Question 1085694
Let the point of intersection be (u,v).
So then,
{{{u^2+v^2+2gu+2fv+c=0}}}
and
{{{lu+mv+n=0}}}
You know that the slope of the tangent line is equal to the value of the derivative at the intersection point. 
Find the derivative using implicit differentiation,
{{{2xdx+2ydy+2gdx+2fdy=0}}}
{{{(2x+2g)dx+(2y+2f)dy=0}}}
{{{(y+f)dy=-(x+g)dx}}}
{{{dy/dx=-(x+g)/(y+f)}}}
So then at (u,v),
{{{m=dy/dx=-(u+g)/(v+f)}}}
Using the point slope form of a line,
{{{y-v=-((u+g)/(v+f))(x-u)}}}
{{{(y-v)(v+f)=-(u+g)(x-u)}}}
{{{(y-v)(v+f)+(u+g)(x-u)=0}}}
{{{vy+fy-v^2-fv+ux-u^2+gx-gu=0}}}
{{{(u+g)x+(v+f)y-(u^2+v^2+gu+fv)=0}}}
Comparing,
{{{l=u+g}}}
{{{m=v+f}}}
{{{n=-(u^2+v^2+gu+fv)}}}