The common tangents to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle Problem : 
Show that the common tangents to circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle. 
Solution : 
Let $C_1 : x^2+y^2+2x=0$ 
here centre of the circle is $(-1,0) $ and radius 1 unit.
$C_2:x^2+y^2-6x=0$
here centre of the circle is $(3,0) $ and radius 3 units. 
But how to proceed to prove that the tangents form equilateral triangle please suggest thanks. 

More systematic and direct method!

Tangent line to $y=f(x)$ at the point $(x_0,y_0)$ is:
  $$y=y_0+y'(x_0)(x-x_0).$$

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the tangent points of the common increasing tangent line to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$, respectively.
Equate the slopes and intercepts of the tangent line:
$$\begin{cases}y'(x_1)=y'(x_2)\\ y_1-x_1y'(x_1)=y_2-x_2y'(x_2)\end{cases} \Rightarrow \begin{cases}\frac{-x_1-1}{\sqrt{-x_1^2-2x_1}}=\frac{-x_2+3}{\sqrt{-x_2^2+6x_2}}\\ \frac{-x_1}{\sqrt{-x_1^2-2x_1}}=\frac{3x_2}{\sqrt{-x_2^2+6x_2}}\end{cases}$$
Divide $(1)$ by $(2)$ to find:
$$x_1=\frac{3x_2}{3-4x_2}$$
Substitute it to $(2)$:
$$\frac{x_1^2}{-x_1^2-2x_1}=\frac{9x_2^2}{-x_2^2+6x_2} \Rightarrow \frac{9x_2^2}{15x_2^2-18x_2}=\frac{9x_2^2}{-x_2^2+6x_2} \Rightarrow x_2=1.5 \Rightarrow x_1=-1.5.$$
Hence, the tangent line is:
$$y=\frac1{\sqrt{3}}x+\sqrt{3} \Rightarrow \tan \alpha=\frac1{\sqrt{3}} \Rightarrow \alpha=30^\circ.$$
Note: Similarly, you can find the decreasing tangent line.
Refer to the diagram (for verifying other given answers and finding more methods):

