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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 $m$.

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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 $r_3$.

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  • $\begingroup$ Hint : expand the first equation so see $(h^2+x^2)$ appearing. Replace this term by the second. Solve for $x$ and so on $\endgroup$ Commented Jul 5, 2023 at 13:10
  • $\begingroup$ @ClaudeLeibovici I tried that but couldn't solve the resulting equation. $\endgroup$
    – Frosty
    Commented Jul 5, 2023 at 13:21
  • $\begingroup$ @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. $\endgroup$
    – sirous
    Commented Jul 5, 2023 at 16:46

3 Answers 3

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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: $$ d=\frac{2 r_2 r_3\left(2 r_1+r_2+r_3 +2\sqrt{r_1 (r_1+r_2+r_3)}\right)}{(r_2+r_3)^2}. $$

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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.

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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.

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