Line tangent to two circles In the following picture, the circle on the left is the unit circle and the circle on the right is centered at (3,0). With calculus, I can determine the coordinates of the points of tangency. But I'm wondering if there is a purely geometric way to get information about them. For instance, is there a purely geometric proof that the distance between the two points where the common tangent line touches is $2\sqrt{2}$?

 A: 
The figure depicts the two circles.  The smaller circle on the left is of radius $1$, and the tangent bigger circle on the right is of radius $2$.  Connect the centers of the two circles to the tangency points, the radii will meet the tangent line at right angles at $C$ and $D$.  Drop a perpendicular (green) from $A$ to $BD$.  Since $AEDC$ is a rectangle, then $AC = ED = 1$ and $CD = AE$.  From the right triangle $ABE$ this is equal to $\sqrt{ (2 + 1)^2 - 1^2 } = \sqrt{8} = 2 \sqrt{2} $
A: The fundamental feature of this figure is that the big circle is the image of the small circle by the homothety (= enlargment) with center $Q$ and ratio 2:1 with the notations of figure below. Using this homothety,  right triangle $PP'Q$ is the image of right triangle $OO'Q$. As a consequence, we have $QP=6$.
The sides of right triangle $PP'Q$ being :
$$QP=6, PP'=2 \ \text{and} \  QP'= x,$$
we can apply it Pythagoras theorem, giving:
$$x^2=6^2-2^2=32$$
As a consequence,
$$P'Q=x=\sqrt{32}=4 \sqrt{2} \ \implies  \ P'O'= x/2=2 \sqrt{2}$$

A: Sure....
Let's call the centers of the circles $O,A$  With $O = (0,0)$ and $A = (3,0)$
And the points of tangency $P,Q$ (with P on the smaller circle and Q on the larger circle).  Construct a line through $P$, parallel to $OA.$  Let $R$ be the point of intersection with $AQ$
$OPQ$ and $PQA$ are right angles.
$OP$ is parallel to $AQ.$
$OARP$ is a parallogram.
$PQR$ is a right triangle with one leg of length $1,$ and hypotenuse $3.$
Applying the Pythagorean theorem, PQ must be of length $\sqrt 8$
A: 
Draw the contacting circles. Recognize the right triangle of sides made up from sum and difference of radii.
If direct tangent length (vertical side) is $d,$ then by Pythagoras thm
$$ d=\sqrt{c^2-a^2}=\sqrt{3^2-1^2}= 2 \sqrt 2. $$
In general if the radii are $(R,r)$ then then direct tangent length is
$$ d= \sqrt{(R+r)(R-r)}=\sqrt{R^2-r^2}~.$$
