Central coordinate C( 1, 1), R= 1.

The AB equation is x/(2a)+y/(2b) = 1, bx+ay-2ab = 0,

(1) When the circle is tangent to AB, the distance from the center C to AB =R= 1,

That is |b+a-2ab|/ radical sign (b 2+a 2) = 1.

Let the p coordinate of the midpoint of the line segment AB be (x, y).

Then there are x = a and y = b.

So the trajectory equation of P is |y+x-2xy|= radical sign (x 2+y 2), (x, y > 1).

(2)

|AB|=2√(a^2+b^2)，

S△ABC=|AB|*R/2=√(a^2+b^2)，

According to the average inequality, a 2+b 2 ≥ 2ab,

∴S△ABC is at least √(2ab),

When 2a=2b, the distance from the origin to the tangent point is √2+ 1, 2a=√ 2(√2+ 1)=2+√2,

a=(2+√2)/2，

The minimum value of S△ABC is √2+ 1.

If S △ AOB = (1/2) * (2+√ 2) * (2+√ 2) = 3+2 √ 2.