Then 3x 2/2-2Rx+R 2- 1 = 0.

Let A(a, r-a), B(b, r-b), b> I,

Then a+b=4r/3 and ab = 2 (r 2- 1)/3.

Because OA⊥OB, then (r-a)(r-b)+ab=0.

That is R2-r (a+b)+2ab = 0 = R2-4R2/3+4 (R2-1)/3,

Then r 2 = 4/3, r >；; 0, then r=2√3/3.

① So l: x+y = 2 √ 3/3.

P(r，0)，r | r-a |/2 = S△oap = 3S△obp = 3r | r-b |/2，

② ① If r-a≥0, r-b≥0, or r-a≤0, r-b≤0,

Then r=(3b-a)/2.

={[2r+√(6-2r^2)]-[2r-√(6-2r^2)]/3}/2

=2{[r+√(6-2r^2)]/3

r=2√(6-2r^2),r^2=8/3,r=2√6/3。

②-②, if r-a≥0 and r-b≤0,

Then r=(3b+a)/4.

={[2r+√(6-2r^2)]+[2r-√(6-2r^2)]/3}/4

=[4r+√(6-2r^2)]/6

2r=√(6-2r^2),r^2= 1,r= 1。