先假設 AP = m,BP = n
則 AB = m + n ---------(1)
又 AP = PC = m,PB = PD = n
所以 CD = n - m ---------(2)
題目要求PD,即為PB = n
外公切線^2 = 連心線^2 - 半徑差^2
內公切線^2 = 連心線^2 - 半徑和^2
AB = 4 根號 13
CD = 8
m + n = 4 根號 13
n - m = 8
因此可求得 n = 2 根號 13 + 4