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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 (x^{2} + y^{2}) = a^{2} + b^{2}

b

4 (x^{2} + y^{2}) = a^{2} - b^{2}

c

4 (x^{2} - y^{2}) = a^{2} - b^{2}

d

x y = a b

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detailed solution

Correct option is C

$L e t t h e c i r c l e b e x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

Length of intercept on x-axis is $= 2 \sqrt{g^{2} - c}$

and on y-axis is $= 2 \sqrt{f^{2} - c}$

$Given that a = 2 \sqrt{g^{2} - c} b = 2 \sqrt{f^{2} - c} N o w a^{2} - b^{2} = 4 (g^{2} - c) - 4 (f^{2} - c) Now locus of centre (- g, - f) is \Rightarrow a^{2} - b^{2} = 4 (x^{2} - c) - 4 (y^{2} - c) = 4 (x^{2} - y^{2}) \Rightarrow a^{2} - b^{2} = 4 (x^{2} - y^{2})$

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