The equation of the normal at $(1,1)$ to the circle $x^2+y^2-4x+6y-4=0$ is

: $(x_1,y_1)=(1,1,)$

   $S=x^2+y^2-4x+6y-4=0$

 Tangent equation is $S_1=0$

 $\begin{array}{l}x{x}_1+y{y}_1-2(x+x_1)+3(y+y_1)-4=0\\ \Rightarrow x\left(1\right)+y\left(1\right)-2(x+1)+3(y+1)-4=0\\ =-x+2y-2+3-4=0\\ =-x+2y-3=0\\ =x-2y+3=0----\left(i\right)\end{array}$

  any line ${\perp }^r$(1) is

 $\begin{array}{l}2x+y+k=0\\ \text{it passes through }(1,1)\\ 2\left(1\right)+1+k=0\\ \Rightarrow k=-3\end{array}$

  $\therefore$ The required normal is $2x+y-3=0$