If the angle between the circles $x^2+y^2-4x-6y-3=0,x^2+y^2+8x-4y+\lambda =0$ is ${60}^{\circ }$, then a positive  value of $\lambda$ is

For the circle, $x^2+y^2-4x-6y-3=0$, centre $C_1=(2,3),$ radius $r_1=\sqrt{4+9+3}=4$

For the circle, $x^2+y^2+8x-4y+\lambda =0\text{, centre }{C}_2=(-4,2),\text{ radius }{r}_2=\sqrt{16+4-\lambda }=\sqrt{20-\lambda }$

$d=C_1{C}_2:=\sqrt{36+1}=\sqrt{37}$

$\begin{array}{l}\cos {60}^{\circ }=\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2}\Rightarrow \frac{1}{2}=\frac{37-16-(20-\lambda )}{2\cdot 4\cdot \sqrt{20-\lambda }}\\ \Rightarrow 4\sqrt{20-\lambda }=I+\lambda \Rightarrow 16(20-\lambda )=(1+\lambda {)}^2\Rightarrow 320-16\lambda ={\lambda }^2+2\lambda +1\\ \Rightarrow {\lambda }^2+18\lambda -319=0\Rightarrow {\lambda }^2+29\lambda -11\lambda -319=0\Rightarrow \lambda (\lambda +29)-11(\lambda +29)=0\\ \Rightarrow (\lambda -11)(\lambda +29)=0\Rightarrow \lambda =11,\lambda =-29.\therefore \lambda =11\end{array}$