设过P点弦为AB,AP=y,BP=x,
作OM⊥AB,垂足M,
在△OAM中,根据勾股定理,
OM^2=OA^2-MA^2,
OA=R,
MA=AB/2=(x+y)/2,
OM^2=R^2-(x+y)^2/4,(1)
在△OMP中,
OM^2=OP^2-PM^2=a^2-PM^2,
PM=PA-AM=y-AB/2=y-(x+y)/2=(y-x)/2,
OM^2=a^2-(y-x)^2/4,(2),
联立(1)和(2)式,
a^2-(y-x)^2/4=R^2-(x+y)^2/4,
xy=R^2-a^2,
∴y=(R^2-a^2)/x.