# 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 $\sqrt{3^2+3}=2\sqrt3$.

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.

Can it be proved by elementary geometry?

Suppose there are two circles $C_1$ and $C_2$ having respective radii $r_1$ and $r_2$, such that $r_1>r_2$, and respective centers $O_1$ and $O_2$.

Draw line $O_1O_2$ and a common external tangent which intersects circle 1 at $P_1$, circle 2 at $P_2$. These intersect at a point $Q$ such that $O_2$ lies between $O_1$ and $Q$. Draw radii $O_1P_1$ and $O_2O_P$ which are perpendicular to $P_1P_2$, thus parallel to one another. Thereby, triangles $O_1P_1Q$ and $O_2P_2Q$ are similar right triangles. From the proportionality of correspondung sides we then have:

$\frac{s+r_1+r_2}{r_1}=\frac{s}{r_2}$

$s=\frac{r_2(r_1+r_2)}{r_1-r_2}$

where $s$ is the length of line segment $O_2Q$.

When we put $r_1=3, r_2=1$ this gives $s=2$ so that the right triangle $O_2P_2Q$ has a hypoteneuse twice as long as one leg. Then angle $O_2QP_2$ opposite this leg measures $30°$.

Now draw all three common tangents. They form an isosceles triangle enclosing the smaller circle, whose apex angle at $Q$ measures twice the angle $O_2QP_2$, or $60°$. The triangle is thereby certified equilateral.

Orange triangle is half of equilateral triangle. Then Blue is equilateral triangle.  The figure will be something like this. Let the external point be $$P$$ and centres be $$c_1$$ and $$c_2$$. We have to prove that triangle $$\Delta PBE$$ is equilateral.

We know that $$PB=PE$$ and thus $$\angle PBE=\angle PEB$$. If we prove that $$\angle EPB = 60^{\circ}$$, then we will be able to prove that $$\Delta PBE$$ is an equilateral triangle.

Join $$AC_2, DC_2, BC_1$$ and $$EC_2$$. By observing carefully, you will be able to see that $$\Delta PAC_2$$ and $$\Delta PBC_1$$ are similar by angle-angle similarity.

So, $$\frac{AC_2}{PC_2}=\frac{BC_1}{PC_1}$$

We can find the radius by using the equation and you know that $$C_1C_2$$ is equal to $$4$$ units from the figure. Now you will be able to find length of $$PC_1$$ which comes out to be $$6$$ units.

Now, let $$\angle BPE=\theta$$. So $$\angle C_1PB=\frac{\theta}{2}$$.

$$\sin \frac{\theta}{2}=\frac{BC_1}{PC_1}=\frac{1}{2} \implies \frac{\theta}{2}=30 \implies \theta = 60$$

This proves that triangle $$\Delta PEB$$ is equilateral triangle.

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): 