Two rods of lengths 'a' and 'b' slide along coordinate axes such that their ends are concyclic. Locus of the center of the circle is.

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

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

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

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x y = a b

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Solution

The correct options are
A $4 (x^{2} + y^{2}) = a^{2} + b^{2}$
C $4 (x^{2} - y^{2}) = a^{2} - b^{2}$
Length of the intercept made by the circle on $x -$ axis $= 2 \sqrt{g^{2} - c}$
Length of intercept by the circle on $y -$ axis $= 2 \sqrt{f^{2} - c}$
$a = 2 \sqrt{g^{2} - c}$
$b = 2 \sqrt{f^{2} - c}$
Let the locus of $P (x, y)$ be
$a^{2} \pm b^{2} = 4 (g^{2} - c) \pm 4 (f^{2} - c)$
$= 4 (x^{2} - c) \pm (f^{2} - c)$
$= 4 (x^{2} \pm y^{2})$ on simplification

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List I	List II
$A)$ The locus of the point $(a sec θ, b tan θ)$	$1)$ $x^{2} - y^{2} = a^{2}$
$B)$ The locus of the point $(c t, \frac{c}{t})$	$2)$ $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$C)$ The locus of the point $(a sec θ, a tan θ)$	$3)$ $x y = c^{2}$
$D)$ The locus of the point $(a cos θ, a sin θ)$	$4) x^{2} + y^{2} - a^{2} = 0$

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