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I have two circles that are tangent to one another and touches one another at exactly one point. The two circles share a common tangent line at the tangent point.

A (x1, y1), A (r1), B (x2, y2), B (r2), and l are known.

Is there an easy geometric way to get (x, y) positions of C and D points by changing the given input (length) l?

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    $\begingroup$ Of course you can. Find coordinates of the tangency point using coordinates of both centers and the slope of the line through the centers. Now find the slope of the line perp to the line through their centers and as you know the coordinates of the point of tangency too, you can find coordinates of both $C$ and $D$. $l$ distance away from the point of tangency. $\endgroup$
    – Math Lover
    Dec 25, 2021 at 17:28
  • $\begingroup$ You did not mark the tangent point. You have the equation of the line perpendicular to the line connecting the two centers through the tangent point. Another equation will be based on distance. Solve for x and y and you get both D and C as solutions. $\endgroup$
    – Moti
    Dec 25, 2021 at 18:42
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    $\begingroup$ A point of terminology: these circles are externally tangent. $\endgroup$
    – Jean Marie
    Dec 27, 2021 at 7:48

1 Answer 1

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The tangency point $T$ is given by

$T = (x_1, y_1) + \dfrac{r_1}{r_1 + r_2} ( (x_2 - x_1, y_2 - y_1) ) $

The normalized vector extending from $(x_1, y_1)$ to $(x_2, y_2)$ is given by

$u = (u_x, u_y) = \dfrac{ ( x_2 - x_1, y_2 - y_1 ) }{ \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } }$

The vector orthogonal to that is given by

$v = (- u_y, u_x) $

Now points $C$ and $D$ can be expressed as follows,

$C = T + L v $

and

$D = T - L v $

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