Question

Two rods of lengths $a$ and $b$ slide along the $x$- and the $y$-axis, respectively, in such a manner that their ends are concyclic. Find the locus of the center of the circle passing through the endpoints.

Answer 1

Let $(h,k)$ be the centre of circle of radius $r$ which intersects the axes. Then $y$ intercept of circle $(y-k{)}^2+(x-h{)}^2=r^2$ is

$y=k\pm \sqrt{r^2-h^2}$

$x$ intercept $x=h\pm \sqrt{r^2-k^2}$

Length of rod $b$ is $b=2\sqrt{r^2-h^2}=y_1-y_2$

Similarly length of rod $a$ is $a=2\sqrt{r^2-k^2}=x_1-x_2$

$b^2+4h^2=a^2+4k^2=4r^{2}4(h^2-k^2)=(a^2-b^2)h^2-k^2=\frac{a^2-b^2}{4}{x}^2-y^2=\frac{a^2-b^2}{4}$