Solution: According to the given conditions, the center coordinate ⊙( 1, 0) and the vertex coordinate of parabola are the origin o (0, 0);

Both parabola and circle are located on the positive semi-axis of X (that is, x≥0) and are symmetrical about X, so only the minimum area of △PBC formed by the intersection of the tangent of the moving point P on the first quadrant parabola and the circle O at point B and point C is needed.

Let the moving point coordinate P(x,

Y), two straight lines tangent to the circle intersect the Y axis at B(0,

B) and C(0,

C) Point, where the straight lines PB and PC are tangent to points A and D respectively, there are ⊙A⊥PB, ⊙D⊥PC, |⊙A|=|⊙D|=|⊙o|= 1.

△⊙oB=△⊙AB，△⊙DP=△⊙AP，△⊙oC=△⊙DC

(All right triangles)

Therefore, the area of △PBC is S=2(△⊙AB area +△⊙AP area +△⊙DC area).

but

2△⊙AP area = |⊙ a ||| AP | = | AP |

2δ⊙AB area = |⊙ A ||| AB | = | AB |

The area of 2 △⊙DC = |⊙ O ||| OC | = | OC |

therefore

S=|AB|+|AP|+|oC|=|PB|+|oC|

You can get the coordinates of tangents A and D, and then get the coordinates of points B and C (related to the coordinates of point P), so as to get the values of |PB| and |oC|, and then get the minimum value of S.

The derivation is complicated, please solve the problem yourself.