If the angle between the circles $x^2+y^2-2x-4y+c=0$  and $\hspace{0.17em}{x}^2+y^2-4x-2y+4=0$ is $60°$ then $c$ is equal to 

$$\begin{gathered} \mathrm{cos\theta }=\frac{\mathrm{c}+{\mathrm{c}}^1-2{\mathrm{gg}}^1-2{\mathrm{ff}}^1}{2\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \sqrt{{\mathrm{g}}^{12}+{\mathrm{f}}^{12}-{\mathrm{c}}^1}} \\ \frac{1}{2}=\frac{\mathrm{c}+4-4}{2\sqrt{5-\mathrm{c}} \sqrt{4+1-4}} \\ \frac{1}{2}=\frac{\mathrm{c}-4}{2\sqrt{5-\mathrm{c}}} \\ \sqrt{5-\mathrm{c}}=\mathrm{c}-4 \\ \mathrm{Squating} \mathrm{on} \mathrm{both} \mathrm{side} \\ 5-\mathrm{c}={\mathrm{c}}^2+16-8\mathrm{c} \\ {\mathrm{c}}^2-7\mathrm{c}+11=0 \\ \mathrm{c}=\frac{7\pm \sqrt{5}}{2} \end{gathered}$$