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. 

The two circles are in homothetic position with respect to the center $H=(-3,0)$. Drawing a tangent $t$ from $H$ to the smaller circle we obtain a right triangle with hypotenuse $2$ and one leg $1$. It follows that $t$, which is a common tangent to both circles, makes an angle $30^\circ$ with the $x$-axis, and so does the other tangent from $H$. The third common tangent is vertical.
