as an extend to previous post, we suppose the 2 circles are labeled by subscript 1 and 2.

lets suppose we have the touch point on circle 1, say \vec{r_0}. thus, the solution of the normal vector of tangent lines are:

\vec{n}\cdot (\vec{r_2}-\vec{r_0}) = l_\pm

(P \cdot \vec{n}) \cdot (\vec{r_2}-\vec{r_0}) = \pm d_2

P = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

where P is the right rotation matrix. l_\pm is the length between 2 touching points.

l_\pm = \sqrt{(r_1-r_2)^2+(d_1\pm d_2)^2}

and the pulse sign for d2 is for the center of the circle is on the right hand size.

the assumed touching point on circle can be expressed by P\cdot n.

\vec{r_0} \pm d_1 (P\cdot n) = \vec{r_1}

sub into the equation,  simplify, we have:

\vec{n}\cdot (\vec{r_2} - \vec{r_1}) = l_\pm

(P\cdot \vec{n}) \cdot (\vec{r_2}-\vec{r_1}) = \pm d_2 \pm (d_1)

again, plus sign for center of circle on left side.

 

 

 

 

 

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