To solve this problem, we need to determine the length of the path between the two cities that lies within the 50-mile radius of the transmitter.

Step 1: Define the Points

• Let the transmitter be located at the origin $(0, 0)$.
• The first city is located 59 miles north of the transmitter, so its coordinates are $(0, 59)$.
• The second city is located 64 miles east of the transmitter, so its coordinates are $(64, 0)$.

Step 2: Find the Equation of the Line Connecting the Two Cities

The line passing through the two points $(0, 59)$ and $(64, 0)$ has a slope $m$ given by: $m = \frac{0 - 59}{64 - 0} = \frac{-59}{64}$ So, the equation of the line in point-slope form is: $y - 59 = \frac{-59}{64}(x - 0)$ Simplifying, the equation becomes: $y = -\frac{59}{64}x + 59$

Step 3: Find Points of Intersection with the Circle

The circle representing the transmitter's range has the equation: $x^2 + y^2 = 50^2 = 2500$ Substitute the equation of the line into the circle's equation: $x^2 + \left(-\frac{59}{64}x + 59\right)^2 = 2500$ Expanding and simplifying: $x^2 + \left(\frac{3481}{4096}x^2 - \frac{6962}{64}x + 3481\right) = 2500$ $\frac{7577}{4096}x^2 - \frac{6962}{64}x + 3481 - 2500 = 0$ Multiplying through by 4096 to clear the fraction: $7577x^2 - 445568x + 3990016 = 10240000$ $7577x^2 - 445568x - 6251136 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting the values: $x = \frac{445568 \pm \sqrt{(-445568)^2 - 4 \times 7577 \times (-6251136)}}{2 \times 7577}$ $x = \frac{445568 \pm \sqrt{198546922624 + 189552103552}}{15154}$ $x = \frac{445568 \pm \sqrt{388099026176}}{15154}$ $x = \frac{445568 \pm 62301}{15154}$

Calculating the roots: $x_1 = \frac{445568 + 62301}{15154} \approx 33.6 \text{ miles}$ $x_2 = \frac{445568 - 62301}{15154} \approx 25.3 \text{ miles}$

Step 4: Find the Length of the Segment within the Circle

The length of the drive within the signal range is the difference between $x_1$ and $x_2$: $\text{Length} = 33.6 - 25.3 = 8.3 \text{ miles}$

Thus, the driver will pick up a signal from the transmitter for approximately 8.3 miles.

Would you like any further details or have any questions?

Related Questions:

1. How would the answer change if the radius of the transmitter's signal was increased to 60 miles?
2. What if the cities were located west and south of the transmitter instead of north and east?
3. How would the calculation change if the transmitter was not at the origin but at some other point $(a, b)$?
4. Could this problem be solved using trigonometry instead of algebra? How?
5. What if the road between the two cities was not a straight line, but a curve?

Tip:

Always double-check the setup of your coordinate system to ensure that all points and distances are correctly placed before starting any calculations.