# Find distance between common tangent to 2 circles and parallel tangent to third circle in terms of radii of the three externally touching circles. .

Three circles $$O_1(r_1)$$, $$O_2(r_2)$$ and $$O_3(r_3)$$ touch each other externally. The line $$l$$ is tangent to $$O_1(r_1)$$ and parallel to the exterior common tangent $$m$$ to $$O_2(r_2)$$ and $$O_3(r_3)$$ which does not intersect $$O_1(r_1)$$. Find the distance between the lines $$l$$ and $$m$$. I tried using Pythagoras theorem and got the following equations $$(h+r_3-r_2)^2 + (2\sqrt{r_2r_3}-x)^2 = (r_1 +r_2) ^2$$ $$h^2+x^2=(r_1+r_3)^2$$ where $$r_1 + r_3 + h$$ is the required length. I don't know how to proceed i.e. how to isolate $$h$$ in terms of $$r_1$$,$$r_2$$ and $$r_3$$.

• Hint : expand the first equation so see $(h^2+x^2)$ appearing. Replace this term by the second. Solve for $x$ and so on Jul 5 at 13:10
• @ClaudeLeibovici I tried that but couldn't solve the resulting equation. Jul 5 at 13:21
• @Frosty, if you have the coordinates of centers of circle , then you can find the coordinates of centroid of triangle the centers make. Centroid is mid point of a line perpendicularly joins two lines. Having the equation of each line you find the distance. Jul 5 at 16:46

The distance we want to compute is $$d=r_1+r_2+CH$$ (see figure below), where $$CH=BC\sin(\alpha+\theta)=(r1+r2)(\sin\alpha\cos\theta+\cos\alpha\sin\theta).$$ From triangle $$ABK$$ we get: $$\sin\theta={r_3-r_2\over r_3+r_2}, \quad \cos\theta={2\sqrt{r_3r_2}\over r_3+r_2}$$ and from the cosine rule applied at triangle $$ABC$$ we obtain: $$\cos\alpha={(r_3+r_2)^2+(r_1+r_2)^2-(r_1+r_3)^2\over2(r_3+r_2)(r_1+r_2)}, \quad \sin\alpha={2\sqrt{ r_1 r_2 r_3 (r_1 + r_2 + r_3)}\over(r_3+r_2)(r_1+r_2)}.$$ Note that $$\sin\theta$$ and $$\cos\alpha$$ can take a negative value in some cases, which is correct.

Plugging these equalities into the expression for $$d$$ and simplifying, we finally get: $$d=\frac{2 r_2 r_3\left(2 r_1+r_2+r_3 +2\sqrt{r_1 (r_1+r_2+r_3)}\right)}{(r_2+r_3)^2}.$$  Let $$x$$ be the distance between the parallel lines, and angles $$\theta$$ and $$\phi$$ as indicated in the diagram.

Using the cosine rule in $$\triangle O_1O_2O_3$$ gives $$(r_1+r_2)^2=(r_1+r_3)^2+(r_2+r_3)^2-2(r_1+r_3)(r_2+r_3)\cos(\pi-\theta-\phi)$$

Which leads to $$\cos(\theta+\phi)=\frac{r_1r_2-r_1r_3-r_2r_3-r_3^2}{(r_1+r_3)(r_2+r_3)}$$

Meanwhile, you also have $$\cos\phi=\frac{r_3-r_2}{r_3+r_2}$$ and $$\cos\theta=\frac{x-r_3-r_1}{r_1+r_3}$$

From this you can extract $$x$$ since you have all the radii.

Using algebra $$(h+r_3-r_2)^2 + (2\sqrt{r_2\,r_3}-x)^2 = (r_1 +r_2) ^2 \tag 1$$ $$h^2+x^2=(r_1+r_3)^2\tag 2$$ Subtract $$(2)$$ from $$(1)$$ : $$h^2$$ and $$x^2$$ disappear and you have a linear equation in $$h$$ and $$x$$. Solve it for $$x$$ $$x=\frac{h (r_3-r_2)+r_1 (r_3-r_2)+r_3(r_2+r_3) }{2\sqrt{r_2\,r_3} }$$ Plug it in $$(2)$$ to get a quadratic equation in $$h$$; select the proper root.