Question

Two rods of lengths $a$ and $b$ slide along the x-axis and y-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is

• A. $4(x^2+y^2)=a^2+b^2$
• B. $x^2+y^2=a^2+b^2$
• C. $4(x^2-y^2)=a^2-b^2$
• D. $x^2-y^2=a^2-b^2$

Answer 1

The correct option is C $4(x^2-y^2)=a^2-b^2$

Let $C(h,k)$ be the center of the circle passing through the endpoints of the rod $AB$ and $PQ$ of lengths $a$ and $b$ respectively,

$CL$ and $CM$ be perpendiculars from $C$ on $AB$ and $PQ$, respectively.

then $AL=\left(\frac{1}{2}\right)AB=\frac{a}{2}$, $PM=\left(\frac{1}{2}\right)PQ=\frac{b}{2}$

and $CA=CP$ ...(radii of the same circle)

$k^2+\frac{a^2}{4}=h^2+\frac{b^2}{4}$

$\Rightarrow 4(h^2-k^2)=a^2-b^2$

So, the locus of $(h,k)$ is $4(x^2-y^2)=a^2-b^2$