解答下这个问题，并罗列下知识点---**Question Stem:** 4、如图，∠PAC=30°，在射线 AC 上依次截取 AD=3cm，DB=10cm，以 DB 为直径作⊙O 交射线 AP 于 E、F 两点，求圆心 O 到 AP 的距离及 EF 的长。 **Geometric Figure Description:** * **Type:** Geometric figure involving a circle, lines, and points. * **Elements:** * A horizontal line segment with points labeled A, D, O, B, C in that order from left to right. O is the center of the circle. DB is a diameter of the circle, implying O is the midpoint of DB. * A ray AP originates from point A and extends upwards and to the right. * A circle with center O. The circle passes through points D and B. * The ray AP intersects the circle at two points, labeled E and F. E is closer to A than F. * Points A, E, F, P are collinear on ray AP. * Points A, D, O, B, C are collinear on line AC. * An angle ∠PAC is indicated, with vertex A and sides along ray AC and ray AP. This angle is given as 30°. * **Relationships:** * D is between A and O on line AC. * O is between D and B on line AC. * B is between O and C on line AC. * E is between A and F on ray AP. * F is between E and P on ray AP. * The circle with diameter DB intersects ray AP at E and F. * AD = 3 cm. * DB = 10 cm. * ∠PAC = 30°. * **Labels:** Points are labeled A, D, O, B, C, E, F, P. The angle ∠PAC is labeled. Lengths AD and DB are given in the text. O is indicated as the center of the circle.