There are two circles with radii 2 and 6. The centers of the circles are 10 units apart. What is the distance between the two tangency point

Here is a representation :

Notice that angles AEC  and BED are equal (vertical angles)

And angle ACE = angle BDE are also equal (right angles)

Then by AA congruency, triangle BED   is similar to  triangle AEC

And  BD = 6  and AC = 2

So  the ratio of the  sides =  3 : 1

Then BE = (3 / (3 + 1)  * AB  =  (3/4) (10) = 7.5

So AE = 10 - 7.5 = 2.5

Then using the Pythagorean Theorem twice,  CD =  sqrt [ 7.5^2 - 6^2 ] + sqrt [ 2.5^2 - 2^2 ] =

sqrt [ 20.25] + sqrt [ 2.25] =

       4.5    +  1.5  =

              6

Given that there are two circles with radii 2 and 6, and their centers are 10 units apart, we want to find the distance between the two tangency points as shown in the diagram.

Let's denote the centers of the circles as \(A\) and \(B\), with radii \(r_1 = 2\) and \(r_2 = 6\) respectively. The distance between the centers is \(AB = 10\).

We can form a right triangle using the centers of the circles and one of the tangency points, let's call it \(T\), as shown below:

      T
       |\
       | \
       |  \   6
     A |   \
       |    \
       |     \
       |______\
         10   B
 

Using the Pythagorean theorem, we can find the distance between the tangency point \(T\) and the center \(A\):

\[AT^2 = AB^2 - BT^2 = 10^2 - r_1^2 = 100 - 4 = 96.\]

Now, we can find the distance between the tangency point \(T\) and the center \(B\):

\[BT^2 = AB^2 - AT^2 = 10^2 - r_2^2 = 100 - 36 = 64.\]

Using the distance formula, we have \(BT = \sqrt{64} = 8\).

So, the distance between the two tangency points as shown in the diagram is \(2 \times 8 = 16\) units.