circlesgeometry
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 fixed circle described in your problem is actually $$(x-2)^2+(y-\frac{\sqrt{5}}{2})^2=\frac{9}{4}$$ This is because its tangent circle is $(x-1)^2+y^2=9$ is centered in $(1,0)$ which lies on the given circle, which means that its diameter is equal to the radius of the tangent circle, so it is $\frac{3}{2}$.
The $x$ coordinate $a$ of the center is 2 because $(1,0)$ and $(3,0)$ lie on the circle.
The $y$ coordinate $b$ of the center can be found by plugging $(1,0)$ into the equation $(x-2)^2+(y-b)^2=\frac{9}{4}$.
The circumcenter of $\triangle PAB$ is the midpoint of $PO$ where $O=(a,b)$ is the center of the given circle. This is because $\angle PAO=\frac{\pi}{2}$.
For the point $P=(x,y=cos(x))$, the circumcenter is $C=(\frac{x+a}{2},\frac{y+b}{2})$ so its locus is a modified cosine line: $$ Y = \frac{\cos(2X-a)+b}{2}$$ where $a=2$ and $b=\frac{\sqrt{5}}{2}$.
By construction you have $$1=\sqrt 3r+r+r+\sqrt 3r\Rightarrow r=\frac{\sqrt3-1}{4}\approx 0.183012701$$ Later add an explanatory figure.
Best Answer
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 $\sqrt{3^2+3}=2\sqrt3$.