Let the radius be x.
Because OE is perpendicular to BC
So BE=CE=4.
therefore
PE=6
In the right triangle BOE, Pythagoras concludes that the square of OE = the square of Bo-the square of Be.
So the square of OE = the square of x-16.
In the Rt triangle POE, Pythagoras obtains: the square of PO = the square of OE+the square of PE.
So the square of PO = X- 16+36 square.
In the Rt triangle APO, Pythagoras obtains: the square of PO = the square of AO+the square of AP.
So the square of X- 16+36 = the square of x+the square of AP.
(The square of the X side of Equation 2 is rounded)
Solution: AP=2 radical number 5