If $2kx+3y-1=0, 2x+y+5=0$ are conjugate lines with respect to the circle $x^2+y^2-2x-4y-4=0,$then k=

Centre of the circle C=(1, 2), radius of the circle $r=\sqrt{1+4+4}=3.$

$2kx+3y-1=0, 2x+y+5=0$are conjugate lines $\Rightarrow r^2\left(l_1{l}_2+m_1{m}_2\right)=L_1\left(C\right) L_2\left(C\right)$

$\Rightarrow 9\left(4k+3\right)=\left(2k+6-1\right)\left(2+2+5\right)\Rightarrow 36k+27=18k+45\Rightarrow 18k=18\Rightarrow k=1.$