# 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 $$2\sqrt{2}$$?  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 $$\sqrt{ (2 + 1)^2 - 1^2 } = \sqrt{8} = 2 \sqrt{2}$$

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 $$\sqrt 8$$

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,

$$P'Q=x=\sqrt{32}=4 \sqrt{2} \ \implies \ P'O'= x/2=2 \sqrt{2}$$  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

$$d= \sqrt{(R+r)(R-r)}=\sqrt{R^2-r^2}~.$$