Line tangent to two circles

Question

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 $2\sqrt{2}$?

Answer 1

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} $

Answer 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 $\sqrt 8$

Answer 3

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,

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

Answer 4

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

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