Question

Two rods whose lengths are $10\text{units}$ and $6\text{units}$ slides along $X$ and $Y$ axes respectively in such a way that their extremities are always concyclic, then the locus of the centre of the circle is

• A. $x^2+y^2=16$
• B. $x^2-y^2=16$
• C. $x^2+y^2=34$
• D. $x^2-y^2=36$

Answer 1

The correct option is B $x^2-y^2=16$


Since extremities of both the rods are concyclic, the length of rods are the intercept made by the variable circle on the coordinate axes
So, the $x-$ intercept is $10$ and $y-$ intercept is $6$
Let the centre of the circle be $(-g,-f)$
Therefore,
$2\sqrt{g^2-c}=10\Rightarrow g^2=25+c$
Similarly,
$2\sqrt{f^2-c}=6\Rightarrow f^2=9+c$
So,
$g^2-f^2=16$

Hence, the locus of the centre is
$x^2-y^2=16$