To solve this problem, we’ll use the tangent-secant theorem. This theorem states that if a tangent segment $PT$ and a secant segment $PA$ intersect outside a circle at point $P$, then:

$PT^2 = PA \times PB$

where:

• $PT$ is the length of the tangent segment from $P$ to the point of tangency $T$,
• $PA$ is the length of the external secant segment (from $P$ to $A$),
• $PB$ is the total length of the secant line from $P$ through $A$ and ending at $B$, which equals $PA + AB$.

Given:

• The width of the river, $AB = 40$ (this is the segment that the secant line crosses within the circle),
• $PA = x$, the external segment length we’re solving for,
• We want to find $PT$, the tangent length.

Using the tangent-secant theorem:

1. Express $PB$ in terms of $PA$: $PB = PA + AB = x + 40$

2. Apply the tangent-secant theorem formula: $PT^2 = PA \times PB$ Substitute $PA = x$ and $PB = x + 40$: $PT^2 = x \times (x + 40)$ $PT^2 = x^2 + 40x$

3. Solve for $PT$: Taking the square root of both sides, we get: $PT = \sqrt{x^2 + 40x}$

So, the length of the tangent segment $PT$ in terms of the external secant segment length $x$ is:

$PT = \sqrt{x^2 + 40x}$

This gives the answer to the question.

Would you like a detailed example or further explanation of the tangent-secant theorem?

Here are five additional questions related to this concept:

1. If $x = 10$, what would be the length of $PT$?
2. How does the tangent-secant theorem apply when the circle is very small compared to $PA$?
3. What would happen to $PT$ if $x$ were zero?
4. Can we derive $PA$ given $PT$ and $AB$?
5. How would this change if $AB$ were variable instead of a constant 40?

Tip: In geometry problems involving circles, recognizing patterns such as tangent-secant, chord-chord, or secant-secant relationships can simplify the solution significantly.