The equation of the smallest circle passing through the intersection of the line $x+y=1$  and the circle $x^2+y^2=9$  is:

$S=x^2+y^2-9=0$ ,

$L=x+y-1=0$ .

$p=1/\sqrt{2}<3=r$    $\therefore$    Line intersects the circle $S$ .

Equation of circle passing through the points of intersections of $S=0$  and $L=0$  is

  $\left(x^2+y^2-9\right)+\lambda \left(x+y-1\right)=0$ .

This circle is smallest if chord $x+y-1=0$

becomes diameter i.e., centre $\left(-\frac{\lambda }{2},-\frac{\lambda }{2}\right)$ lies on it.

$\therefore$     $-\frac{\lambda }{2}-\frac{\lambda }{2}-1=0\Rightarrow \lambda =-1$ .

$\therefore$    required circle is $x^2+y^2-x-y-8=0$ .