, 2015

Solution:
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:

$\Phi(\{(x, y): f(x, y)=c\})=\{\Phi(x, y): f(x, y)=c\}=\{(x', y'): f(\Phi^{-1}(x', y'))=c\}$.

Our line $\ell$ has the form $y-4=-m x$ for some $m>0$. We have $\Phi(G)=\{(x-3)^{2}+(y-4)^{2}=1: 0 \leq y \leq 4\}$ and $\Phi(\{4-y=m x: 0 \leq y \leq 4\})=\{\sqrt{16-y^{2}}=m x: 0 \leq y \leq 4\}$.

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 $(u, v)=\left(\frac{12}{5}, \frac{8}{5}\right)$.