II. (Full marks 50 points) Given two points $A$ and $B$ on the same side of a line $l$, with a distance of $15 \mathrm{~cm}$ between them, and the distances from $A$ and $B$ to the line $l$ are $4 \mathrm{~cm}$ and $16 \mathrm{~cm}$, respectively. Can a circle be constructed that passes through points $A$ and $B$ and is tangent to the line $l$? If so, please describe the construction method, draw a diagram, and calculate the radius of the circle. If not, explain the reason.
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\begin{array}{l}
\text { Question: } \odot \mathrm{O}_{2} \text { is also a circle that meets the conditions of the problem, } \\
\text { Solution 1: As shown in Figure 5. } \\
\text { Draw perpendiculars from } A \text { and } B \text { to } l \text {, } \\
\text { with feet at } C \text { and } D \text {, respectively, and connect } A N_{1} \text { and } B N_{1} \text {. } \\
\text { Then } \\
\begin{array}{l}
C D= \\
\sqrt{A B^{2}-(B D-A C)^{2}}
\end{array} \\
=\sqrt{15^{2}-(16-4)^{2}} \\
=9(\mathrm{~cm}) \text {. } \\
\because A C / / B D \text {, } \\
\therefore \frac{A M}{M B}=\frac{A C}{B D}=\frac{M C}{M D} \text {, } \\
\frac{A M}{A M+15}=\frac{4}{16}=\frac{M C}{M C+9} \text {, } \\
\therefore A M=5, M C=3 \text {. } \\
\text { By the tangent-secant theorem, we get } \\
M N_{1}=\sqrt{A M \cdot M B}=\sqrt{5 \times 20}=10(\mathrm{~cm}) \text {. } \\
\therefore C N_{1}=M N_{1}-M C=7 \mathrm{~cm} \text {, } \\
N_{1} D=C D-C N_{1}=9-7=2(\mathrm{~cm}) \text {. } \\
\therefore A N_{1}=\sqrt{A C^{2}+C N_{1}^{2}}=\sqrt{65} \mathrm{~cm} \text {, } \\
B N_{1}=\sqrt{B D^{2}+N_{1} D^{2}}=2 \sqrt{65} \mathrm{~cm} \text {. } \\
\text { In } \triangle A B N_{1} \text {, by the cosine rule, we get } \\
A B^{2}=A N_{1}^{2}+B N_{1}^{2}-2 A N_{1} \cdot B N_{1} \cos \angle A N_{1} B \text {. } \\
15^{2}=\left(\sqrt{65}^{2}+(2 \sqrt{65})^{2}\right. \\
-2 \sqrt{65} \times 2 \sqrt{65} \cos \angle A N_{1} B \text {. } \\
\therefore \cos \angle A N_{1} B=\frac{5}{13} \text {. } \\
\text { Also, } \sin \angle A N_{1} B=\sqrt{1-\cos ^{2} \angle A N_{1} B}=\frac{12}{13} \text {, } \\
\end{array}
$$
$$
\begin{array}{l}
\text { By the sine rule, we get } \\
\frac{A B}{\sin \angle A N_{1} B}=2 R_{1} \text {, so } R_{1}=\frac{65}{8} \mathrm{~cm} . \\
\text { As shown in Figure 6. Connect }
\end{array}
$$
$\mathrm{AN}_{2}, \mathrm{BN}_{2}$
$\because M N_{2}=M N_{1}$
$$
\begin{aligned}
& =10 \mathrm{~cm}, \\
\therefore & A N_{2} \\
= & \sqrt{N_{2} C^{2}+A C^{2}} \\
= & \sqrt{(10+3)^{2}+4^{2}} \\
= & \sqrt{185(\mathrm{~cm}),} \\
B N_{2} & =\sqrt{N_{2} D^{2}+B D^{2}} \\
& =\sqrt{\left(N_{2} M+M D\right)^{2}+B D^{2}} \\
& =\sqrt{(10+12)^{2}+16^{2}}=2 \sqrt{185} \mathrm{~cm} .
\end{aligned}
$$

In $\triangle A B N_{2}$, 
$\cos \angle A N_{2} B=\frac{35}{37}$,
$$
\therefore \sin \angle A N_{2} B=\frac{12}{37} \text {. }
$$

By the sine rule, we get
$\frac{A B}{\sin \angle A N_{2} B}=2 R_{2}$, so $R_{2}=\frac{185}{8} \mathrm{~cm}$.
Therefore, the radii of the required circles are $\frac{65}{8} \mathrm{~cm}$ or $\frac{185}{8} \mathrm{~cm}$.
Solution 2: As shown in Figure 7. Take the line
$l$ as the $x$-axis, and the line perpendicular to $l$ through $A$ as the $y$-axis to establish a Cartesian coordinate system, with $B C \perp l$ at $C$. Then $O C=$
9, so $A(0,4)$ and $B(9,16)$.
Let the coordinates of the center $O_{1}$

of the required circle be $(a, b)$, then the radius $R$ is $b$.
$\therefore O_{1} A=O_{1} B=b$.
$\therefore\left\{\begin{array}{l}a^{2}+(b-4)^{2}=b^{2}, \\ (a-9)^{2}+(b-16)^{2}=b^{2} .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}a=7 \mathrm{~cm}, \\ b=\frac{65}{8} \mathrm{~cm}\end{array}\right.$ or $\left\{\begin{array}{l}a=-13 \mathrm{~cm}, \\ b=\frac{185}{8} \mathrm{~cm} .\end{array}\right.$
Therefore, the radii of the required circles are $\frac{65}{8} \mathrm{~cm}$ or $\frac{185}{8} \mathrm{~cm}$.
Note: This problem can also be solved by first finding the values of $M N_{1}$ and $M N_{2}$, and then constructing circles with $A_{1}, B_{2}, N_{i}$.
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