The locus of the centre of the circle which cuts off intercepts of length 2a and 2b on X-axis and Y-axis respectively, is

x + y = a + b

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Solution

The correct option is C
$2 \sqrt{g^{2} - c} = 2 a \dots \dots \dots \dots (i)$
$2 \sqrt{f^{2} - c} = 2 b \dots \dots \dots \dots \dots (i i)$
On squaring (i) and (ii) and then subtracting (ii) from (i), we get $g^{2} - f^{2} = a^{2} - b^{2}$ .
Hence the locus is $x^{2} - y^{2} = a^{2} - b^{2} .$

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