Plotting the diagram of the given data we have,

Now let the circle $S_1\equiv x^2+y^2-2x-1=0$ whose radius is $\sqrt{2}$ and centre $C\left(1,0\right)$,

And circle $S_2\equiv x^2+y^2-2x=0$ whose radius is $1$ and centre $C\left(1,0\right)$ now we can see that radius of $S_1$ is $\sqrt{2}$ times of radius of $S_2$, so $S_1$ will be director circle where we have perpendicular tangents,

So $△CAB$ is right angle triangle right angled at $C$, now we know that circum centre will lie on midpoint of hypotenuse,

So circumcentre $\left(h,k\right)$ is midpoint of $AB$,

Now let point point $P$ be $\left(\sqrt{2}\cos \theta +1,\sqrt{2}\sin \theta \right)$,

Now we know that in square midpoint of both diagonal are same, so by midpoint formula we get,

$h=\frac{\sqrt{2}\cos \theta +1+1}{2} \& k=\frac{\sqrt{2}\sin \theta +0}{2}$

$\Rightarrow 2h-2=\sqrt{2}\cos \theta \& 2k=\sqrt{2}\sin \theta$

Now by squaring and adding we get,

$\Rightarrow {\left(2h-2\right)}^2+4k^2=2$

$\Rightarrow 4h^2+4k^2-8h+2=0$

$\Rightarrow 2x^2+2y^2-4x+1=0$