$13 \cdot 27$ If $x^{2}+y^{2}=14 x+6 y+6$, then the maximum value of $3 x+4 y$ is
(A) 72.
(B) 73.
(C) 74.
(D) 75.
(E) 76.
(47th American High School Mathematics Examination, 1996)
[Sol] $x^{2}+y^{2}=14 x+6 y+6$ can be transformed into
$$
(x-7)^{2}+(y-3)^{2}=8^{2}.
$$

It is a circle with center at $Q(7,3)$ and radius 8.
The direction of $3 x+4 y=k$ is as shown in the figure. Clearly, to find the maximum value of $3 x+4 y$ within the known circle, it is achieved at the tangent point $P$ (the value of $k$ increases as the line $3 x+4 y=k$ moves upwards to the right) where the line is tangent to $\odot Q$.

The coordinates of point $P$ can also be found by solving the intersection of $P Q$ (perpendicular to the tangent line) and $\odot P$:
$$
\left\{\begin{array}{l}
y-3=\frac{4}{3}(x-7), \\
(x-7)^{2}+(y-3)^{2}=8^{2}
\end{array}\right.
$$

We get $x=11 \frac{4}{5}, y=9 \frac{2}{5}$, at this point $(3 x+4 y)_{\text {max }}=73$. Therefore, the answer is $(B)$.