If $x+2y+3=0,2x-3y+k=0$  are conjugate with respect to the circle $x^2+y^2=16$, then $\text{'}k\text{'}=$

$\begin{array}{l}x+2y+3=0\to \left(1\right)\\ x{x}_1+y{y}_1-16=0\to \left(2\right)\end{array}$

(1) and (2) represents the same line then

$\begin{array}{l}\frac{1}{x_1}=\frac{2}{y_1}=\frac{3}{-16}\\ x_1=\frac{-16}{3}\\ \frac{2}{y_1}=\frac{3}{-16}\\ \frac{y_1}{2}=\frac{-16}{3}\\ y_1=\frac{-32}{3}\\ 2x-3y+k=0\\ 2\left(\frac{-16}{3}\right)-3\left(\frac{-32}{3}\right)+k=0\\ -32+96+3k=0\\ \therefore k=\frac{-64}{3}\end{array}$