题目（九年级 · 解析几何）：已知在平面直角坐标系中, 点 $A(3,0), B(-3,0), C(-3,8)$, 以线段 $B C$ 为直径作圆, 圆心为 $E$,直线 $A C$ 交 $\odot E$ 于点 $D$, 连接 $O D$.

(1)求证: 直线 $O D$ 是 $\odot E$ 的切线;

(2)点 $F$ 为 $x$ 轴上任意一动点, 连接 $C F$ 交 $\odot E$ 于点 $G$, 连接 $B G ;$

(1) 当 $\tan \angle A C F=\frac{1}{7}$ 时, 求所有 $F$ 点的坐标 \$ \qquad \$ (直接写出);



(2) 求 $\frac{B G}{C F}$ 的最大值.
选项：
答案：解: (1)证明: 如图 1, 连接 $D E$,

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图1

$\because B C$ 为圆的直径,

$\therefore \angle B D C=90^{\circ}$,

$$
\begin{gathered}
\therefore \angle B D A=90^{\circ} \\
\because O A=O B \\
\therefore O D=O B=O A \\
\therefore \angle O B D=\angle O D B \\
\because E B=E D
\end{gathered}
$$

$$
\begin{gathered}
\therefore \angle E B D=\angle E D B \\
\therefore \angle E B D+\angle O B D=\angle E D B+\angle O D B
\end{gathered}
$$

即: $\angle E B O=\angle E D O$

$\because C B \perp x$ 轴

$$
\begin{aligned}
& \therefore \angle E B O=90^{\circ} \\
& \therefore \angle E D O=90^{\circ}
\end{aligned}
$$

$\because$ 点 $D$ 在 $\odot E$ 上

$\therefore$ 直线 $O D$ 为 $\odot E$ 的切线.

(2)(1)如图 2, 当 $F$ 位于 $A B$ 上时, 过 $F_{1}$ 作 $F_{1} N \perp A C$ 于 $N$,

<ImageHere>

图2

$$
\begin{gathered}
\because F_{1} N \perp A C \\
\therefore \angle A N F_{1}=\angle A B C=90^{\circ}
\end{gathered}
$$

$\therefore \triangle A N F_{1} \sim \triangle A B C$

$$
\therefore \frac{A N}{A B}=\frac{N F_{1}}{B C}=\frac{A F_{1}}{A C}
$$

$\because A B=6, B C=8$,

$\therefore A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{6^{2}+8^{2}}=10$, 即 $A B: B C: A C=6: 8: 10=3: 4: 5$

$\therefore$ 设 $A N=3 k$, 则 $N F_{1}=4 k, A F_{1}=5 k$

$$
\therefore C N=C A-A N=10-3 k
$$

$\therefore \tan \angle A C F_{1}=\frac{F_{1} N}{C N}=\frac{4 k}{10-3 k}=\frac{1}{7}$, 解得: $k=\frac{10}{31}$,

经检验 $k=\frac{10}{31}$ 是原方程的解,

$$
\begin{gathered}
\therefore A F_{1}=5 k=\frac{50}{31} \\
O F_{1}=3-\frac{50}{31}=\frac{43}{31}
\end{gathered}
$$

即 $F_{1}\left(\frac{43}{31}, 0\right)$
如图 3, 当 $F$ 位于 $B A$ 的延长线上时, 过 $F_{2}$ 作 $F_{2} M \perp C A$ 于 $M$,

<ImageHere>

$\because \triangle A M F_{2} \sim \triangle A B C$

$\therefore$ 设 $A M=3 k$, 则 $M F_{2}=4 k, A F_{2}=5 k$

$$
\begin{gathered}
\therefore C M=C A+A M=10+3 k \\
\therefore \tan \angle A C F_{2}=\frac{F_{2} M}{C M}=\frac{4 k}{10+3 k}=\frac{1}{7}
\end{gathered}
$$

解得: $k=\frac{2}{5}$,

经检验 $k=\frac{2}{5}$ 是原方程的解,

$$
\begin{aligned}
& \therefore A F_{2}=5 k=2 \\
& O F_{2}=3+2=5
\end{aligned}
$$

即 $F_{2}(5,0)$

故答案为: $F_{1}\left(\frac{43}{31}, 0\right), F_{2}(5,0)$.

(2)如图 $4, \because C B$ 为直径,

<ImageHere>

图4

$$
\therefore \angle C G B=\angle C B F=90^{\circ}
$$

$\therefore \triangle C B G \sim \triangle C F B$

$$
\begin{gathered}
\therefore \frac{B G}{B F}=\frac{B C}{C F}=\frac{C G}{B C} \\
\therefore B C^{2}=C G \cdot C F \\
C F=\frac{B C^{2}}{C G}
\end{gathered}
$$

$\because C G^{2}+B G^{2}=B C^{2}$

$$
\begin{aligned}
& \quad \therefore B G^{2}=B C^{2}-C G^{2} \\
& \therefore \frac{B G^{2}}{C F^{2}}=\frac{B C^{2}-C G^{2}}{\frac{B C^{4}}{C G^{2}}}=\frac{\left(64-C G^{2}\right) \cdot C G^{2}}{64^{2}} \\
& \quad \therefore \frac{B G}{C F}=\frac{\sqrt{C G^{2}\left(64-C G^{2}\right)}}{64} \\
& \text { 令 } y=C G^{2}\left(64-C G^{2}\right)=-C G^{4}+64 C G^{2}=-\left[\left(C G^{2}-32\right)^{2}-32^{2}\right]=-\left(C G^{2}-32\right)^{2}+32^{2} \\
& \therefore \text { 当 } C G^{2}=32 \text { 时, } y_{\text {最大值 }}=32^{2} \\
& \text { 此时 } C G=4 \sqrt{2} \\
& \left(\frac{B G}{C F}\right)_{\text {最大值 }}=\frac{32}{64}=\frac{1}{2} .
\end{aligned}
$$
解析：解: (1)证明: 如图 1, 连接 $D E$,



