Question 315355
<pre><b>
{{{drawing(800,11200/29,-4,25,-6,8,

circle(0,0,3), circle(17,0,5), locate(.8,1.3,3),
locate(0,0,A), locate(17,0,B), locate(24/17-.2,45/17+.7,C), locate(249/17,-75/17,D),triangle(0,0,51/8,0,24/17,45/17),locate(16,-2,5),

triangle(51/8,0,249/17,-75/17,17,0) )}}}

Given AC=3, BD=5, AB=17

Need to find internal tangent CD

Extend radius AC by 5 units to E, where CE is 5 units long.
then Draw BE, so that CDBE is a rectangle.

{{{drawing(800,11200/29,-4,25,-6,8, locate(.8,1.3,3),locate(16,-2,5),
triangle(0,0,64/17,120/17,24/17,45/17), locate(64/17-.5,120/17+.4,E),
locate(2.7,5,5),
circle(0,0,3), circle(17,0,5), triangle(17,0,17,0,65/17,120/17),
locate(0,0,A), locate(17,0,B), locate(24/17-.2,45/17+.7,C), locate(249/17,-75/17,D),triangle(0,0,51/8,0,24/17,45/17),

triangle(51/8,0,249/17,-75/17,17,0) )}}}

Now triangle ABE is a right triangle, with shorter leg AE = 3+5 = 8,
and hypotenuse AB which is given to be 17.

So by the Pythagorean theorem,

{{{AE^2+EB^2=AB^2}}}
{{{8^2+EB^2=17^2}}}
{{{64+EB^2=289}}}
{{{EB^2=225}}}
{{{EB=sqrt(225)}}}
{{{EB=15}}}

And since CDBE is a rectangle, CD = EB = 15

So the internal tangent, CD, is 15 units.

Edwin</pre>