$27 \cdot 7$ Among the real number pairs $(x, y)$ that satisfy the equation
$(x-3)^{2}+(y-3)^{2}=6$, the maximum value of $\frac{y}{x}$ is
(A) $3+2 \sqrt{2}$.
(B) $2+\sqrt{3}$.
(C) $3 \sqrt{3}$.
(D) 6 .
(E) $6+2 \sqrt{3}$.
(35th American High School Mathematics Examination, 1984)
[Solution 1] In the Cartesian coordinate plane, let $P=(x, y), A=(0,0), C=(3, 3)$, and $B$ be any point on the positive $x$-axis. The trajectory of point $P$ is a circle centered at $C$ with a radius of $\sqrt{6}$ (as shown in the figure). The ratio $\frac{y}{x}$ is the ratio of the $y$-coordinate to the $x$-coordinate of point $P$. If we connect $A P$, then $\frac{y}{x}$ can be seen as the slope of the line $A P$. Clearly, as $P$ moves along the circle $C$, $\frac{y}{x}$ changes with the position of $P$. Point $P$ moves in the first quadrant (note that $\sqrt{6}<3$). Therefore, from the figure, it is intuitively clear that when $A P$ is tangent to the circle on the left side, this slope is maximized. To calculate its value, we can connect $A C$, since $C=(3,3)$, so $\angle C A B = 45^{\circ}$.
Let $\angle P A C = \alpha$, then at this time
$$
\frac{y}{x} = \tan(\alpha + 45^{\circ}) = \frac{\tan \alpha + \tan 45^{\circ}}{1 - \tan 45^{\circ} \tan \alpha} = \frac{\tan \alpha + 1}{1 - \tan \alpha}.
$$

In $\triangle A P C$, since $\angle A P C = 90^{\circ}$, then $\tan \alpha = \frac{P C}{P A}$.
By the Pythagorean theorem, $P A = \sqrt{(A C)^{2} - (P C)^{2}} = 2 \sqrt{3}$.
Thus, $\tan \alpha = \frac{\sqrt{6}}{2 \sqrt{3}} = \frac{1}{\sqrt{2}}$.
Therefore,
$$
\frac{y}{x} = \frac{\tan \alpha + 1}{1 - \tan \alpha} = 3 + 2 \sqrt{2}.
$$

Hence, the answer is $(A)$.
[Solution 2] Consider the maximum value of $\frac{y}{x}$, which is the maximum slope of the line passing through the origin and intersecting the circle $(x-3)^{2} + (y-3)^{2} = 6$. Clearly, when this slope is maximized, the line is tangent to the circle and is the steeper of the two tangents. Therefore,
(1) $y = m x$ is a tangent to the circle;
(2) The system of equations $\left\{\begin{array}{l}y = m x, \\ (x-3)^{2} + (y-3)^{2} = 6,\end{array}\right.$ has only one solution $(x, y)$;
(3) The quadratic equation $(x-3)^{2} + (m x-3)^{2} = 6$,
i.e., $\left(m^{2} + 1\right) x^{2} - 6(m + 1) x + 12 = 0$ has a pair of repeated roots;
(4) The discriminant is zero, i.e., $36(m + 1)^{2} - 48\left(m^{2} + 1\right) = 0$.
Solving the last equation, we get $m = 3 \pm 2 \sqrt{2}$, choosing the larger one, i.e.,
$$
m = 3 + 2 \sqrt{2}.
$$

Hence, the answer is $(A)$.
[Solution 3] Let $\frac{y}{x} = k$, then $y = k x$, so the maximum value of $\frac{y}{x}$ is the maximum slope of the line passing through a point on the circle and the origin.
Thus, $\sqrt{6} = \frac{|3 - 3k|}{\sqrt{1 + k^{2}}}$, solving this gives $k = 3 \pm 2 \sqrt{2}$.

Therefore, the maximum value of $\frac{y}{x}$ is $3 + 2 \sqrt{2}$.
Hence, the answer is $(A)$.