We start with the given circles O and P.
Note: If you are not given the center of a circle, you can find it using the method shown in
Finding the center of a circle with compass and straightedge. 

In steps 14, we construct a circle, center O, whose radius is the sum of the radii of the two given circles. 
1. Draw a line through the center points of both circles. 

2. Set the compass width to the radius of smaller circle P.


3.
Move the compasses to left edge of circle O and mark the point S outside the circle where it crosses the line PO.


4.
With the compasses on O, set the compass width to S and draw a circle.
This circle has a radius which is the sum of the radii of circles O and P


In steps 57, we construct the tangent from P to the outer circle O.
For more on this see
Tangents through an external point.
(Since it is not necessary, we do not actually draw this tangent.
It would be the line from P to J). 
5. Construct the perpendicular bisector of OP to find the midpoint M.
For more on this see Constructing a perpendicular at a point on a line.


6.
Set the compass on M and set its width to O. 

7.
Make an arc across the new outer circle, creating the point of tangency J. 

8. Draw a line from O through J, creating point L on the inner, given circle O.
This will become the point of tangency for the desired tangent line. 

We next construct a radius of circle P which is parallel to the line OL.
This is done using the "angle copy" method of
constructing a tangent through an external point.

9.
Set the compasses width to OJ, and draw an arc from P, creating point Z where it crosses the line OP. 

10.
Set the compass width to SJ. 

11.
Draw an arc from Z across the previous arc, creating point T. 

12.
Draw a line from P through T, creating point F where it crosses the given circle P.
F will become the point of tangency for the desired tangent line.


13.
Draw a line through F and L 

Done.
FL is one of the two internal tangents common to the given circles.
The other tangent is on the opposite side of the circles and is constructed in a similar way.

