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. 

Clearly the $Y$ axis (i.e. $x=0$) is a common tangent.  Let $y=mx+c$ be the equation of the other common tangent(s).  Then we need both the quadratics
$$x^2+(mx+c)^2+2x=0 ; \qquad  x^2+(mx+c)^2-6x=0$$
to have zero discriminant (why?).  This gives us the conditions
$$(cm+1)^2=(m^2+1)c^2; \qquad (cm-3)^2=(m^2+1)c^2$$
Solving these, we have $\pm\sqrt3 y=x+3$ as the other tangents.  Quite obviously the intersection points are then $(-3, 0), (0, \pm \sqrt3)$ and it easily follows that the distance between any two vertices is $\sqrt{3^2+3}=2\sqrt3$.
