This page shows how to draw one of the two possible internal tangents common to two given circles with compass and straightedge or ruler. This construction assumes you are already familiar with Constructing the Perpendicular Bisector of a Line Segment.
As shown below, there are two such tangents, the other one is constructed the same way but on the other half of the circles.
The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.
This is the same drawing as the last step in the above animation with line PJ added.
Argument | Reason | |
---|---|---|
1 | PJ is a tangent to the outer circle O at J. | By construction. See Constructing the tangent through an external point for method and proof. | 2 | FP is parallel to LJ | By construction. See Constructing a parallel (angle copy method) for method and proof. | 3 | FP = LJ | QS was set from the radius of circle P in construction steps 2 and 3. | 4 | FPJL is a rectangle |
FP is parallel to and equal to LJ from (2) and (3). ∠FLJ = ∠FLO = 90° (a tangent is at right angles to radius) |
5 | ∠PFL = ∠FLO = 90° | Interior angles of rectangles are 90° (4) |
6 | FL is a tangent to circle O and P | Touches each circle at one place (F and L), and is at right angles to the radius at the point of contact, (5) |