f(x)=r02−x2,g(x)=−xr02−x2,a=3−2r02f(x)=\sqrt{r_0^2-x^2},g(x)=-\frac{x}{\sqrt{r_0^2-x^2}},a=\frac{3-2r_0}{2}f(x)=r02−x2,g(x)=−r02−x2x,a=23−2r0
已知
r02+(g(a)(r0−a)+f(a))2=81r_0^2+(g(a)(r_0-a)+f(a))^2=81r02+(g(a)(r0−a)+f(a))2=81
,求 r0r_0r0 的值。