## Calculus (3rd Edition)

$(1,8)$
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 $(1,8)$