The centre of the required circle is at a distance ofOriginally Posted byHelder Barros

from , and also at a distance of from .

Hence the centre is at the point of intersection of:

and

(note there can be zero one or two solutions to this, also this

is for exterior tangency, if is large enough we

can also have interior tangency which would require a change to

the RHS of the equations to and respectivly.

Now I think of it you can also have mixed interior

and exterior tangency)

RonL