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See my problem on Desmos

enter image description here


There is a circle with radius $R$ at center $(a_1, b_1)$ with the following equation

$(x-a_1)^2+(y-b_1)^2=R^2$

Then a tangent circle with radius $r$ rotated by $\theta_0$ would have the following coordinates.

$a_2 = (R+r)cos(\theta_0)+a_1$

$b_2 = (R+r)sin(\theta_0)+b_1$

But what would be the coordinates for third circle with same radius $r$ that is tangent to both circles?


What i have tried

let $t$ be the angle from central circle to the arc that ends on two tangent points. (sorry for bad english)

$t= \arcsin(\frac{2r}{R+r})+\theta_0$

the coordinates for third tangent circle with radius $r$ would be

$a_3 = (R+r)cos(t)+a_1$

$b_3 = (R+r)sin(t)+b_1$


Problem

Although the third circle (green) is tangent to central circle but its not tangent to the other circle (blue). what did i do wrong?

Also i get undefined values if $r$ is bigger than some value.

See my problem on Desmos

enter image description here

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1 Answer 1

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You are almost there. You computed the angle $t$ incorrectly. Specifically, it should be (why?) $$t = \theta_0 \pm 2\arcsin\left(\frac{r}{R+r}\right).$$

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  • $\begingroup$ Oh. I thought when two circles are tangent then the distance between their center would be sum of their radius, in this case 2r. Now i can see why 2arcsin(r/(R+r)) works but i dont know why arcsin(2r/(R+r)) doesn't work $\endgroup$ Commented May 9, 2018 at 18:52
  • $\begingroup$ @M.kazemAkhgary Try to find the angle opposite to the base of an isosceles triangles with the side lengths $R+r$, $R+r$ and $2r$. You would realize why the given formula works. For example, if $R=r$ then this angle should be $\pi/3$. What would your formula compute? $\endgroup$
    – Math Lover
    Commented May 9, 2018 at 18:57
  • $\begingroup$ Thanks, my formula calculates $\pi/2$, I did a few experiments and i realized my mistake. I cant explain it well but i have it in my head . you know :) $\endgroup$ Commented May 9, 2018 at 19:42

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