Tangents to circles

From the first equation:- $x^{2} + y^{2} + 2x=0$ $(x+1)^{2} + y^{2} = 1$ This is the equation of a circle of radius $1$ centered at $(-1,0)$ .

From second equation:- $x^{2}+y^{2}-6x=0$ $(x-3)^{2} + y^{2} = 9$ This is the equation of a circle of radius $3$ centered at $(3,0)$ .

These two circles have $3$ common tangents. One of them is $y-axis$ and the other two are shown in the figure.

In the figure let $EF = x$ and let $\angle AEF = \theta$ .

Now in $\Delta AED$ :- $\sin\theta = \dfrac{1}{1+x} \ldots (1)$ In $\Delta BEC$ :- $\sin\theta = \dfrac{3}{5+x} \ldots (2)$ From $(1)$ and $(2)$ $\dfrac{1}{1+x} = \dfrac{3}{5+x}$ Simplifying above we get:- $x =1$ Hence $\sin\theta = \dfrac{1}{2}$ $\theta = 30^{\circ}$ The figure has symmetry about $x-axis$ . So we can say that the triangle so formed is equilateral.