As shown in the figure, circle O and circle P intersect at two points A.B. Circle P passes through the center O, and point C is any point on the optimal arc AB of circle P, even AB. AC，BC，OC .

(1) indicates an angle equal to the angle ACO in the graph;

∠ACO=∠BCO

This involves a theorem: the straight line connecting the centers of two intersecting circles bisects the common chord vertically.

That is the problem: if the circle O and the circle P intersect at two points A and B, then: OP bisects AB vertically.

(the proof method is: connect OA, OB, PA, PB.

Because OA=OB, PA=PB, PO * * *

So, △ pao △ PBO (SSS)

So ∠APO=∠BPO.

In △PAB, PA=PB.

So PO is divided vertically into AB.

From the previous proof process, we know that ∠APO=∠BPO.

Besides,

∠APO=2∠ACO (the central angle of the same arc is twice the circumferential angle)

∠BPO=2∠BCO

So: ∠ACO=∠BCO

(2) When point C is in the position of circle P, is the straight line CA tangent to circle O? Explain why.

Let the straight line CA be tangent to the circle O, and the straight line CA and the circle P intersect at the point C' (the red line in the figure).

Connecting OA

Because C 1A is tangent to the circle o, so: OA⊥C'A

That is, ∠ OAC' = 90.

Then, in the circle P, the chord subtended by ∠OAC' is the diameter of the circle P.

That is, C'O is the diameter of circle p.

That is, when the straight line CA is tangent to the circle O, the point C is the intersection of the extension line of OP and the circle P (or: the point C is the symmetrical point of the center O about the center P).

(3) What is the relationship between the radii of two circles when the angle ACB = 60°? Explain why.

Known ∠ ACB = 60.

And, from the conclusion of (1), ∠ACO=∠BC0.

Therefore ∠ ACO = ∠ BC0 = 30.

And, ∠ACO=∠AC'O

Therefore, ∠ AC 'o = 30.

Moreover, △AC'O is a right triangle.

Therefore, C'O=2AO.

However, C'O=2PO.

So: PO=AO

That is, circle p and circle o have the same radius.