First, we find the center of the circle by solving the system of equations: \(x+2y=8\)

\(2x-3y=-5\)

Multiply (1) by (2) to obtain (3):

\(2x+4y=16\)

Subtract each side of (2) adn (3) to eliminate 2x and solve for y:

\(-7y=-21\)

\(y=3\)

Solve for x using (1):

\(x+2(3)=8\)

\(x+6=8\)

\(x=2\)

The point of intersection is the center:\((h,k)=(2,3)\)

The equation of the circle with center (h,k) and radius r is given by:

\(\displaystyle{\left({x}-{h}\right)}^{{2}}+{\left({y}-{k}\right)}^{{2}}={r}^{{2}}\)

\(\displaystyle{\left({x}-{2}\right)}^{{2}}+{\left({y}-{3}\right)}^{{2}}={8}^{{2}}\)

\(\displaystyle{\left({x}-{2}\right)}^{{2}}+{\left({y}-{3}\right)}^{{2}}={64}\)