Locus of centre of circle passing through concyclic points 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 $4(x^2-y^2)=a^2-b^2$.
 A: 
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,

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

A: 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.
A: 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:
$$\left(\frac{a}{2}\right)^2+y^2=x^2+\left(\frac{b}{2}\right)^2 \Rightarrow 4(x^2-y^2)=a^2-b^2$$
A: 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)=(p+a2,q+b2)
and so
x−p=a2andy−q=b2(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+y2=x2+(y−q)2(2)
and replacing (1) in (2) we get:
(a2)2+y2=x2+(b2)2⇒4(x2−y2)=a2−b2
