The normal to a circle at any point passes through the centre. This is because the radius is perpendicular to the tangent at the point of tangency. So the equation of the normal at (x1,y1) on x^{2+y}^{2+2gx+2fy+c}=0 is the line through (x1,y1) and (-g,-f): $\frac{y-y1}{(x-x1)}$ = $\frac{y1+f}{(x1+g)}$.