2+\frac{a^2}{4}=h^2+\frac{b^2}{4} }
\mathrm{ \Rightarrow \quad 4\left(h^2-k^2\right)=a^2-b^2 }
so that locus of \mathrm{ (h, k) \: is \: 4\left(x^2-y^2\right)=a^2-b^2 }

Hence option 3 is correct.

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Nehul

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