Given two non-overlapping circles, $R_1$ and $R_2$. The radii of $R_1$ and $R_2$ may be different. The distance between the centers of $R_1$ and $R_2$ is defined as $x$.
Draw the four tangents between $R_1$ and $R_2$. There will be two tangents that cross between $R_1$ and $R_2$ and two tangents that do not cross between $R_1$ and $R_2$.
Call the two tangents that cross inner tangents and the two tangents that do not cross outer tangents.
I assert that there are two concentric circles that can be drawn, $C_1$ and $C_2$. $C_1$ will have the four tangents points of the inner tangents on its circumference and $C_2$ will have the four tangent points of the outer tangents on its circumference.
I remember solving this problem using high school geometry, basic algebra and some trig, but that was over $20$ years ago.
Is my assertion correct? If so, what is the solution?
I vaguely remember that one key point was noting that radii that intersect at tangent points are perpendicular to the tangent line.
Best Answer
For unequal radii it is equivalent to the perpendicular bisectors of all the inner and outer tangent segments intersecting at one point. By symmetry this point would be on the line of centers and one need only check it for one inner and one outer segment. In fact it seems the point should be the midpoint of the line segment joining the centers, and once you have observed that it is easy to prove, because the perpendicular bisector is parallel to the radii connecting the centers to the tangent points, and halfway between the two lines extending those radii. This argument holds also in the case of equal radii.