4
$\begingroup$

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.

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I understand how the perpendicular bisectors of the tangents form circles. What I can't seem to prove is how the two circles are co-centric. I see that chords between tangent points share the same perpendicular bisector and a perpendicular bisector goes through the center of a circle, but I can't seem to show how both centers are the same point. $\endgroup$
    – KeithSmith
    Commented Jun 12, 2013 at 13:20
  • 1
    $\begingroup$ I figured it out. My solution is to construct a trapezoid with an outer tangent, one radius from each circle and the line segment connecting the centers. The perpendicular bisector of the tangent also divides the line segment connecting the centers in half. Take an inner tangent, one radius from each circle and the line segment connecting the centers. Extend one of the radii in a colinear direction on the opposite side of the center line. A new trapezoid is formed with a tangent opposite the center line. The perpendicular bisector divides the center line. Both circles have the same center. $\endgroup$
    – KeithSmith
    Commented Jun 12, 2013 at 20:51
  • 1
    $\begingroup$ For any of the tangent segments, the radii (the segments joining the center to the points of tangency) and the perpendicular bisector of the tangent segment are all perpendicular to the tangent line, so the perpendicular bisector is parallel to the radii. In the family of all lines parallel to the radii, distance along any line transverse to the family serves as a coordinate. In any such coordinate (here we use the one from the line of centers) the perpendicular bisecting line is halfway between the radial lines, and thus it goes through the midpoint of the line of centers. $\endgroup$
    – zyx
    Commented Jun 13, 2013 at 2:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .