Solution: Solution: (1) Because the quadrilateral ABCD is inscribed in a circle,

So ∠ ABC+∠ ADC = 180, connecting communication, by cosine theorem:

AC2=42+62-2×4×6×cos∠ABC

=42+22-2×2×4cos∠ADC，

So cos∠ABC= 12, ∫∠ABC ∈( 0, π),

Therefore ∠ ABC = 60.

S quadrilateral ABCD =12× 4× 6× sin60+12× 2× 4× sin120.

=83 (ten thousand square meters).

In △ABC, through cosine theorem:

AC2=AB2+BC2-2AB？ BC? cos∠ABC

= 16+36-2×4×6× 12.

AC=27。

According to sine theorem asinA=bsinB=2R,

∴2R=ACsin∠ABC=2732=42 13，

∴R=22 13 (ten thousand meters).

(2)∫S quadrilateral APCD=S△ADC+S△APC,

S△ADC= 12AD？ CD？ sin 120 =23，

Let AP=x and CP = y.

So S△APC= 12xy? sin60 =34xy。

And cosine theorem ac2 = x2+y2-2xycos 60.

=x2+y2-xy=28。

∴x2+y2-xy≥2xy-xy=xy.

∴xy≤28, equals sign if and only if x = y.

∴S quadrilateral apcd = 23+34xy ≤ 23+34x28 = 93,

The largest area is 930,000 square meters.