#### Answer

$(1,8)$

#### Work Step by Step

Given $$f(x)= 9-x^2$$
Consider that the coordinate of $P$ is $(a,f(a))$. Since $f(x)$ has a tangent at $x=a$, then
\begin{align*}
f'(x)&= -2x\\
f'(a)&=-2a
\end{align*}
Since
\begin{align*}
y-y_1&=f'(a) (x-x_1)\\
y-\left(9-a^{2}\right)&=-2 a(x-a)\\
\end{align*}
Since the line passes through $(5,0)$, then
\begin{aligned}
0-\left(9-a^{2}\right)&=-2 a(5-a)\\
9-a^{2}&=10 a-2 a^{2}\\
a^2-10a+9&=0
\end{aligned}
Then $a=1$ or $a=9$. We reject $a=9$ and the coordinates of the point $P $ are $(1,8)$