5. A circle with center $O$ and diameter $|A B|=4 \text{ cm}$ is given. Draw three tangents to this circle, two at points $A$ and $B$, and the third at a point such that the segment $\overline{C D}$ between the first two tangents is $5 \text{ cm}$ long. Explain and describe the procedure.
First method: Algebraic approach.

It is necessary to determine the distance of point $C$ from line $A B$.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_c165b3842bf96502feaeg-25.jpg?height=486&width=423&top_left_y=445&top_left_x=311)

Draw a parallel line with the diameter $\overline{A B}$ through $C$ and let $x=|A C|=|B E|$.

We obtain a right triangle $C E D$ with the hypotenuse $\overline{C D}$ of length $5 \mathrm{~cm}$ and the leg $\overline{C E}$ of length $4 \mathrm{~cm}$. By applying the Pythagorean theorem, we calculate the length of the other leg $\overline{E D}$, so $|D E|=3 \mathrm{~cm}$. The length of $\overline{C F}$ is $x$ because triangles $A O C$ and $F O C$ are congruent. (Right triangles over the same hypotenuse $\overline{O C}$ - congruent two corresponding sides and congruent angles opposite the larger side.) The length of $\overline{F D}$ is $x+3$ because triangles $B O D$ and $F O D$ are congruent. (Right triangles over the same hypotenuse $\overline{O D}$ - congruent two corresponding sides and congruent angles opposite the larger side.) From the above two statements, it follows that $x+x+3=5$ or $2 x+3=5$ so $x=1 \mathrm{~cm}$.

Since the desired distance can be applied in four different ways, there are four solutions for

![](https://cdn.mathpix.com/cropped/2024_05_30_c165b3842bf96502feaeg-25.jpg?height=725&width=857&top_left_y=1822&top_left_x=288)

the tangent.