A normal is drawn at a point   on a curve.  It meets the x-axis and the y-axis at $A$ and $B$ respectively such that $\frac{1}{OA}+\frac{1}{OB}=1,$ where $O$ is origin, then the radius of the director circle of the curve passing through $(3,3)$ is__________  

The equation of normal at $(x,y)$ is $Y-y=\frac{-1}{\frac{dy}{dx}}(X-x)+(Y-y)\frac{dy}{dx}=0$
$OA=x+y.\frac{dy}{dx}$ and $OB=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$  
Given $\frac{1}{OA}+\frac{1}{OB}=1$  
$1+\frac{dy}{dx}=x+y\frac{dy}{dx}$ 
$(y-1)dy+(x-1)dx=0$ 
Integrating, we get ${(y-1)}^2+{(x-1)}^2=C$  
$(3,3)$ lies on this curve,
$\therefore C=8$ 
Hence curve is circle ${(x-1)}^2+{(y-1)}^2={\left(2\sqrt{2}\right)}^2$ 
The director circle of ${(x-1)}^2+{(y-1)}^2={\left(2\sqrt{2}\right)}^2$