1):pc^2=pe*po=>； PC/PE = PO/PC = & gt； Triangle PCO is similar to triangle PEC= > angle OCP= angle CEP=90 degrees = & gtPC is tangent to circle O ..

2):OC is the radius, OA is the radius = & gtOC = OA = & gtOE:OC= 1:2.

When the angle COP is * * * and the angle OCP= the angle CEO=90 degrees, the triangle OCP is similar to the triangle CEO. Co: OE = OP: Co, OP = AO+PA = OA+6, OA = OC = R (R is the radius).

Simultaneous expression: R:(0.5R)=(R+6):R solution, R=6.

3) To sum up: AE=OE= 1/2 OA, CE is perpendicular to OA, and CE is both the center line and the vertical line. Then the triangle COA is an equilateral triangle and the angle OCA is 60 degrees, then the angle PCA= angle OCP- angle OCA=90 degrees -60 degrees =30 degrees, and SINPCA= 1/2.