(1) if the intersection o of AO is extended to s, then Tana = PS: PA = 1/2, AS=2*3=6, PA⊥PS, then the solution is AP= 12/√5.

(2) The connecting line OP has ∠OPA=∠PQO=∠PAQ, so the triangle OPQ∠ the triangle APO.

Therefore, OP:PQ=AP:OP, that is, OP*OP=PQ*AP, OP=3, PQ=y, AP=x, then y = 9/X.

(3) According to Tana =4/3, we can get AP=3.6, and then get QP=9/3.6=2.5 from (2).

Let OQ be perpendicular to L, and the intersection point O is at D and E, then point M is at L. Let's take point D as an example (point E is similar to point D).

Join OQ, QM, QM and cycle Q to cross N.

OQ=QN=2.5, let the radius r of the circle be m.

According to the conditions, there is QM = qn+Mn = 2.5+R.

OM=OD-MD=3-r

Because of the straight line L⊥OQ, the triangle QOM is a right triangle.

So MQ*MQ=OQ*OQ+OM*OM, that is, (2.5+R) * (2.5+R) = (3-R) * (3-R)+2.5 *.

Solution, r=9/ 1 1.

That is, the radius of circle m is 9/ 1 1.