Let the circle be #x^2+y^2+2gx+2fy+c=0#,
let us seek normal and tangent at #(x_1,y_1)#
(note that #x_1^2+y_1^2+2gx_1+2fy_1+c=0#) ...(A)
as the center of the circle is #(-g,-f)#,
as normal joins #(-g,-f)# and #(x_1,y_1)# its slope is #(y_1+f)/(x_1+g)#
and equation of normal is #y-y_1=(y_1+f)/(x_1+g)(x-x_1)#
i.e. #(x_1+g)y-x_1y_1-gy_1=(y_1+f)x-x_1y_1-fx_1#
or #(y_1+f)x-(x_1+g)y-fx_1+gy_1=0#
or #(y_1+f)x-(x_1+g)y-fx_1+gy_1=0#
and as product of slopes of normal and tangent is #-1#,
slope of tangent is #-(x_1+g)/(y_1+f)# and its equation will be
#y-y_1=-(x_1+g)/(y_1+f)(x-x_1)#
or #(y-y_1)(y_1+f)+(x-x_1)(x_1+g)=0#
or #yy_1-y_1^2+fy-fy_1+x x_1-x_1^2+gx-gx_1=0#
or #x x_1+yy_1+gx+fy-(x_1^2+y_1^2+fy_1+gx_1)=0#
from (A) #x_1^2+y_1^2+fy_1+gx_1=-fy_1-gx_1-c#
Hence equation of tangent is #x x_1+yy_1+gx+fy+fy_1+gx_1+c=0#
or #x x_1+yy_1+g(x+x_1)+f(y+y_1)+c=0#
