smc

The equations of the two circles are $(x+10{)}^2+(y+4{)}^2=169$ and $(x-3{)}^2+(y-9{)}^2=65$. Rearrange them to $(x+10{)}^2+(y+4{)}^2-169=0$ and $(x-3{)}^2+(y-9{)}^2-65=0$, respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10{)}^2+(y+4{)}^2-169=(x-3{)}^2+(y-9{)}^2-65$. We can simplify this like the following. $(x+10{)}^2+(y+4{)}^2-169=(x-3{)}^2+(y-9{)}^2-65\to (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104\to 26x+26y+26=104\to 26x+26y=78\to x+y=3$. Thus, $c=\boxed{3}$.