If two circles ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-6\mathrm{y}+13=0$ and ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{y}+9=0$ intersect at A and B, then the focus of the parabola whose directrix is the line AB and vertex is the point 0(0,0) is

Two circles are intersects radical axis is the equation of the chord

then chord AB is $-3x+y+2=0$

m is the foot of perpendicular from vertex to the line $-3x+y+2=0$

$\frac{x-0}{-3}=\frac{y-0}{1}=-\frac{-3\left(0\right)+0+2}{(-3{)}^2+1^2}$

$\frac{x}{-3}=\frac{y}{1}=-\frac{2}{9+1}$

$\text{ so, }m(3/5,-1/5)$

Focus is the mid point of point m and vertex then

$\frac{\alpha +3/5}{2}=0,\frac{\beta -1/5}{2}=0$

$\text{ So, S}(-3/5,1/5)$