The minimum area bounded by circle $x^2+y^2=9$ and straight line $x+y=3$ is:

Area of circle $=\pi (3^2)=9\pi$.The line $x+y=3$ cuts the circle in the first quadrant at $(3,0)$ and $(0,3)$.Area of the sector in the 1st quadrant $=\frac{1}{4}\times 9\pi =\frac{9\pi }{4}$.Area of the triangle formed by origin and intercepts $=\frac{1}{2}\times 3\times 3=\frac{9}{2}$.The segment area (minimum area) $=$ Area of sector $-$ Area of triangle.Area $=\frac{9\pi }{4}-\frac{9}{2}=\frac{9}{2}(\frac{\pi }{2}-1)$.