图1

$\because B C$ 为圆的直径,

$\therefore \angle B D C=90^{\circ}$,

$$
\begin{gathered}
\therefore \angle B D A=90^{\circ} \\
\because O A=O B \\
\therefore O D=O B=O A \\
\therefore \angle O B D=\angle O D B \\
\because E B=E D
\end{gathered}
$$

$$
\begin{gathered}
\therefore \angle E B D=\angle E D B \\
\therefore \angle E B D+\angle O B D=\angle E D B+\angle O D B
\end{gathered}
$$

即: $\angle E B O=\angle E D O$

$\because C B \perp x$ 轴

$$
\begin{aligned}
& \therefore \angle E B O=90^{\circ} \\
& \therefore \angle E D O=90^{\circ}
\end{aligned}
$$

$\because$ 点 $D$ 在 $\odot E$ 上

$\therefore$ 直线 $O D$ 为 $\odot E$ 的切线.

(2)(1)如图 2, 当 $F$ 位于 $A B$ 上时, 过 $F_{1}$ 作 $F_{1} N \perp A C$ 于 $N$,



图2

$$
\begin{gathered}
\because F_{1} N \perp A C \\
\therefore \angle A N F_{1}=\angle A B C=90^{\circ}
\end{gathered}
$$

$\therefore \triangle A N F_{1} \sim \triangle A B C$

$$
\therefore \frac{A N}{A B}=\frac{N F_{1}}{B C}=\frac{A F_{1}}{A C}
$$

$\because A B=6, B C=8$,

$\therefore A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{6^{2}+8^{2}}=10$, 即 $A B: B C: A C=6: 8: 10=3: 4: 5$

$\therefore$ 设 $A N=3 k$, 则 $N F_{1}=4 k, A F_{1}=5 k$

$$
\therefore C N=C A-A N=10-3 k
$$

$\therefore \tan \angle A C F_{1}=\frac{F_{1} N}{C N}=\frac{4 k}{10-3 k}=\frac{1}{7}$, 解得: $k=\frac{10}{31}$,

经检验 $k=\frac{10}{31}$ 是原方程的解,

$$
\begin{gathered}
\therefore A F_{1}=5 k=\frac{50}{31} \\
O F_{1}=3-\frac{50}{31}=\frac{43}{31}
\end{gathered}
$$

即 $F_{1}\left(\frac{43}{31}, 0\right)$
如图 3, 当 $F$ 位于 $B A$ 的延长线上时, 过 $F_{2}$ 作 $F_{2} M \perp C A$ 于 $M$,



$\because \triangle A M F_{2} \sim \triangle A B C$

$\therefore$ 设 $A M=3 k$, 则 $M F_{2}=4 k, A F_{2}=5 k$

$$
\begin{gathered}
\therefore C M=C A+A M=10+3 k \\
\therefore \tan \angle A C F_{2}=\frac{F_{2} M}{C M}=\frac{4 k}{10+3 k}=\frac{1}{7}
\end{gathered}
$$

解得: $k=\frac{2}{5}$,

经检验 $k=\frac{2}{5}$ 是原方程的解,

$$
\begin{aligned}
& \therefore A F_{2}=5 k=2 \\
& O F_{2}=3+2=5
\end{aligned}
$$

即 $F_{2}(5,0)$

故答案为: $F_{1}\left(\frac{43}{31}, 0\right), F_{2}(5,0)$.

(2)如图 $4, \because C B$ 为直径,



图4

$$
\therefore \angle C G B=\angle C B F=90^{\circ}
$$

$\therefore \triangle C B G \sim \triangle C F B$

$$
\begin{gathered}
\therefore \frac{B G}{B F}=\frac{B C}{C F}=\frac{C G}{B C} \\
\therefore B C^{2}=C G \cdot C F \\
C F=\frac{B C^{2}}{C G}
\end{gathered}
$$

$\because C G^{2}+B G^{2}=B C^{2}$

$$
\begin{aligned}
& \quad \therefore B G^{2}=B C^{2}-C G^{2} \\
& \therefore \frac{B G^{2}}{C F^{2}}=\frac{B C^{2}-C G^{2}}{\frac{B C^{4}}{C G^{2}}}=\frac{\left(64-C G^{2}\right) \cdot C G^{2}}{64^{2}} \\
& \quad \therefore \frac{B G}{C F}=\frac{\sqrt{C G^{2}\left(64-C G^{2}\right)}}{64} \\
& \text { 令 } y=C G^{2}\left(64-C G^{2}\right)=-C G^{4}+64 C G^{2}=-\left[\left(C G^{2}-32\right)^{2}-32^{2}\right]=-\left(C G^{2}-32\right)^{2}+32^{2} \\
& \therefore \text { 当 } C G^{2}=32 \text { 时, } y_{\text {最大值 }}=32^{2} \\
& \text { 此时 } C G=4 \sqrt{2} \\
& \left(\frac{B G}{C F}\right)_{\text {最大值 }}=\frac{32}{64}=\frac{1}{2} .
\end{aligned}
$$