Locus of centre of circle passing through concyclic points

Question

Two rods of length $a$ and $b$ slide along the coordinate axes in a manner that their ends are always concyclic. Find the locus of the centre of the circle passing through these ends.

Apart from figuring out a probable geometry given in the problem... I havent been able to go any further in the problem. Any clues or hints will be quite helpful...

Thanks in advance!!...

The answer given in the key is $4(x^2-y^2)=a^2-b^2$.

Answer 1

By the Power of a Point Theorem, we have $$|\overline{OA}||\overline{OA^\prime}| = |\overline{OB}||\overline{OB^\prime}| \qquad\to\qquad \left(x - \frac{a}{2}\right)\left(x + \frac{a}{2}\right) = \left(y - \frac{b}{2}\right)\left(y + \frac{b}{2}\right)$$

Thus,

$$x^2 - y^2 = \frac{1}{4}\left(a^2-b^2\right)$$

Answer 2

Suppose that the point $(X,Y)$ is the centre of a circle of radius $r$ which intersects the axes. Then the $y$-intercepts would satisfy $(x-X)^2 + (y-Y)^2 =r^2$ and $x=0$, or that $(y-Y)^2 =r^2 -(x-X)^2$. The $y$- intercepts will be $y=Y\pm \sqrt{r^2-(x-X)^2}$. Since this difference is the length of the rod $b$, we can easily see that $b=2\sqrt{r^2-X^2}$. Similarly we get that $a=2\sqrt{r^2-Y^2}$, and thus the centre of the circle satisfies $4(r^2-Y^2) -4(r^2-X^2) =a^2-b^2$ or that $$4(X^2-Y^2) = a^2-b^2$$ Hope it helps.

EDIT: This is how the rods need to be arranged. I have taken this example to prove my case.

Answer 3

For the rod of length $b$ the extremes points $(0,q)$ and $(0,q+b)$ and for the rod of length $a$ the extremes points $(p,0)$ and $(p,p+a)$. The center of the circle will be the intersection between the perpendicular bisector of the rods with lengths $a$ and $b$. And taking to account the points coorcinates above is easy conclude that the center

$$(x,y)=\left(p+\frac{a}{2},q+\frac{b}{2} \right)$$

and so

$$x-p=\frac{a}{2} \quad \quad and \quad \quad y-q=\frac{b}{2} \quad (1)$$

Now we just need to calculate the radius. But it is the distance between $(x,y)$ and $(p,0)$ and also the distance between $(x,y)$ and $(0,q)$, so

$$(x-p)^2+y^2=x^2+(y-q)^2 \quad(2)$$

and replacing $(1)$ in $(2)$ we get:

$$\left(\frac{a}{2}\right)^2+y^2=x^2+\left(\frac{b}{2}\right)^2 \Rightarrow 4(x^2-y^2)=a^2-b^2$$