First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:
$$\angle BAP = 30^{\circ}$$
Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:
$$\tan \angle BAP = \frac{PD}{AD}$$
$$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$
$$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$
$$ PD = \frac{1}{2\sqrt{3}}$$
Now we have: $PO = OD + PD$, i.e.:
$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{4}{2\sqrt{3}}$$
$$PO = \frac{2}{\sqrt{3}} \approx 1.154$$
So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:
$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$
$$x^2 + y^2 = \frac 43$$
So the locus of the point $P$ is:
$$x^2 + y^2 = \frac 43$$
You know it's a circle with center $O$, so you just need a point to find the radius.
You can take a tangent the the first circle: $y=1$ and a perpendicular tangent to the other circle: $x=\sqrt{7}$ , the intersection $(1,\sqrt{7})$ is on the circle: the radius is
$$\sqrt{1^2+\sqrt{7}^2}=\sqrt{8}=2\sqrt{2}$$
Best Answer
There're two pairs of tangents for the two circles are perpendicular. Let the contact points on the circle be $\begin{pmatrix} a+b\cos t \\ b\sin t \end{pmatrix}$, $\begin{pmatrix} -a-c\sin t \\ c\cos t \end{pmatrix}$, $\begin{pmatrix} a-b\cos t \\ -b\sin t \end{pmatrix}$ and $\begin{pmatrix} -a+c\sin t \\ -c\cos t \end{pmatrix}$.
The tangent from the first point: $$(a+b\cos t)x-a(x+a+b\cos t)+a^2+by\sin t=b^2$$
The tangent from the second point: $$(-a-c\sin t)x+a(x-a-c\sin t)+a^2+cy\cos t=c^2$$
On solving,
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a\cos 2t+b\cos t-c\sin t \\ a\sin 2t+b\sin t+c\cos t \end{pmatrix}$$
Similarly, we have another branch $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a\cos 2t+b\cos t+c\sin t \\ a\sin 2t+b\sin t-c\cos t \end{pmatrix}$$
Each branch correspond to one diagonal of the rectangle.