Question

Two straight rods of lengths $2a$ and $2b$ move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centres of such circles is

• A. $2 (x^2 + y^2) = a^2 + b^2$
• B. $2 (x^2 - y^2) = a^2 + b^2$
• C. $ x^2 + y^2 = a^2 + b^2$
• D. $ x^2 - y^2 = a^2 - b^2$

Answer 1

Answer: (D)

According to given information, if we draw the figure.

Let the equation of circle is
$x^{2}+y^{2}+2 g x+2f y +c=0$
$\because 2 \sqrt{g^{2}-c}=2 a$
and $2 \sqrt{f^{2}-c}=2 b$
then $g^{2}-a^{2}=0$ and $f^{2}-b^{2}=0$
so, $g^{2}-a^{2}=f^{2}-b^{2}$
$\Rightarrow g^{2}-f^{2}=a^{2}-b^{2}$
On taking locus of the centre $(-g,-f)$, we get
$x^{2}-y^{2}=a^{2}-b^{2}$