For the circles $x^2+y^2+6x=0$ and $x^2+y^2-2x=0$, find the area of triangle formed by the common tangents.

Question

I know that the triangle formed is an equilateral triangle, so that part is cleared.

I am not able to find the sides of the triangle however, not geometrically at least.

I would prefer to do it geometrically, but I also tried it analytically. To find the equation of the common tangents to the circles I used the condition of tangency

$$y=m(x+3)\pm 3\sqrt {1+m^2}$$ And $$y=m(x-1)\pm \sqrt {1+m^2}$$

So $$3m\pm 3\sqrt{1+m^2}=-m\pm \sqrt {1+m^2}$$

$$m=\pm \frac{1}{\sqrt 3}$$

Then the tangents are

$$y=\pm \frac{1}{\sqrt 3}(x-1)\pm \sqrt {1+m^2}$$ The POI was (3,0)

From there I found the side to be $2\sqrt 3$

So the area is $3\sqrt 3$

But as I said, I want to do it geometrically. Is there a way to do that?

Answer 1

Let $O_1(1,0)$ and $O_2(-3,0)$ be centers of our circles.

Also, let $QPK$ be a common tangent to our circles, where $Q$ and $P$ are touching points to circles $O_2$ and $O_1$ respective and $K$ be a common point of the tangent and the $x$- axis.

Also, $K(k,0)$.

Thus, since $\Delta KO_1P\sim\Delta KO_2Q$, we obtain: $$\frac{KO_1}{KO_2}=\frac{O_1P}{O_2Q}$$ or $$\frac{k-1}{k+3}=\frac{1}{3}$$ or $$k=3,$$ which gives $$K(3,0)$$ and since $$KO_1=2=2\cdot1=2O_1P,$$ we see that $$\measuredangle O_1KP=30^{\circ}$$ and the rest is smooth.

Answer 2

HINT.

Exploit the similarity between the triangles formed by one of the external tangents, $x$ axis and the radii to tangency points.

Then use the similarity between the smaller of the triangles described above and the triangle forme by the external tangent with $x$ and $y$ axes.

Answer 3

This is a graph that shows your problem:

If we know that the formed triangle is equilateral, we can do this.

First, we can simply show that the triangle $AHI$ and $GHI$ are congruent. They have $HI$ is common, so all the sides are congruent and for the $3$rd cryteria they are conguent. So $\angle{GHI}=\angle{IHA}=60°$.

The angle $AHC$ measures: $\angle{ACH}=180°-2\cdot\angle{GHI}=60°$ and you know also the radius $R=3$, so we can say: $$\overline{CH}=\overline{HO}=\frac{R}{\sin(60°)}\cdot\cos(60°)=\frac{3}{\tan(60°)}=\sqrt{3}$$

Let $O(0,0)$. Being $\overline{HO}=\overline{OI}$ because tangents and circle are simmetric in respect to $y=0$, we have that: $$\overline{HI}=\sqrt{3}$$ And from here the area is: $$A=\frac{1}{2}\cdot\overline{HI}^2\cdot\sin(60°)=\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\cdot 12=3\sqrt{3}$$