Answer: $\left(\frac{12}{5}, \frac{8}{5}\right)$ Let $G$ be $\mathcal{G}$ restricted to the strip of plane $0 \leq y \leq 4$ (we only care about this region since $\ell$ has negative slope going down from $(0,4))$. By completing the square, the original equation rearranges to $(x-3)^{2}+(\sqrt{y(8-y)}-4)^{2}=1$. One could finish the problem in a completely standard way via the single-variable parameterization $(x, \sqrt{y(8-y)})=(3+\cos t, 4+\sin t)$ on the appropriate interval of $t$-just take derivatives with respect to $t$ to find slopes (the computations would probably not be too bad)—but we will present a slightly cleaner solution.
Consider the bijective plane transformation $\Phi:(x, y) \mapsto(x, \sqrt{y(8-y)})$, with inverse $\Phi^{-1}:(x, y) \mapsto$ $\left(x, 4-\sqrt{16-y^{2}}\right)$. In general, $\Phi$ maps curves as follows: $\Phi(\{(x, y): f(x, y)=c\})=\{\Phi(x, y): f(x, y)=$ $c\}=\left\{\left(x^{\prime}, y^{\prime}\right): f\left(\Phi^{-1}\left(x^{\prime}, y^{\prime}\right)\right)=c\right\}$.
Our line $\ell$ has the form $y-4=-m x$ for some $m>0$. We have $\Phi(G)=\left\{(x-3)^{2}+(y-4)^{2}=1: 0 \leq\right.$ $y \leq 4\}$ and $\Phi(\{4-y=m x: 0 \leq y \leq 4\})=\left\{\sqrt{16-y^{2}}=m x: 0 \leq y \leq 4\right\}$. Since $\ell$ is unique, $m$ must
also be. But it's easy to see that $m=1$ gives a tangency point, so if our original tangency point was $(u, v)$, then our new tangency point is $(u, \sqrt{v(8-v)})=\frac{4}{5}(3,4)=\left(\frac{12}{5}, \frac{16}{5}\right)$, and so $(u, v)=\left(\frac{12}{5}, \frac{8}{5}\right)$.
Remark. To see what $\mathcal{G}$ looks like, see Wolfram Alpha using the plotting/graphing commands.
10. Let $\mathcal{G}$ be the set of all points $(x, y)$ in the Cartesian plane such that $0 \leq y \leq 8$ and
$$
(x-3)^{2}+31=(y-4)^{2}+8 \sqrt{y(8-y)} .
$$

There exists a unique line $\ell$ of negative slope tangent to $\mathcal{G}$ and passing through the point $(0,4)$. Suppose $\ell$ is tangent to $\mathcal{G}$ at a unique point $P$. Find the coordinates $(\alpha, \beta)$ of $P$.