Solution: ∵ straight line PQ and ⊙O 1 tangent to point P ⊙O2 tangent to point Q,

∴CB？ CA=PC2=CQ2，

∠∠CPB =∠ Pabu, ∠PBC =∠ Paga ∠ Pabu,

∴∠PBC=∠CPB+∠APB=∠CPA，

∴ ①, ③ Correct,

When the radii of two circles are equal, the graph is symmetrical with respect to the straight line where AC is located.

∴ ② Error.

So choose C.

According to the tangents of PQ and ⊙O 1 to point P and ⊙O2 to point Q, the tangent theorem and the tangent angle theorem, it can be proved that ∠PBC=∠CPB+∠APB=∠CPA, so ① and ③ are correct; Because the radii of two circles are not necessarily equal, the relationship between arc PB and arc BQ is unclear. When the radii of two circles are equal, the graph is symmetrical about the straight line where AC is located, so ② is wrong; So option c is correct.