Locus of the point of intersection of tangents to the circle $x^2+y^2+2x+4y-1=0$ which include an angle of ${60}^0$ is 
 

$$\begin{gathered} For the given circle \\ \mathrm{c}=(-1, -2), \mathrm{r}=\sqrt{1+4+1}=\sqrt{6} \\ Let P(x_1,y_1) be locus of point \\ If \theta is angle between \tan gents then \\ \tan \left(\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\raisebox{1ex}{$\mathrm{r}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{{\mathrm{S}}_{11}}$}\right. \\ \tan {30}^0=\frac{\sqrt{6}}{\sqrt{{\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{x}}_1+4{\mathrm{y}}_1-1}} \\ {\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{x}}_1+4{\mathrm{y}}_1-1=18 \\ locus of P is \\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}+4\mathrm{y}-19=0 \end{gathered}$$